Morphisms of Schemes

Introduction

In this chapter we introduce some types of morphisms of schemes. A basic reference is [EGA].

Closed immersions

In this section we elucidate some of the results obtained previously on closed immersions of schemes. Recall that a morphism of schemes \(i : Z \to X\) is defined to be a closed immersion if (a) \(i\) induces a homeomorphism onto a closed subset of \(X\), (b) \(i^\sharp : \mathcal{O}_X \to i_*\mathcal{O}_Z\) is surjective, and (c) the kernel of \(i^\sharp\) is locally generated by sections, see Schemes, Definitions 01IO and 01HK. It turns out that, given that \(Z\) and \(X\) are schemes, there are many different ways of characterizing a closed immersion.

Lemma

Let \(i : Z \to X\) be a morphism of schemes. The following are equivalent:

The morphism \(i\) is a closed immersion.

For every affine open \(\Spec(R) = U \subset X\), there exists an ideal \(I \subset R\) such that \(i^{-1}(U) = \Spec(R/I)\) as schemes over \(U = \Spec(R)\).

There exists an affine open covering \(X = \bigcup_{j \in J} U_j\), \(U_j = \Spec(R_j)\) and for every \(j \in J\) there exists an ideal \(I_j \subset R_j\) such that \(i^{-1}(U_j) = \Spec(R_j/I_j)\) as schemes over \(U_j = \Spec(R_j)\).

The morphism \(i\) induces a homeomorphism of \(Z\) with a closed subset of \(X\) and \(i^\sharp : \mathcal{O}_X \to i_*\mathcal{O}_Z\) is surjective.

The morphism \(i\) induces a homeomorphism of \(Z\) with a closed subset of \(X\), the map \(i^\sharp : \mathcal{O}_X \to i_*\mathcal{O}_Z\) is surjective, and the kernel \(\Ker(i^\sharp)\subset \mathcal{O}_X\) is a quasi-coherent sheaf of ideals.

The morphism \(i\) induces a homeomorphism of \(Z\) with a closed subset of \(X\), the map \(i^\sharp : \mathcal{O}_X \to i_*\mathcal{O}_Z\) is surjective, and the kernel \(\Ker(i^\sharp)\subset \mathcal{O}_X\) is a sheaf of ideals which is locally generated by sections.

Proof

Condition (6) is our definition of a closed immersion, see Schemes, Definitions 01HK and 01IO. So (6) \(\Leftrightarrow\) (1). We have (1) \(\Rightarrow\) (2) by Schemes, Lemma 01IN. Trivially (2) \(\Rightarrow\) (3).

Assume (3). Each of the morphisms \(\Spec(R_j/I_j) \to \Spec(R_j)\) is a closed immersion, see Schemes, Example 01IG. Hence \(i^{-1}(U_j) \to U_j\) is a homeomorphism onto its image and \(i^\sharp|_{U_j}\) is surjective. Hence \(i\) is a homeomorphism onto its image and \(i^\sharp\) is surjective since this may be checked locally. We conclude that (3) \(\Rightarrow\) (4).

The implication (4) \(\Rightarrow\) (1) is Schemes, Lemma 01LD. The implication (5) \(\Rightarrow\) (6) is trivial. And the implication (6) \(\Rightarrow\) (5) follows from Schemes, Lemma 01IN.

Lemma

Let \(X\) be a scheme. Let \(i : Z \to X\) and \(i' : Z' \to X\) be closed immersions and consider the ideal sheaves \(\mathcal{I} = \Ker(i^\sharp)\) and \(\mathcal{I}' = \Ker((i')^\sharp)\) of \(\mathcal{O}_X\).

The morphism \(i : Z \to X\) factors as \(Z \to Z' \to X\) for some \(a : Z \to Z'\) if and only if \(\mathcal{I}' \subset \mathcal{I}\). If this happens, then \(a\) is a closed immersion.

We have \(Z \cong Z'\) over \(X\) if and only if \(\mathcal{I} = \mathcal{I}'\).

Proof

This follows from our discussion of closed subspaces in Schemes, Section 01HJ especially Schemes, Lemmas 01HO and 01HP. It also follows in a straightforward way from characterization (3) in Lemma 01QO above.

Lemma

Let \(X\) be a scheme. Let \(\mathcal{I} \subset \mathcal{O}_X\) be a sheaf of ideals. The following are equivalent:

\(\mathcal{I}\) is locally generated by sections as a sheaf of \(\mathcal{O}_X\)-modules,

\(\mathcal{I}\) is quasi-coherent as a sheaf of \(\mathcal{O}_X\)-modules, and

there exists a closed immersion \(i : Z \to X\) of schemes whose corresponding sheaf of ideals \(\Ker(i^\sharp)\) is equal to \(\mathcal{I}\).

Proof

The equivalence of (1) and (2) is immediate from Schemes, Lemma 01IN. If (1) holds, then there is a closed subspace \(i : Z \to X\) with \(\mathcal{I} = \Ker(i^\sharp)\) by Schemes, Definition 01HN and Example 01HM. By Schemes, Lemma 01IN this is a closed immersion of schemes and (3) holds. Conversely, if (3) holds, then (2) holds by Schemes, Lemma 01IN (which applies because a closed immersion of schemes is a fortiori a closed immersion of locally ringed spaces).

Lemma

The base change of a closed immersion is a closed immersion.

Proof

See Schemes, Lemma 01JY.

Lemma

A composition of closed immersions is a closed immersion.

Proof

We have seen this in Schemes, Lemma 02V0, but here is another proof. Namely, it follows from the characterization (3) of closed immersions in Lemma 01QO. Since if \(I \subset R\) is an ideal, and \(\overline{J} \subset R/I\) is an ideal, then \(\overline{J} = J/I\) for some ideal \(J \subset R\) which contains \(I\) and \((R/I)/\overline{J} = R/J\).

Lemma

A closed immersion is quasi-compact.

Proof

This lemma is a duplicate of Schemes, Lemma 01K7.

Lemma

A closed immersion is separated.

Proof

This lemma is a special case of Schemes, Lemma 01L7.

Immersions

In this section we collect some facts on immersions.

Lemma

Let \(Z \to Y \to X\) be morphisms of schemes.

If \(Z \to X\) is an immersion, then \(Z \to Y\) is an immersion.

If \(Z \to X\) is a quasi-compact immersion and \(Y \to X\) is quasi-separated, then \(Z \to Y\) is a quasi-compact immersion.

If \(Z \to X\) is a closed immersion and \(Y \to X\) is separated, then \(Z \to Y\) is a closed immersion.

Proof

In each case the proof is to contemplate the commutative diagram \[\xymatrix{ Z \ar[r] \ar[rd] & Y \times_X Z \ar[r] \ar[d] & Z \ar[d] \\ & Y \ar[r] & X }\] where the composition of the top horizontal arrows is the identity. Let us prove (1). The first horizontal arrow is a section of \(Y \times_X Z \to Z\), whence an immersion by Schemes, Lemma 01KT. The arrow \(Y \times_X Z \to Y\) is a base change of \(Z \to X\) hence an immersion (Schemes, Lemma 01JY). Finally, a composition of immersions is an immersion (Schemes, Lemma 02V0). This proves (1). The other two results are proved in exactly the same manner.

Lemma

Let \(h : Z \to X\) be an immersion. If \(h\) is quasi-compact, then we can factor \(h = i \circ j\) with \(j : Z \to \overline{Z}\) an open immersion and \(i : \overline{Z} \to X\) a closed immersion.

Proof

Note that \(h\) is quasi-compact and quasi-separated (see Schemes, Lemma 01L7). Hence \(h_*\mathcal{O}_Z\) is a quasi-coherent sheaf of \(\mathcal{O}_X\)-modules by Schemes, Lemma 01LC. This implies that \(\mathcal{I} = \Ker(\mathcal{O}_X \to h_*\mathcal{O}_Z)\) is a quasi-coherent sheaf of ideals, see Schemes, Section 01LA. Let \(\overline{Z} \subset X\) be the closed subscheme corresponding to \(\mathcal{I}\), see Lemma 01QQ. By Schemes, Lemma 01HP the morphism \(h\) factors as \(h = i \circ j\) where \(i : \overline{Z} \to X\) is the inclusion morphism. To see that \(j\) is an open immersion, choose an open subscheme \(U \subset X\) such that \(h\) induces a closed immersion of \(Z\) into \(U\). Then it is clear that \(\mathcal{I}|_U\) is the sheaf of ideals corresponding to the closed immersion \(Z \to U\). Hence we see that \(Z = \overline{Z} \cap U\).

Lemma

Let \(h : Z \to X\) be an immersion. If \(Z\) is reduced, then we can factor \(h = i \circ j\) with \(j : Z \to \overline{Z}\) an open immersion and \(i : \overline{Z} \to X\) a closed immersion.

Proof

Let \(\overline{Z} \subset X\) be the closure of \(h(Z)\) with the reduced induced closed subscheme structure, see Schemes, Definition 01J4. By Schemes, Lemma 0356 the morphism \(h\) factors as \(h = i \circ j\) with \(i : \overline{Z} \to X\) the inclusion morphism and \(j : Z \to \overline{Z}\). From the definition of an immersion we see there exists an open subscheme \(U \subset X\) such that \(h\) factors through a closed immersion into \(U\). Hence \(\overline{Z} \cap U\) and \(h(Z)\) are reduced closed subschemes of \(U\) with the same underlying closed set. Hence by the uniqueness in Schemes, Lemma 01J3 we see that \(h(Z) \cong \overline{Z} \cap U\). So \(j\) induces an isomorphism of \(Z\) with \(\overline{Z} \cap U\). In other words \(j\) is an open immersion.

Example

Here is an example of an immersion which is not a composition of an open immersion followed by a closed immersion. Let \(k\) be a field. Let \(X = \Spec(k[x_1, x_2, x_3, \ldots])\). Let \(U = \bigcup_{n = 1}^{\infty} D(x_n)\). Then \(U \to X\) is an open immersion. Consider the ideals \[I_n = (x_1^n, x_2^n, \ldots, x_{n - 1}^n, x_n - 1, x_{n + 1}, x_{n + 2}, \ldots) \subset k[x_1, x_2, x_3, \ldots][1/x_n].\] Note that \(I_n k[x_1, x_2, x_3, \ldots][1/x_nx_m] = (1)\) for any \(m \not = n\). Hence the quasi-coherent ideals \(\widetilde I_n\) on \(D(x_n)\) agree on \(D(x_nx_m)\), namely \(\widetilde I_n|_{D(x_nx_m)} = \mathcal{O}_{D(x_n x_m)}\) if \(n \not = m\). Hence these ideals glue to a quasi-coherent sheaf of ideals \(\mathcal{I} \subset \mathcal{O}_U\). Let \(Z \subset U\) be the closed subscheme corresponding to \(\mathcal{I}\). Thus \(Z \to X\) is an immersion.

We claim that we cannot factor \(Z \to X\) as \(Z \to \overline{Z} \to X\), where \(\overline{Z} \to X\) is closed and \(Z \to \overline{Z}\) is open. Namely, \(\overline{Z}\) would have to be defined by an ideal \(I \subset k[x_1, x_2, x_3, \ldots]\) such that \(I_n = I k[x_1, x_2, x_3, \ldots][1/x_n]\). But the only element \(f \in k[x_1, x_2, x_3, \ldots]\) which ends up in all \(I_n\) is \(0\)! Hence \(I\) does not exist.

Lemma

Let \(f : Y \to X\) be a morphism of schemes. If for all \(y \in Y\) there is an open subscheme \(f(y) \in U \subset X\) such that \(f|_{f^{-1}(U)} : f^{-1}(U) \to U\) is an immersion, then \(f\) is an immersion.

Proof

This statement follows readily from the discussion of closed subschemes at the end of Schemes, Section 01IM but we will also give a detailed proof. Let \(Z \subset X\) be the closure of \(f(Y)\). Since taking closures commutes with restricting to opens, we see from the assumption that \(f(Y) \subset Z\) is open. Hence \(Z' = Z \setminus f(Y)\) is closed. Hence \(X' = X \setminus Z'\) is an open subscheme of \(X\) and \(f\) factors as \(f : Y \to X'\) followed by the inclusion. If \(y \in Y\) and \(U \subset X\) is as in the statement of the lemma, then \(U' = X' \cap U\) is an open neighbourhood of \(f'(y)\) such that \((f')^{-1}(U') \to U'\) is an immersion (Lemma 07RK) with closed image. Hence it is a closed immersion, see Schemes, Lemma 01IQ. Since being a closed immersion is local on the target (for example by Lemma 01QO) we conclude that \(f'\) is a closed immersion as desired.

Closed immersions and quasi-coherent sheaves

The following lemma finally does for quasi-coherent sheaves on schemes what Modules, Lemma 01AX does for abelian sheaves. See also the discussion in Modules, Section 01C1.

Lemma

Let \(i : Z \to X\) be a closed immersion of schemes. Let \(\mathcal{I} \subset \mathcal{O}_X\) be the quasi-coherent sheaf of ideals cutting out \(Z\). The functor \[i_* : \QCoh(\mathcal{O}_Z) \longrightarrow \QCoh(\mathcal{O}_X)\] is exact, fully faithful, with essential image those quasi-coherent \(\mathcal{O}_X\)-modules \(\mathcal{G}\) such that \(\mathcal{I}\mathcal{G} = 0\).

Proof

A closed immersion is quasi-compact and separated, see Lemmas 01QT and 01QU. Hence Schemes, Lemma 01LC applies and the pushforward of a quasi-coherent sheaf on \(Z\) is indeed a quasi-coherent sheaf on \(X\).

By Modules, Lemma 08KS the functor \(i_*\) is fully faithful.

Now we turn to the description of the essential image of the functor \(i_*\). We have \(\mathcal{I}(i_*\mathcal{F}) = 0\) for any quasi-coherent \(\mathcal{O}_Z\)-module, for example by Modules, Lemma 08KS. Next, suppose that \(\mathcal{G}\) is any quasi-coherent \(\mathcal{O}_X\)-module such that \(\mathcal{I}\mathcal{G} = 0\). It suffices to show that the canonical map \[\mathcal{G} \longrightarrow i_* i^*\mathcal{G}\] is an isomorphism¹. In the case of schemes and quasi-coherent modules, working affine locally on \(X\) and using Lemma 01QO and Schemes, Lemma 01I9 it suffices to prove the following algebraic statement: Given a ring \(R\), an ideal \(I\) and an \(R\)-module \(N\) such that \(IN = 0\) the canonical map \[N \longrightarrow N \otimes_R R/I,\quad n \longmapsto n \otimes 1\] is an isomorphism of \(R\)-modules. Proof of this easy algebra fact is omitted.

Let \(i : Z \to X\) be a closed immersion. Because of the lemma above we often, by abuse of notation, denote \(\mathcal{F}\) the sheaf \(i_*\mathcal{F}\) on \(X\).

Lemma

Let \(X\) be a scheme. Let \(\mathcal{F}\) be a quasi-coherent \(\mathcal{O}_X\)-module. Let \(\mathcal{G} \subset \mathcal{F}\) be a \(\mathcal{O}_X\)-submodule. There exists a unique quasi-coherent \(\mathcal{O}_X\)-submodule \(\mathcal{G}' \subset \mathcal{G}\) with the following property: For every quasi-coherent \(\mathcal{O}_X\)-module \(\mathcal{H}\) the map \[\Hom_{\mathcal{O}_X}(\mathcal{H}, \mathcal{G}') \longrightarrow \Hom_{\mathcal{O}_X}(\mathcal{H}, \mathcal{G})\] is bijective. In particular \(\mathcal{G}'\) is the largest quasi-coherent \(\mathcal{O}_X\)-submodule of \(\mathcal{F}\) contained in \(\mathcal{G}\).

Proof

Let \(\mathcal{G}_a\), \(a \in A\) be the set of quasi-coherent \(\mathcal{O}_X\)-submodules contained in \(\mathcal{G}\). Then the image \(\mathcal{G}'\) of \[\bigoplus\nolimits_{a \in A} \mathcal{G}_a \longrightarrow \mathcal{F}\] is quasi-coherent as the image of a map of quasi-coherent sheaves on \(X\) is quasi-coherent and since a direct sum of quasi-coherent sheaves is quasi-coherent, see Schemes, Section 01LA. The module \(\mathcal{G}'\) is contained in \(\mathcal{G}\). Hence this is the largest quasi-coherent \(\mathcal{O}_X\)-module contained in \(\mathcal{G}\).

To prove the formula, let \(\mathcal{H}\) be a quasi-coherent \(\mathcal{O}_X\)-module and let \(\alpha : \mathcal{H} \to \mathcal{G}\) be an \(\mathcal{O}_X\)-module map. The image of the composition \(\mathcal{H} \to \mathcal{G} \to \mathcal{F}\) is quasi-coherent as the image of a map of quasi-coherent sheaves. Hence it is contained in \(\mathcal{G}'\). Hence \(\alpha\) factors through \(\mathcal{G}'\) as desired.

Lemma

Let \(i : Z \to X\) be a closed immersion of schemes. There is a functor² \(i^! : \QCoh(\mathcal{O}_X) \to \QCoh(\mathcal{O}_Z)\) which is a right adjoint to \(i_*\). (Compare Modules, Lemma 01AZ.)

Proof

Given quasi-coherent \(\mathcal{O}_X\)-module \(\mathcal{G}\) we consider the subsheaf \(\mathcal{H}_Z(\mathcal{G})\) of \(\mathcal{G}\) of local sections annihilated by \(\mathcal{I}\). By Lemma 01QZ there is a canonical largest quasi-coherent \(\mathcal{O}_X\)-submodule \(\mathcal{H}_Z(\mathcal{G})'\). By construction we have \[\Hom_{\mathcal{O}_X}(i_*\mathcal{F}, \mathcal{H}_Z(\mathcal{G})') = \Hom_{\mathcal{O}_X}(i_*\mathcal{F}, \mathcal{G})\] for any quasi-coherent \(\mathcal{O}_Z\)-module \(\mathcal{F}\). Hence we can set \(i^!\mathcal{G} = i^*(\mathcal{H}_Z(\mathcal{G})')\). Details omitted.

Using the \(1\)-to-\(1\) corresponding between quasi-coherent sheaves of ideals and closed subschemes (see Lemma 01QQ) we can define scheme theoretic intersections and unions of closed subschemes.

Definition

Let \(X\) be a scheme. Let \(Z, Y \subset X\) be closed subschemes corresponding to quasi-coherent ideal sheaves \(\mathcal{I}, \mathcal{J} \subset \mathcal{O}_X\). The scheme theoretic intersection of \(Z\) and \(Y\) is the closed subscheme of \(X\) cut out by \(\mathcal{I} + \mathcal{J}\). The scheme theoretic union of \(Z\) and \(Y\) is the closed subscheme of \(X\) cut out by \(\mathcal{I} \cap \mathcal{J}\).

Lemma

Let \(X\) be a scheme. Let \(Z, Y \subset X\) be closed subschemes. Let \(Z \cap Y\) be the scheme theoretic intersection of \(Z\) and \(Y\). Then \(Z \cap Y \to Z\) and \(Z \cap Y \to Y\) are closed immersions and \[\xymatrix{ Z \cap Y \ar[r] \ar[d] & Z \ar[d] \\ Y \ar[r] & X }\] is a cartesian diagram of schemes, i.e., \(Z \cap Y = Z \times_X Y\).

Proof

The morphisms \(Z \cap Y \to Z\) and \(Z \cap Y \to Y\) are closed immersions by Lemma 01QP. Let \(U = \Spec(A)\) be an affine open of \(X\) and let \(Z \cap U\) and \(Y \cap U\) correspond to the ideals \(I \subset A\) and \(J \subset A\). Then \(Z \cap Y \cap U\) corresponds to \(I + J \subset A\). Since \(A/I \otimes_A A/J = A/(I + J)\) we see that the diagram is cartesian by our description of fibre products of schemes in Schemes, Section 01JO.

Lemma

Let \(S\) be a scheme. Let \(X, Y \subset S\) be closed subschemes. Let \(X \cup Y\) be the scheme theoretic union of \(X\) and \(Y\). Let \(X \cap Y\) be the scheme theoretic intersection of \(X\) and \(Y\). Then \(X \to X \cup Y\) and \(Y \to X \cup Y\) are closed immersions, there is a short exact sequence \[0 \to \mathcal{O}_{X \cup Y} \to \mathcal{O}_X \times \mathcal{O}_Y \to \mathcal{O}_{X \cap Y} \to 0\] of \(\mathcal{O}_S\)-modules, and the diagram \[\xymatrix{ X \cap Y \ar[r] \ar[d] & X \ar[d] \\ Y \ar[r] & X \cup Y }\] is cocartesian in the category of schemes, i.e., \(X \cup Y = X \amalg_{X \cap Y} Y\).

Proof

The morphisms \(X \to X \cup Y\) and \(Y \to X \cup Y\) are closed immersions by Lemma 01QP. In the short exact sequence we use the equivalence of Lemma 01QY to think of quasi-coherent modules on closed subschemes of \(S\) as quasi-coherent modules on \(S\). For the first map in the sequence we use the canonical maps \(\mathcal{O}_{X \cup Y} \to \mathcal{O}_X\) and \(\mathcal{O}_{X \cup Y} \to \mathcal{O}_Y\) and for the second map we use the canonical map \(\mathcal{O}_X \to \mathcal{O}_{X \cap Y}\) and the negative of the canonical map \(\mathcal{O}_Y \to \mathcal{O}_{X \cap Y}\). Then to check exactness we may work affine locally. Let \(U = \Spec(A)\) be an affine open of \(S\) and let \(X \cap U\) and \(Y \cap U\) correspond to the ideals \(I \subset A\) and \(J \subset A\). Then \((X \cup Y) \cap U\) corresponds to \(I \cap J \subset A\) and \(X \cap Y \cap U\) corresponds to \(I + J \subset A\). Thus exactness follows from the exactness of \[0 \to A/I \cap J \to A/I \times A/J \to A/(I + J) \to 0\] To show the diagram is cocartesian, suppose we are given a scheme \(T\) and morphisms of schemes \(f : X \to T\), \(g : Y \to T\) agreeing as morphisms \(X \cap Y \to T\). Goal: Show there exists a unique morphism \(h : X \cup Y \to T\) agreeing with \(f\) and \(g\). To construct \(h\) we may work affine locally on \(X \cup Y\), see Schemes, Section 01JA. If \(s \in X\), \(s \not \in Y\), then \(X \to X \cup Y\) is an isomorphism in a neighbourhood of \(s\) and it is clear how to construct \(h\). Similarly for \(s \in Y\), \(s \not \in X\). For \(s \in X \cap Y\) we can pick an affine open \(V = \Spec(B) \subset T\) containing \(f(s) = g(s)\). Then we can choose an affine open \(U = \Spec(A) \subset S\) containing \(s\) such that \(f(X \cap U)\) and \(g(Y \cap U)\) are contained in \(V\). The morphisms \(f|_{X \cap U}\) and \(g|_{Y \cap V}\) into \(V\) correspond to ring maps \[B \to A/I \quad\text{and}\quad B \to A/J\] which agree as maps into \(A/(I + J)\). By the short exact sequence displayed above there is a unique lift of these ring homomorphism to a ring map \(B \to A/I \cap J\) as desired.

Supports of modules

In this section we collect some elementary results on supports of quasi-coherent modules on schemes. Recall that the support of a sheaf of modules has been defined in Modules, Section 01AS. On the other hand, the support of a module was defined in Algebra, Section 00KY. These match.

Lemma

Let \(X\) be a scheme. Let \(\mathcal{F}\) be a quasi-coherent sheaf on \(X\). Let \(\Spec(A) = U \subset X\) be an affine open, and set \(M = \Gamma(U, \mathcal{F})\). Let \(x \in U\), and let \(\mathfrak p \subset A\) be the corresponding prime. The following are equivalent

\(\mathfrak p\) is in the support of \(M\), and

\(x\) is in the support of \(\mathcal{F}\).

Proof

This follows from the equality \(\mathcal{F}_x = M_{\mathfrak p}\), see Schemes, Lemma 01HV and the definitions.

Lemma

Let \(X\) be a scheme. Let \(\mathcal{F}\) be a quasi-coherent sheaf on \(X\). The support of \(\mathcal{F}\) is closed under specialization.

Proof

If \(x' \leadsto x\) is a specialization and \(\mathcal{F}_x = 0\) then \(\mathcal{F}_{x'}\) is zero, as \(\mathcal{F}_{x'}\) is a localization of the module \(\mathcal{F}_x\). Hence the complement of \(\text{Supp}(\mathcal{F})\) is closed under generalization.

For finite type quasi-coherent modules the support is closed, can be checked on fibres, and commutes with base change.

Lemma

Let \(\mathcal{F}\) be a finite type quasi-coherent module on a scheme \(X\). Then

The support of \(\mathcal{F}\) is closed.

For \(x \in X\) we have \[x \in \text{Supp}(\mathcal{F}) \Leftrightarrow \mathcal{F}_x \not = 0 \Leftrightarrow \mathcal{F}_x \otimes_{\mathcal{O}_{X, x}} \kappa(x) \not = 0.\]

For any morphism of schemes \(f : Y \to X\) the pullback \(f^*\mathcal{F}\) is of finite type as well and we have \(\text{Supp}(f^*\mathcal{F}) = f^{-1}(\text{Supp}(\mathcal{F}))\).

Proof

Part (1) is a reformulation of Modules, Lemma 01BA. You can also combine Lemma 056I, Properties, Lemma 01PB, and Algebra, Lemma 00L2 to see this. The first equivalence in (2) is the definition of support, and the second equivalence follows from Nakayama’s lemma, see Algebra, Lemma 00DV. Let \(f : Y \to X\) be a morphism of schemes. Note that \(f^*\mathcal{F}\) is of finite type by Modules, Lemma 01B6. For the final assertion, let \(y \in Y\) with image \(x \in X\). Recall that \[(f^*\mathcal{F})_y = \mathcal{F}_x \otimes_{\mathcal{O}_{X, x}} \mathcal{O}_{Y, y},\] see Sheaves, Lemma 0098. Hence \((f^*\mathcal{F})_y \otimes \kappa(y)\) is nonzero if and only if \(\mathcal{F}_x \otimes \kappa(x)\) is nonzero. By (2) this implies \(x \in \text{Supp}(\mathcal{F})\) if and only if \(y \in \text{Supp}(f^*\mathcal{F})\), which is the content of assertion (3).

Lemma

Let \(\mathcal{F}\) be a finite type quasi-coherent module on a scheme \(X\). There exists a smallest closed subscheme \(i : Z \to X\) such that there exists a quasi-coherent \(\mathcal{O}_Z\)-module \(\mathcal{G}\) with \(i_*\mathcal{G} \cong \mathcal{F}\). Moreover:

If \(\Spec(A) \subset X\) is any affine open, and \(\mathcal{F}|_{\Spec(A)} = \widetilde{M}\) then \(Z \cap \Spec(A) = \Spec(A/I)\) where \(I = \text{Ann}_A(M)\).

The quasi-coherent sheaf \(\mathcal{G}\) is unique up to unique isomorphism.

The quasi-coherent sheaf \(\mathcal{G}\) is of finite type.

The support of \(\mathcal{G}\) and of \(\mathcal{F}\) is \(Z\).

Proof

Suppose that \(i' : Z' \to X\) is a closed subscheme which satisfies the description on open affines from the lemma. Then by Lemma 01QY we see that \(\mathcal{F} \cong i'_*\mathcal{G}'\) for some unique quasi-coherent sheaf \(\mathcal{G}'\) on \(Z'\). Furthermore, it is clear that \(Z'\) is the smallest closed subscheme with this property (by the same lemma). Finally, using Properties, Lemma 01PB and Algebra, Lemma 0560 it follows that \(\mathcal{G}'\) is of finite type. We have \(\text{Supp}(\mathcal{G}') = Z\) by Algebra, Lemma 00L2. Hence, in order to prove the lemma it suffices to show that the characterization in (1) actually does define a closed subscheme. And, in order to do this it suffices to prove that the given rule produces a quasi-coherent sheaf of ideals, see Lemma 01QQ. This comes down to the following algebra fact: If \(A\) is a ring, \(f \in A\), and \(M\) is a finite \(A\)-module, then \(\text{Ann}_A(M)_f = \text{Ann}_{A_f}(M_f)\). We omit the proof.

Definition

Let \(X\) be a scheme. Let \(\mathcal{F}\) be a quasi-coherent \(\mathcal{O}_X\)-module of finite type. The scheme theoretic support of \(\mathcal{F}\) is the closed subscheme \(Z \subset X\) constructed in Lemma 05JU.

In this situation we often think of \(\mathcal{F}\) as a quasi-coherent sheaf of finite type on \(Z\) (via the equivalence of categories of Lemma 01QY).

Scheme theoretic image

Caution: Some of the material in this section is ultra-general and behaves differently from what you might expect.

Lemma

Let \(f : X \to Y\) be a morphism of schemes. There exists a closed subscheme \(Z \subset Y\) such that \(f\) factors through \(Z\) and such that for any other closed subscheme \(Z' \subset Y\) such that \(f\) factors through \(Z'\) we have \(Z \subset Z'\).

Proof

Let \(\mathcal{I} = \Ker(\mathcal{O}_Y \to f_*\mathcal{O}_X)\). If \(\mathcal{I}\) is quasi-coherent then we just take \(Z\) to be the closed subscheme determined by \(\mathcal{I}\), see Lemma 01QQ. This works by Lemma 01QP. In general the same lemma requires us to show that there exists a largest quasi-coherent sheaf of ideals \(\mathcal{I}'\) contained in \(\mathcal{I}\). This follows from Lemma 01QZ.

Definition

Let \(f : X \to Y\) be a morphism of schemes. The scheme theoretic image of \(f\) is the smallest closed subscheme \(Z \subset Y\) through which \(f\) factors, see Lemma 01R6 above.

For a morphism \(f : X \to Y\) of schemes with scheme theoretic image \(Z\) we often denote \(f : X \to Z\) the factorization of \(f\) through its scheme theoretic image. If the morphism \(f\) is not quasi-compact, then (in general)

the set theoretic inclusion \(\overline{f(X)} \subset Z\) is not an equality, i.e., \(f(X) \subset Z\) is not a dense subset, and
the construction of the scheme theoretic image does not commute with restriction to open subschemes to \(Y\).

In Examples, Section 0GIK the reader finds an example for both phenomena. These phenomena can arise even for immersions, see Examples, Section 078B. However, the next lemma shows that both disasters are avoided when the morphism is quasi-compact.

Lemma

Let \(f : X \to Y\) be a morphism of schemes. Let \(Z \subset Y\) be the scheme theoretic image of \(f\). If \(f\) is quasi-compact then

the sheaf of ideals \(\mathcal{I} = \Ker(\mathcal{O}_Y \to f_*\mathcal{O}_X)\) is quasi-coherent,

the scheme theoretic image \(Z\) is the closed subscheme determined by \(\mathcal{I}\),

for any open \(U \subset Y\) the scheme theoretic image of \(f|_{f^{-1}(U)} : f^{-1}(U) \to U\) is equal to \(Z \cap U\), and

the image \(f(X) \subset Z\) is a dense subset of \(Z\), in other words the morphism \(X \to Z\) is dominant (see Definition 01RJ).

Proof

Part (4) follows from part (3). To show (3) it suffices to prove (1) since the formation of \(\mathcal{I}\) commutes with restriction to open subschemes of \(Y\). And if (1) holds then in the proof of Lemma 01R6 we showed (2). Thus it suffices to prove that \(\mathcal{I}\) is quasi-coherent. Since the property of being quasi-coherent is local we may assume \(Y\) is affine. As \(f\) is quasi-compact, we can find a finite affine open covering \(X = \bigcup_{i = 1, \ldots, n} U_i\). Denote \(f'\) the composition \[X' = \coprod U_i \longrightarrow X \longrightarrow Y.\] Then \(f_*\mathcal{O}_X\) is a subsheaf of \(f'_*\mathcal{O}_{X'}\), and hence \(\mathcal{I} = \Ker(\mathcal{O}_Y \to f'_*\mathcal{O}_{X'})\). By Schemes, Lemma 01LC the sheaf \(f'_*\mathcal{O}_{X'}\) is quasi-coherent on \(Y\). Hence we win.

Example

If \(A \to B\) is a ring map with kernel \(I\), then the scheme theoretic image of \(\Spec(B) \to \Spec(A)\) is the closed subscheme \(\Spec(A/I)\) of \(\Spec(A)\). This follows from Lemma 01R8.

If the morphism is quasi-compact, then the scheme theoretic image only adds points which are specializations of points in the image.

Lemma

Let \(f : X \to Y\) be a quasi-compact morphism. Let \(Z\) be the scheme theoretic image of \(f\). Let \(z \in Z\)³. There exists a valuation ring \(A\) with fraction field \(K\) and a commutative diagram \[\xymatrix{ \Spec(K) \ar[rr] \ar[d] & & X \ar[d] \ar[ld] \\ \Spec(A) \ar[r] & Z \ar[r] & Y }\] such that the closed point of \(\Spec(A)\) maps to \(z\). In particular any point of \(Z\) is the specialization of a point of \(f(X)\).

Proof

Let \(z \in \Spec(R) = V \subset Y\) be an affine open neighbourhood of \(z\). By Lemma 01R8 the intersection \(Z \cap V\) is the scheme theoretic image of \(f^{-1}(V) \to V\). Hence we may replace \(Y\) by \(V\) and assume \(Y = \Spec(R)\) is affine. In this case \(X\) is quasi-compact as \(f\) is quasi-compact. Say \(X = U_1 \cup \ldots \cup U_n\) is a finite affine open covering. Write \(U_i = \Spec(A_i)\). Let \(I = \Ker(R \to A_1 \times \ldots \times A_n)\). By Lemma 01R8 again we see that \(Z\) corresponds to the closed subscheme \(\Spec(R/I)\) of \(Y\). If \(\mathfrak p \subset R\) is the prime corresponding to \(z\), then we see that \(I_{\mathfrak p} \subset R_{\mathfrak p}\) is not an equality. Hence (as localization is exact, see Algebra, Proposition 00CS) we see that \(R_{\mathfrak p} \to (A_1)_{\mathfrak p} \times \ldots \times (A_n)_{\mathfrak p}\) is not zero. Hence one of the rings \((A_i)_{\mathfrak p}\) is not zero. Hence there exists an \(i\) and a prime \(\mathfrak q_i \subset A_i\) lying over a prime \(\mathfrak p_i \subset \mathfrak p\). By Algebra, Lemma 00IA we can choose a valuation ring \(A \subset K = \kappa(\mathfrak q_i)\) dominating the local ring \(R_{\mathfrak p}/\mathfrak p_iR_{\mathfrak p} \subset \kappa(\mathfrak q_i)\). This gives the desired diagram. Some details omitted.

Lemma

Let \[\xymatrix{ X_1 \ar[d] \ar[r]_{f_1} & Y_1 \ar[d] \\ X_2 \ar[r]^{f_2} & Y_2 }\] be a commutative diagram of schemes. Let \(Z_i \subset Y_i\), \(i = 1, 2\) be the scheme theoretic image of \(f_i\). Then the morphism \(Y_1 \to Y_2\) induces a morphism \(Z_1 \to Z_2\) and a commutative diagram \[\xymatrix{ X_1 \ar[r] \ar[d] & Z_1 \ar[d] \ar[r] & Y_1 \ar[d] \\ X_2 \ar[r] & Z_2 \ar[r] & Y_2 }\]

Proof

The scheme theoretic inverse image of \(Z_2\) in \(Y_1\) is a closed subscheme of \(Y_1\) through which \(f_1\) factors. Hence \(Z_1\) is contained in this. This proves the lemma.

Lemma

Let \(f : X \to Y\) be a morphism of schemes. If \(X\) is reduced, then the scheme theoretic image of \(f\) is the reduced induced scheme structure on \(\overline{f(X)}\).

Proof

This is true because the reduced induced scheme structure on \(\overline{f(X)}\) is clearly the smallest closed subscheme of \(Y\) through which \(f\) factors, see Schemes, Lemma 0356.

Lemma

Let \(f : X \to Y\) be a separated morphism of schemes. Let \(V \subset Y\) be a retrocompact open. Let \(s : V \to X\) be a morphism such that \(f \circ s = \text{id}_V\). Let \(Y'\) be the scheme theoretic image of \(s\). Then \(Y' \to Y\) is an isomorphism over \(V\).

Proof

The assumption that \(V\) is retrocompact in \(Y\) (Topology, Definition 005A) means that \(V \to Y\) is a quasi-compact morphism. By Schemes, Lemma 03GI the morphism \(s : V \to X\) is quasi-compact. Hence the construction of the scheme theoretic image \(Y'\) of \(s\) commutes with restriction to opens by Lemma 01R8. In particular, we see that \(Y' \cap f^{-1}(V)\) is the scheme theoretic image of a section of the separated morphism \(f^{-1}(V) \to V\). Since a section of a separated morphism is a closed immersion (Schemes, Lemma 01KT), we conclude that \(Y' \cap f^{-1}(V) \to V\) is an isomorphism as desired.

Scheme theoretic closure and density

We take the following definition from [EGA, IV, Definition 11.10.2].

Definition

Let \(X\) be a scheme. Let \(U \subset X\) be an open subscheme.

The scheme theoretic image of the morphism \(U \to X\) is called the scheme theoretic closure of \(U\) in \(X\).

We say \(U\) is scheme theoretically dense in \(X\) if for every open \(V \subset X\) the scheme theoretic closure of \(U \cap V\) in \(V\) is equal to \(V\).

With this definition it is not the case that \(U\) is scheme theoretically dense in \(X\) if and only if the scheme theoretic closure of \(U\) is \(X\), see Example 01RC. This is somewhat inelegant; but see Lemmas 01RD and 056D below. On the other hand, with this definition \(U\) is scheme theoretically dense in \(X\) if and only if for every \(V \subset X\) open the ring map \(\mathcal{O}_X(V) \to \mathcal{O}_X(U \cap V)\) is injective, see Lemma 01RE below. In particular we see that scheme theoretically dense implies dense which is pleasing.

Example

Here is an example where scheme theoretic closure being \(X\) does not imply dense for the underlying topological spaces. Let \(k\) be a field. Set \(A = k[x, z_1, z_2, \ldots]/(x^n z_n)\) Set \(I = (z_1, z_2, \ldots) \subset A\). Consider the affine scheme \(X = \Spec(A)\) and the open subscheme \(U = X \setminus V(I)\). Since \(A \to \prod_n A_{z_n}\) is injective we see that the scheme theoretic closure of \(U\) is \(X\). Consider the morphism \(X \to \Spec(k[x])\). This morphism is surjective (set all \(z_n = 0\) to see this). But the restriction of this morphism to \(U\) is not surjective because it maps to the point \(x = 0\). Hence \(U\) cannot be topologically dense in \(X\).

Lemma

Let \(X\) be a scheme. Let \(U \subset X\) be an open subscheme. If the inclusion morphism \(U \to X\) is quasi-compact, then \(U\) is scheme theoretically dense in \(X\) if and only if the scheme theoretic closure of \(U\) in \(X\) is \(X\).

Proof

Follows from Lemma 01R8 part (3).

Example

Let \(A\) be a ring and \(X = \Spec(A)\). Let \(f_1, \ldots, f_n \in A\) and let \(U = D(f_1) \cup \ldots \cup D(f_n)\). Let \(I = \Ker(A \to \prod A_{f_i})\). Then the scheme theoretic closure of \(U\) in \(X\) is the closed subscheme \(\Spec(A/I)\) of \(X\). Note that \(U \to X\) is quasi-compact. Hence by Lemma 01RD we see \(U\) is scheme theoretically dense in \(X\) if and only if \(I = 0\).

Lemma

Let \(j : U \to X\) be an open immersion of schemes. Then \(U\) is scheme theoretically dense in \(X\) if and only if \(\mathcal{O}_X \to j_*\mathcal{O}_U\) is injective.

Proof

If \(\mathcal{O}_X \to j_*\mathcal{O}_U\) is injective, then the same is true when restricted to any open \(V\) of \(X\). Hence the scheme theoretic closure of \(U \cap V\) in \(V\) is equal to \(V\), see proof of Lemma 01R6. Conversely, suppose that the scheme theoretic closure of \(U \cap V\) is equal to \(V\) for all opens \(V\). Suppose that \(\mathcal{O}_X \to j_*\mathcal{O}_U\) is not injective. Then we can find an affine open, say \(\Spec(A) = V \subset X\) and a nonzero element \(f \in A\) such that \(f\) maps to zero in \(\Gamma(V \cap U, \mathcal{O}_X)\). In this case the scheme theoretic closure of \(V \cap U\) in \(V\) is clearly contained in \(\Spec(A/(f))\) a contradiction.

Lemma

Let \(X\) be a scheme. If \(U\), \(V\) are scheme theoretically dense open subschemes of \(X\), then so is \(U \cap V\).

Proof

Let \(W \subset X\) be any open. Consider the map \(\mathcal{O}_X(W) \to \mathcal{O}_X(W \cap V) \to \mathcal{O}_X(W \cap V \cap U)\). By Lemma 01RE both maps are injective. Hence the composite is injective. Hence by Lemma 01RE \(U \cap V\) is scheme theoretically dense in \(X\).

Lemma

Let \(h : Z \to X\) be an immersion. Assume either \(h\) is quasi-compact or \(Z\) is reduced. Let \(\overline{Z} \subset X\) be the scheme theoretic image of \(h\). Then the morphism \(Z \to \overline{Z}\) is an open immersion which identifies \(Z\) with a scheme theoretically dense open subscheme of \(\overline{Z}\). Moreover, \(Z\) is topologically dense in \(\overline{Z}\).

Proof

By Lemma 01QV or Lemma 03DQ we can factor \(Z \to X\) as \(Z \to \overline{Z}_1 \to X\) with \(Z \to \overline{Z}_1\) open and \(\overline{Z}_1 \to X\) closed. On the other hand, let \(Z \to \overline{Z} \subset X\) be the scheme theoretic image of \(Z \to X\). We conclude that \(\overline{Z} \subset \overline{Z}_1\). Since \(Z\) is an open subscheme of \(\overline{Z}_1\) it follows that \(Z\) is an open subscheme of \(\overline{Z}\) as well. In the case that \(Z\) is reduced we know that \(Z \subset \overline{Z}_1\) is topologically dense by the construction of \(\overline{Z}_1\) in the proof of Lemma 03DQ. Hence \(\overline{Z}_1\) and \(\overline{Z}\) have the same underlying topological spaces. Thus \(\overline{Z} \subset \overline{Z}_1\) is a closed immersion into a reduced scheme which induces a bijection on underlying topological spaces, and hence it is an isomorphism. In the case that \(Z \to X\) is quasi-compact we argue as follows: The assertion that \(Z\) is scheme theoretically dense in \(\overline{Z}\) follows from Lemma 01R8 part (3). The last assertion follows from Lemma 01R8 part (4).

Lemma

Let \(X\) be a reduced scheme and let \(U \subset X\) be an open subscheme. Then the following are equivalent

\(U\) is topologically dense in \(X\),

the scheme theoretic closure of \(U\) in \(X\) is \(X\), and

\(U\) is scheme theoretically dense in \(X\).

Proof

This follows from Lemma 01RG and the fact that a closed subscheme \(Z\) of \(X\) whose underlying topological space equals \(X\) must be equal to \(X\) as a scheme.

Lemma

Let \(X\) be a scheme and let \(U \subset X\) be a reduced open subscheme. Then the following are equivalent

the scheme theoretic closure of \(U\) in \(X\) is \(X\), and

\(U\) is scheme theoretically dense in \(X\).

If this holds then \(X\) is a reduced scheme.

Proof

This follows from Lemma 01RG and the fact that the scheme theoretic closure of \(U\) in \(X\) is reduced by Lemma 056B.

Lemma

Let \(S\) be a scheme. Let \(X\), \(Y\) be schemes over \(S\). Let \(f, g : X \to Y\) be morphisms of schemes over \(S\). Let \(U \subset X\) be an open subscheme such that \(f|_U = g|_U\). If the scheme theoretic closure of \(U\) in \(X\) is \(X\) and \(Y \to S\) is separated, then \(f = g\).

Proof

Follows from the definitions and Schemes, Lemma 01KM.

Dominant morphisms

The definition of a morphism of schemes being dominant is a little different from what you might expect if you are used to the notion of a dominant morphism of varieties.

Definition

A morphism \(f : X \to S\) of schemes is called dominant if the image of \(f\) is a dense subset of \(S\).

So for example, if \(k\) is an infinite field and \(\lambda_1, \lambda_2, \ldots\) is a countable collection of distinct elements of \(k\), then the morphism \[\coprod\nolimits_{i = 1, 2, \ldots } \Spec(k) \longrightarrow \Spec(k[x])\] with \(i\)th factor mapping to the point \(x = \lambda_i\) is dominant.

Lemma

Let \(f : X \to S\) be a morphism of schemes. If every generic point of every irreducible component of \(S\) is in the image of \(f\), then \(f\) is dominant.

Proof

This is a topological fact which follows directly from the fact that the topological space underlying a scheme is sober, see Schemes, Lemma 01IS, and that every point of \(S\) is contained in an irreducible component of \(S\), see Topology, Lemma 004W.

The expectation that morphisms are dominant only if generic points of the target are in the image does hold if the morphism is quasi-compact.

Lemma

Let \(f : X \to S\) be a quasi-compact morphism of schemes. Then \(f\) is dominant if and only if for every irreducible component \(Z \subset S\) the generic point of \(Z\) is in the image of \(f\).

Proof

Let \(V \subset S\) be an affine open. Because \(f\) is quasi-compact we may choose finitely many affine opens \(U_i \subset f^{-1}(V)\), \(i = 1, \ldots, n\) covering \(f^{-1}(V)\). Consider the morphism of affines \[f' : \coprod\nolimits_{i = 1, \ldots, n} U_i \longrightarrow V.\] A disjoint union of affines is affine, see Schemes, Lemma 01I5. Generic points of irreducible components of \(V\) are exactly the generic points of the irreducible components of \(S\) that meet \(V\). Also, \(f\) is dominant if and only if \(f'\) is dominant no matter what choices of \(V, n, U_i\) we make above. Thus we have reduced the lemma to the case of a morphism of affine schemes. The affine case is Algebra, Lemma 00FL.

Lemma

Let \(f : X \to S\) be a quasi-compact dominant morphism of schemes. Let \(g : S' \to S\) be a morphism of schemes, and let \(f' : X' \to S'\) be the base change of \(f\) by \(g\). If generalizations lift along \(g\), then \(f'\) is dominant.

We will see that generalizations lift along \(g\) if \(g\) is open (Lemma 040F) or flat (Lemma 03HV).

Proof

Observe that \(f'\) is quasi-compact by Schemes, Lemma 01K5. Let \(\eta' \in S'\) be the generic point of an irreducible component of \(S'\). If generalizations lift along \(g\), then \(\eta = g(\eta')\) is the generic point of an irreducible component of \(S\). By Lemma 01RL we see that \(\eta\) is in the image of \(f\). Hence \(\eta'\) is in the image of \(f'\) by Schemes, Lemma 01JT. It follows that \(f'\) is dominant by Lemma 01RL.

Part of this holds for arbitrary (not necessarily quasi-compact) dominant morphisms.

Lemma

Let \(f : X \to S\) be a dominant morphism of schemes. Let \(g : S' \to S\) be an open morphism of schemes, and let \(f' : X' \to S'\) be the base change of \(f\) by \(g\). Then \(f'\) is dominant.

Proof

If \(f'\) is not dominant, then \(f'(X')\) is contained in some closed subset \(B\) not equal to \(S'\). Then \(S'-B\) is open in \(S'\) and nonempty. Since \(g\) is open, \(g(S'-B)\) is nonempty and open in \(S\). Since \(f(X)\) is dense in \(S\), there is a point \(x\in X\) with \(f(x) \in g(S'-B)\). Let \(W = \Spec(\kappa (f(x)))\), and write \(X_W\), \(S_W\), \(S'_W\) for the pullbacks of these schemes from \(S\) to \(W\). Then \(X_W\) and \((S'-B)_W\) are nonempty, and so \(X_W \times_W (S'-B)_W\) is nonempty, as \(W\) is the spectrum of a field. This contradicts that \(f'(X \times_S S')\) is contained in \(B \subset S'\). So in fact \(f'\) is dominant.

Lemma

Let \(f : X \to S\) be a quasi-compact morphism of schemes. Let \(\eta \in S\) be a generic point of an irreducible component of \(S\). If \(\eta \not \in f(X)\) then there exists an open neighbourhood \(V \subset S\) of \(\eta\) such that \(f^{-1}(V) = \emptyset\).

Proof

Let \(Z \subset S\) be the scheme theoretic image of \(f\). We have to show that \(\eta \not \in Z\). This follows from Lemma 02JQ but can also be seen as follows. By Lemma 01R8 the morphism \(X \to Z\) is dominant, which by Lemma 01RL means all the generic points of all irreducible components of \(Z\) are in the image of \(X \to Z\). By assumption we see that \(\eta \not \in Z\) since \(\eta\) would be the generic point of some irreducible component of \(Z\) if it were in \(Z\).

There is another case where dominant is the same as having all generic points of irreducible components in the image.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Suppose that \(X\) has finitely many irreducible components. Then \(f\) is dominant (if and) only if for every irreducible component \(Z \subset S\) the generic point of \(Z\) is in the image of \(f\). If so, then \(S\) has finitely many irreducible components as well.

Proof

Assume \(f\) is dominant. Say \(X = Z_1 \cup Z_2 \cup \ldots \cup Z_n\) is the decomposition of \(X\) into irreducible components. Let \(\xi_i \in Z_i\) be its generic point, so \(Z_i = \overline{\{\xi_i\}}\). Note that \(f(Z_i)\) is an irreducible subset of \(S\). Hence \[S = \overline{f(X)} = \bigcup \overline{f(Z_i)} = \bigcup \overline{\{f(\xi_i)\}}\] is a finite union of irreducible subsets whose generic points are in the image of \(f\). The lemma follows.

Lemma

Let \(f : X \to Y\) be a morphism of integral schemes. The following are equivalent

\(f\) is dominant,

\(f\) maps the generic point of \(X\) to the generic point of \(Y\),

for some nonempty affine opens \(U \subset X\) and \(V \subset Y\) with \(f(U) \subset V\) the ring map \(\mathcal{O}_Y(V) \to \mathcal{O}_X(U)\) is injective,

for all nonempty affine opens \(U \subset X\) and \(V \subset Y\) with \(f(U) \subset V\) the ring map \(\mathcal{O}_Y(V) \to \mathcal{O}_X(U)\) is injective,

for some \(x \in X\) with image \(y = f(x) \in Y\) the local ring map \(\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}\) is injective, and

for all \(x \in X\) with image \(y = f(x) \in Y\) the local ring map \(\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}\) is injective.

Proof

The equivalence of (1) and (2) follows from Lemma 01RM. Let \(U \subset X\) and \(V \subset Y\) be nonempty affine opens with \(f(U) \subset V\). Recall that the rings \(A = \mathcal{O}_X(U)\) and \(B = \mathcal{O}_Y(V)\) are integral domains. The morphism \(f|_U : U \to V\) corresponds to a ring map \(\varphi : B \to A\). The generic points of \(X\) and \(Y\) correspond to the prime ideals \((0) \subset A\) and \((0) \subset B\). Thus (2) is equivalent to the condition \((0) = \varphi^{-1}((0))\), i.e., to the condition that \(\varphi\) is injective. In this way we see that (2), (3), and (4) are equivalent. Similarly, given \(x\) and \(y\) as in (5) the local rings \(\mathcal{O}_{X, x}\) and \(\mathcal{O}_{Y, y}\) are domains and the prime ideals \((0) \subset \mathcal{O}_{X, x}\) and \((0) \subset \mathcal{O}_{Y, y}\) correspond to the generic points of \(X\) and \(Y\) (via the identification of the spectrum of the local ring at \(x\) with the set of points specializing to \(x\), see Schemes, Lemma 01J7). Thus we can argue in the exact same manner as above to see that (2), (5), and (6) are equivalent.

Surjective morphisms

Definition

A morphism of schemes is said to be surjective if it is surjective on underlying topological spaces.

Lemma

The composition of surjective morphisms is surjective.

Proof

Omitted.

Lemma

Let \(X\) and \(Y\) be schemes over a base scheme \(S\). Given points \(x \in X\) and \(y \in Y\), there is a point of \(X \times_S Y\) mapping to \(x\) and \(y\) under the projections if and only if \(x\) and \(y\) lie above the same point of \(S\).

Proof

The condition is obviously necessary, and the converse follows from the proof of Schemes, Lemma 01JT.

Lemma

The base change of a surjective morphism is surjective.

Proof

Let \(f: X \to Y\) be a morphism of schemes over a base scheme \(S\). If \(S' \to S\) is a morphism of schemes, let \(p: X_{S'} \to X\) and \(q: Y_{S'} \to Y\) be the canonical projections. The commutative square \[\xymatrix{ X_{S'} \ar[d]_{f_{S'}} \ar[r]_p & X \ar[d]^{f} \\ Y_{S'} \ar[r]^{q} & Y. }\] identifies \(X_{S'}\) as a fibre product of \(X \to Y\) and \(Y_{S'} \to Y\). Let \(Z\) be a subset of the underlying topological space of \(X\). Then \(q^{-1}(f(Z)) = f_{S'}(p^{-1}(Z))\), because \(y' \in q^{-1}(f(Z))\) if and only if \(q(y') = f(x)\) for some \(x \in Z\), if and only if, by Lemma 0495, there exists \(x' \in X_{S'}\) such that \(f_{S'}(x') = y'\) and \(p(x') = x\). In particular taking \(Z = X\) we see that if \(f\) is surjective so is the base change \(f_{S'}: X_{S'} \to Y_{S'}\).

Example

Bijectivity is not stable under base change, and so neither is injectivity. For example consider the bijection \(\Spec(\mathbf{C}) \to \Spec(\mathbf{R})\). The base change \(\Spec(\mathbf{C} \otimes_{\mathbf{R}} \mathbf{C}) \to \Spec(\mathbf{C})\) is not injective, since there is an isomorphism \(\mathbf{C} \otimes_{\mathbf{R}} \mathbf{C} \cong \mathbf{C} \times \mathbf{C}\) (the decomposition comes from the idempotent \(\frac{1 \otimes 1 + i \otimes i}{2}\)) and hence \(\Spec(\mathbf{C} \otimes_{\mathbf{R}} \mathbf{C})\) has two points.

Lemma

Let \[\xymatrix{ X \ar[rr]_f \ar[rd]_p & & Y \ar[dl]^q \\ & Z }\] be a commutative diagram of morphisms of schemes. If \(f\) is surjective and \(p\) is quasi-compact, then \(q\) is quasi-compact.

Proof

Let \(W \subset Z\) be a quasi-compact open. By assumption \(p^{-1}(W)\) is quasi-compact. Hence by Topology, Lemma 04Z9 the inverse image \(q^{-1}(W) = f(p^{-1}(W))\) is quasi-compact too. This proves the lemma.

Radicial and universally injective morphisms

In this section we define what it means for a morphism of schemes to be radicial and what it means for a morphism of schemes to be universally injective. We then show that these notions agree. The reason for introducing both is that in the case of algebraic spaces there are corresponding notions which may not always agree.

Definition

Let \(f : X \to S\) be a morphism.

We say that \(f\) is universally injective if and only if for any morphism of schemes \(S' \to S\) the base change \(f' : X_{S'} \to S'\) is injective (on underlying topological spaces).

We say \(f\) is radicial if \(f\) is injective as a map of topological spaces, and for every \(x \in X\) the field extension \(\kappa(x)/\kappa(f(x))\) is purely inseparable.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent:

For every field \(K\) the induced map \(\Mor(\Spec(K), X) \to \Mor(\Spec(K), S)\) is injective.

The morphism \(f\) is universally injective.

The morphism \(f\) is radicial.

The diagonal morphism \(\Delta_{X/S} : X \longrightarrow X \times_S X\) is surjective.

Proof

Let \(K\) be a field, and let \(s : \Spec(K) \to S\) be a morphism. Giving a morphism \(x : \Spec(K) \to X\) such that \(f \circ x = s\) is the same as giving a section of the projection \(X_K = \Spec(K) \times_S X \to \Spec(K)\), which in turn is the same as giving a point \(x \in X_K\) whose residue field is \(K\). Hence we see that (2) implies (1).

Conversely, suppose that (1) holds. Assume that \(x, x' \in X_{S'}\) map to the same point \(s' \in S'\). Choose a commutative diagram \[\xymatrix{ K & \kappa(x) \ar[l] \\ \kappa(x') \ar[u] & \kappa(s') \ar[l] \ar[u] }\] of fields. By Schemes, Lemma 01J9 we get two morphisms \(a, a' : \Spec(K) \to X_{S'}\). One corresponding to the point \(x\) and the embedding \(\kappa(x) \subset K\) and the other corresponding to the point \(x'\) and the embedding \(\kappa(x') \subset K\). Also we have \(f' \circ a = f' \circ a'\). Condition (1) now implies that the compositions of \(a\) and \(a'\) with \(X_{S'} \to X\) are equal. Since \(X_{S'}\) is the fibre product of \(S'\) and \(X\) over \(S\) we see that \(a = a'\). Hence \(x = x'\). Thus (1) implies (2).

If there are two different points \(x, x' \in X\) mapping to the same point of \(s\) then (2) is violated. If for some \(s = f(x)\), \(x \in X\) the field extension \(\kappa(x)/\kappa(s)\) is not purely inseparable, then we may find a field extension \(K/\kappa(s)\) such that \(\kappa(x)\) has two \(\kappa(s)\)-homomorphisms into \(K\). By Schemes, Lemma 01J9 this implies that the map \(\Mor(\Spec(K), X) \to \Mor(\Spec(K), S)\) is not injective, and hence (1) is violated. Thus we see that the equivalent conditions (1) and (2) imply \(f\) is radicial, i.e., they imply (3).

Assume (3). By Schemes, Lemma 01J9 a morphism \(\Spec(K) \to X\) is given by a pair \((x, \kappa(x) \to K)\). Property (3) says exactly that associating to the pair \((x, \kappa(x) \to K)\) the pair \((s, \kappa(s) \to \kappa(x) \to K)\) is injective. In other words (1) holds. At this point we know that (1), (2) and (3) are all equivalent.

Finally, we prove the equivalence of (4) with (1), (2) and (3). A point of \(X \times_S X\) is given by a quadruple \((x_1, x_2, s, \mathfrak p)\), where \(x_1, x_2 \in X\), \(f(x_1) = f(x_2) = s\) and \(\mathfrak p \subset \kappa(x_1) \otimes_{\kappa(s)} \kappa(x_2)\) is a prime ideal, see Schemes, Lemma 01JT. If \(f\) is universally injective, then by taking \(S'=X\) in the definition of universally injective, \(\Delta_{X/S}\) must be surjective since it is a section of the injective morphism \(X \times_S X \longrightarrow X\). Conversely, if \(\Delta_{X/S}\) is surjective, then always \(x_1 = x_2 = x\) and there is exactly one such prime ideal \(\mathfrak p\), which means that \(\kappa(s) \subset \kappa(x)\) is purely inseparable. Hence \(f\) is radicial. Alternatively, if \(\Delta_{X/S}\) is surjective, then for any \(S' \to S\) the base change \(\Delta_{X_{S'}/S'}\) is surjective which implies that \(f\) is universally injective. This finishes the proof of the lemma.

Lemma

A universally injective morphism is separated.

Proof

Combine Lemma 01S4 with the remark that \(X \to S\) is separated if and only if the image of \(\Delta_{X/S}\) is closed in \(X \times_S X\), see Schemes, Definition 01KK and the discussion following it.

Lemma

A base change of a universally injective morphism is universally injective.

Proof

This is formal.

Lemma

A composition of radicial morphisms is radicial, and so the same holds for the equivalent condition of being universally injective.

Proof

Omitted.

Affine morphisms

Definition

A morphism of schemes \(f : X \to S\) is called affine if the inverse image of every affine open of \(S\) is an affine open of \(X\).

Lemma

An affine morphism is separated and quasi-compact.

Proof

Let \(f : X \to S\) be affine. Quasi-compactness is immediate from Schemes, Lemma 01K4. We will show \(f\) is separated using Schemes, Lemma 01KP. Let \(x_1, x_2 \in X\) be points of \(X\) which map to the same point \(s \in S\). Choose any affine open \(W \subset S\) containing \(s\). By assumption \(f^{-1}(W)\) is affine. Apply the lemma cited with \(U = V = f^{-1}(W)\).

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent

The morphism \(f\) is affine.

There exists an affine open covering \(S = \bigcup W_j\) such that each \(f^{-1}(W_j)\) is affine.

There exists a quasi-coherent sheaf of \(\mathcal{O}_S\)-algebras \(\mathcal{A}\) and an isomorphism \(X \cong \underline{\Spec}_S(\mathcal{A})\) of schemes over \(S\). See Constructions, Section 01LQ for notation.

Moreover, in this case \(X = \underline{\Spec}_S(f_*\mathcal{O}_X)\).

Proof

It is obvious that (1) implies (2).

Assume \(S = \bigcup_{j \in J} W_j\) is an affine open covering such that each \(f^{-1}(W_j)\) is affine. By Schemes, Lemma 01K4 we see that \(f\) is quasi-compact. By Schemes, Lemma 01KO we see the morphism \(f\) is quasi-separated. Hence by Schemes, Lemma 01LC the sheaf \(\mathcal{A} = f_*\mathcal{O}_X\) is a quasi-coherent sheaf of \(\mathcal{O}_S\)-algebras. Thus we have the scheme \(g : Y = \underline{\Spec}_S(\mathcal{A}) \to S\) over \(S\). The identity map \(\text{id} : \mathcal{A} = f_*\mathcal{O}_X \to f_*\mathcal{O}_X\) provides, via the definition of the relative spectrum, a morphism \(can : X \to Y\) over \(S\), see Constructions, Lemma 01LY. By assumption and the lemma just cited the restriction \(can|_{f^{-1}(W_j)} : f^{-1}(W_j) \to g^{-1}(W_j)\) is an isomorphism. Thus \(can\) is an isomorphism. We have shown that (2) implies (3).

Assume (3). By Constructions, Lemma 01LX we see that the inverse image of every affine open is affine, and hence the morphism is affine by definition.

Remark

We can also argue directly that (2) implies (1) in Lemma 01S8 above as follows. Assume \(S = \bigcup W_j\) is an affine open covering such that each \(f^{-1}(W_j)\) is affine. First argue that \(\mathcal{A} = f_*\mathcal{O}_X\) is quasi-coherent as in the proof above. Let \(\Spec(R) = V \subset S\) be affine open. We have to show that \(f^{-1}(V)\) is affine. Set \(A = \mathcal{A}(V) = f_*\mathcal{O}_X(V) = \mathcal{O}_X(f^{-1}(V))\). By Schemes, Lemma 01I1 there is a canonical morphism \(\psi : f^{-1}(V) \to \Spec(A)\) over \(\Spec(R) = V\). By Schemes, Lemma 01IX there exists an integer \(n \geq 0\), a standard open covering \(V = \bigcup_{i = 1, \ldots, n} D(h_i)\), \(h_i \in R\), and a map \(a : \{1, \ldots, n\} \to J\) such that each \(D(h_i)\) is also a standard open of the affine scheme \(W_{a(i)}\). The inverse image of a standard open under a morphism of affine schemes is standard open, see Algebra, Lemma 00E2. Hence we see that \(f^{-1}(D(h_i))\) is a standard open of \(f^{-1}(W_{a(i)})\), in particular that \(f^{-1}(D(h_i))\) is affine. Because \(\mathcal{A}\) is quasi-coherent we have \(A_{h_i} = \mathcal{A}(D(h_i)) = \mathcal{O}_X(f^{-1}(D(h_i)))\), so \(f^{-1}(D(h_i))\) is the spectrum of \(A_{h_i}\). It follows that the morphism \(\psi\) induces an isomorphism of the open \(f^{-1}(D(h_i))\) with the open \(\Spec(A_{h_i})\) of \(\Spec(A)\). Since \(f^{-1}(V) = \bigcup f^{-1}(D(h_i))\) and \(\Spec(A) = \bigcup \Spec(A_{h_i})\) we win.

Lemma

Let \(S\) be a scheme. There is an anti-equivalence of categories \[\begin{matrix} \text{Schemes affine} \\ \text{over }S \end{matrix} \longleftrightarrow \begin{matrix} \text{quasi-coherent sheaves} \\ \text{of }\mathcal{O}_S\text{-algebras} \end{matrix}\] which associates to \(f : X \to S\) the sheaf \(f_*\mathcal{O}_X\). Moreover, this equivalence is compatible with arbitrary base change.

Proof

The functor from right to left is given by \(\underline{\Spec}_S\). The two functors are mutually inverse by Lemma 01S8 and Constructions, Lemma 01LX part (3). The final statement is Constructions, Lemma 01LX part (2).

Lemma

Let \(S\) be a scheme and let \(\mathcal{A}\) be a quasi-coherent \(\mathcal{O}_S\)-algebra. An \(\mathcal{A}\)-module is quasi-coherent as an \(\mathcal{O}_S\)-module if and only if it is quasi-coherent as an \(\mathcal{A}\)-module.

Proof

Let \(\mathcal{F}\) be an \(\mathcal{A}\)-module. If \(\mathcal{F}\) is quasi-coherent as an \(\mathcal{A}\)-module, then for every \(s \in S\) there exists an open neighbourhood \(U\) of \(s\) and an exact sequence \[\bigoplus\nolimits_J\mathcal{A}|_U\to \bigoplus\nolimits_I\mathcal{A}|_U\to \mathcal{F}|_U\] of \(\mathcal{A}|_U\)-modules. Then this is also an exact sequence of \(\mathcal{O}_U\)-modules. Hence \(\mathcal{F}|_U\) is quasi-coherent as the cokernel of a morphism of quasi-coherent \(\mathcal{O}_U\)-modules on a scheme. It follows that \(\mathcal{F}\) is quasi-coherent as an \(\mathcal{O}_X\)-module.

Conversely, assume \(\mathcal{F}\) is quasi-coherent as an \(\mathcal{O}_X\)-module. Pick an open affine \(\Spec(R) = U \subset S\). We have isomorphisms of \(\mathcal{O}_U\)-modules \(\mathcal{A}|_U \cong \widetilde{A}\) and \(\mathcal{F}|_U \cong \widetilde{M}\), for some \(R\)-algebra \(A\) and some \(R\)-module \(M\). The \(\mathcal{A}\)-module structure on \(\mathcal{F}\) translates into an \(A\)-module structure on \(M\) compatible with the given \(R\)-module structure (details omitted). Choose an exact sequence \[\bigoplus\nolimits_J A \to \bigoplus\nolimits_I A \to M \to 0\] of \(A\)-modules. Since the functor \(\widetilde{\ }\) is exact, this produces an exact sequence \[\bigoplus\nolimits_J \widetilde{A}\to \bigoplus\nolimits_I \widetilde{A}\to \widetilde{M} \to 0\] of \(\widetilde{A}\)-modules. This means that \(\mathcal{F}\) is quasi-coherent as an \(\mathcal{A}\)-module.

Lemma

Let \(f : X \to S\) be an affine morphism of schemes. Let \(\mathcal{A} = f_*\mathcal{O}_X\). The functor \(\mathcal{F} \mapsto f_*\mathcal{F}\) induces an equivalence of categories \[\left\{ \begin{matrix} \text{category of quasi-coherent}\\ \mathcal{O}_X\text{-modules} \end{matrix} \right\} \longrightarrow \left\{ \begin{matrix} \text{category of quasi-coherent}\\ \mathcal{A}\text{-modules} \end{matrix} \right\}\] Moreover, an \(\mathcal{A}\)-module is quasi-coherent as an \(\mathcal{O}_S\)-module if and only if it is quasi-coherent as an \(\mathcal{A}\)-module.

Proof

The final statement is Lemma 0H88. By Lemma 01S7 and Schemes, Lemma 01LC the pushforward \(f_*\mathcal{F}\) of a quasi-coherent \(\mathcal{O}_X\)-module \(\mathcal{F}\) is a quasi-coherent \(\mathcal{O}_S\)-module. Hence a functor as in the statement of the lemma.

We will construct an quasi-inverse \(h\) to this functor. Let \(\mathcal{G}\) be a quasi-coherent \(\mathcal{A}\)-module. Then we set \[h(\mathcal{G}) = f^*\mathcal{G} \otimes_{f^*\mathcal{A}} \mathcal{O}_X = f^*\mathcal{G} \otimes_{f^*f_*\mathcal{O}_X} \mathcal{O}_X\] Elucidation: the pullback \(f^*\mathcal{A} = f^*f_*\mathcal{O}_X\) is an \(\mathcal{O}_X\)-algebra, the adjunction map \(f^*f_*\mathcal{O}_X \to \mathcal{O}_X\) is an algebra homomorphism, and the pullback \(f^*\mathcal{G}\) is an \(f^*\mathcal{A}\)-module. Observe that \(h(\mathcal{G})\) is quasi-coherent as quasi-coherence is preserved by pullbacks and change of rings. Observe that there is a functorial map \[h(f_*\mathcal{F}) = f^*f_*\mathcal{F} \otimes_{f^*f_*\mathcal{O}_X} \mathcal{O}_X \to \mathcal{F}\] coming from the adjunction map \(f^*f_*\mathcal{F} \to \mathcal{F}\) and a functorial map \[\mathcal{G} \to f_*h(\mathcal{G}) \quad\text{adjoint to the map}\quad f^*\mathcal{G} \to f^*\mathcal{G} \otimes_{f^*f_*\mathcal{O}_X} \mathcal{O}_X\] which sends a local section \(s\) of \(f^*\mathcal{F}\) to \(s \otimes 1\). To finish the proof it suffices to show that these maps are isomorphisms for \(\mathcal{F}\) and \(\mathcal{G}\) as above. This may be checked on the members of an affine covering, i.e., when \(X\) and \(S\) are affine.

The key algebra observation which makes this work is the following: Let \(R \to A\) be a ring map. Let \(N\) be an \(A\)-module. Then \[(N \otimes_R A) \otimes_{(A \otimes_R A)} A = N\] Namely, the left hand side of this equality is the effect of applying \(h\) to the quasi-coherent \(\widetilde{A}\)-module \(\widetilde{N}\) on \(\Spec(R)\). We omit the details.

Lemma

The composition of affine morphisms is affine.

Proof

Let \(f : X \to Y\) and \(g : Y \to Z\) be affine morphisms. Let \(U \subset Z\) be affine open. Then \(g^{-1}(U)\) is affine by assumption on \(g\). Whereupon \(f^{-1}(g^{-1}(U))\) is affine by assumption on \(f\). Hence \((g \circ f)^{-1}(U)\) is affine.

Lemma

The base change of an affine morphism is affine.

Proof

Let \(f : X \to S\) be an affine morphism. Let \(S' \to S\) be any morphism. Denote \(f' : X_{S'} = S' \times_S X \to S'\) the base change of \(f\). For every \(s' \in S'\) there exists an open affine neighbourhood \(s' \in V \subset S'\) which maps into some open affine \(U \subset S\). By assumption \(f^{-1}(U)\) is affine. By the material in Schemes, Section 01JO we see that \(f^{-1}(U)_V = V \times_U f^{-1}(U)\) is affine and equal to \((f')^{-1}(V)\). This proves that \(S'\) has an open covering by affines whose inverse image under \(f'\) is affine. We conclude by Lemma 01S8 above.

Lemma

A closed immersion is affine.

Proof

The first indication of this is Schemes, Lemma 01IH. See Schemes, Lemma 01IN for a complete statement.

Lemma

Let \(X\) be a scheme. Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. Let \(s \in \Gamma(X, \mathcal{L})\). The inclusion morphism \(j : X_s \to X\) is affine.

Proof

This follows from Properties, Lemma 01PV and the definition.

Lemma

Suppose \(g : X \to Y\) is a morphism of schemes over \(S\).

If \(X\) is affine over \(S\) and \(\Delta : Y \to Y \times_S Y\) is affine, then \(g\) is affine.

If \(X\) is affine over \(S\) and \(Y\) is separated over \(S\), then \(g\) is affine.

A morphism from an affine scheme to a scheme with affine diagonal is affine.

A morphism from an affine scheme to a separated scheme is affine.

Proof

Proof of (1). The base change \(X \times_S Y \to Y\) is affine by Lemma 01SD. The morphism \((1, g) : X \to X \times_S Y\) is the base change of \(Y \to Y \times_S Y\) by the morphism \(X \times_S Y \to Y \times_S Y\). Hence it is affine by Lemma 01SD. The composition of affine morphisms is affine (see Lemma 01SC) and (1) follows. Part (2) follows from (1) as a closed immersion is affine (see Lemma 01SE) and \(Y/S\) separated means \(\Delta\) is a closed immersion. Parts (3) and (4) are special cases of (1) and (2).

Lemma

A morphism between affine schemes is affine.

Proof

Immediate from Lemma 01SG with \(S = \Spec(\mathbf{Z})\). It also follows directly from the equivalence of (1) and (2) in Lemma 01S8.

Lemma

Let \(S\) be a scheme. Let \(A\) be an Artinian ring. Any morphism \(\Spec(A) \to S\) is affine.

Proof

Omitted.

Lemma

Let \(j : Y \to X\) be an immersion of schemes. Assume there exists an open \(U \subset X\) with complement \(Z = X \setminus U\) such that

\(U \to X\) is affine,

\(j^{-1}(U) \to U\) is affine, and

\(j(Y) \cap Z\) is closed.

Then \(j\) is affine. In particular, if \(X\) is affine, so is \(Y\).

Proof

By Schemes, Definition 01IO there exists an open subscheme \(W \subset X\) such that \(j\) factors as a closed immersion \(i : Y \to W\) followed by the inclusion morphism \(W \to X\). Since a closed immersion is affine (Lemma 01SE), we see that for every \(x \in W\) there is an affine open neighbourhood of \(x\) in \(X\) whose inverse image under \(j\) is affine. If \(x \in U\), then the same thing is true by assumption (2). Finally, assume \(x \in Z\) and \(x \not \in W\). Then \(x \not \in j(Y) \cap Z\). By assumption (3) we can find an affine open neighbourhood \(V \subset X\) of \(x\) which does not meet \(j(Y) \cap Z\). Then \(j^{-1}(V) = j^{-1}(V \cap U)\) which is affine by assumptions (1) and (2). It follows that \(j\) is affine by Lemma 01S8.

Families of ample invertible modules

A short section on the notion of a family of ample invertible modules.

Definition

Let \(X\) be a scheme. Let \(\{\mathcal{L}_i\}_{i \in I}\) be a family of invertible \(\mathcal{O}_X\)-modules. We say \(\{\mathcal{L}_i\}_{i \in I}\) is an ample family of invertible modules on \(X\) if

\(X\) is quasi-compact, and

for every \(x \in X\) there exists an \(i \in I\), an \(n \geq 1\), and \(s \in \Gamma(X, \mathcal{L}_i^{\otimes n})\) such that \(x \in X_s\) and \(X_s\) is affine.

If \(\{\mathcal{L}_i\}_{i \in I}\) is an ample family of invertible modules on a scheme \(X\), then there exists a finite subset \(I' \subset I\) such that \(\{\mathcal{L}_i\}_{i \in I'}\) is an ample family of invertible modules on \(X\) (follows immediately from quasi-compactness). A scheme having an ample family of invertible modules has an affine diagonal by the next lemma and hence is a fortiori quasi-separated.

Lemma

Let \(X\) be a scheme such that for every point \(x \in X\) there exists an invertible \(\mathcal{O}_X\)-module \(\mathcal{L}\) and a global section \(s \in \Gamma(X, \mathcal{L})\) such that \(x \in X_s\) and \(X_s\) is affine. Then the diagonal of \(X\) is an affine morphism.

Proof

Given invertible \(\mathcal{O}_X\)-modules \(\mathcal{L}\), \(\mathcal{M}\) and global sections \(s \in \Gamma(X, \mathcal{L})\), \(t \in \Gamma(X, \mathcal{M})\) such that \(X_s\) and \(X_t\) are affine we have to prove \(X_s \cap X_t\) is affine. Namely, then Lemma 01S8 applied to \(\Delta : X \to X \times X\) and the fact that \(\Delta^{-1}(X_s \times X_t) = X_s \cap X_t\) shows that \(\Delta\) is affine. The fact that \(X_s \cap X_t\) is affine follows from Properties, Lemma 01PV.

Remark

In Properties, Lemma 01PY we see that a scheme which has an ample invertible module is separated. This is wrong for schemes having an ample family of invertible modules. Namely, let \(X\) be as in Schemes, Example 01JD with \(n = 1\), i.e., the affine line with zero doubled. We use the notation of that example except that we write \(x\) for \(x_1\) and \(y\) for \(y_1\). There is, for every integer \(n\), an invertible sheaf \(\mathcal{L}_n\) on \(X\) which is trivial on \(X_1\) and \(X_2\) and whose transition function \(U_{12} \to U_{21}\) is \(f(x) \mapsto y^n f(y)\). The global sections of \(\mathcal{L}_n\) are pairs \((f(x), g(y)) \in k[x] \oplus k[y]\) such that \(y^n f(y) = g(y)\). The sections \(s = (1, y)\) of \(\mathcal{L}_1\) and \(t = (x, 1)\) of \(\mathcal{L}_{-1}\) determine an open affine cover because \(X_s = X_1\) and \(X_t = X_2\). Therefore \(X\) has an ample family of invertible modules but it is not separated.

Quasi-affine morphisms

Recall that a scheme \(X\) is called quasi-affine if it is quasi-compact and isomorphic to an open subscheme of an affine scheme, see Properties, Definition 01P6.

Definition

A morphism of schemes \(f : X \to S\) is called quasi-affine if the inverse image of every affine open of \(S\) is a quasi-affine scheme.

Lemma

A quasi-affine morphism is separated and quasi-compact.

Proof

Let \(f : X \to S\) be quasi-affine. Quasi-compactness is immediate from Schemes, Lemma 01K4. Let \(U \subset S\) be an affine open. If we can show that \(f^{-1}(U)\) is a separated scheme, then \(f\) is separated (Schemes, Lemma 01KP shows that being separated is local on the base). By assumption \(f^{-1}(U)\) is isomorphic to an open subscheme of an affine scheme. An affine scheme is separated and hence every open subscheme of an affine scheme is separated as desired.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent

The morphism \(f\) is quasi-affine.

There exists an affine open covering \(S = \bigcup W_j\) such that each \(f^{-1}(W_j)\) is quasi-affine.

There exists a quasi-coherent sheaf of \(\mathcal{O}_S\)-algebras \(\mathcal{A}\) and a quasi-compact open immersion \[\xymatrix{ X \ar[rr] \ar[rd] & & \underline{\Spec}_S(\mathcal{A}) \ar[dl] \\ & S & }\] over \(S\).

Same as in (3) but with \(\mathcal{A} = f_*\mathcal{O}_X\) and the horizontal arrow the canonical morphism of Constructions, Lemma 01LY.

Proof

It is obvious that (1) implies (2) and that (4) implies (3).

Assume \(S = \bigcup_{j \in J} W_j\) is an affine open covering such that each \(f^{-1}(W_j)\) is quasi-affine. By Schemes, Lemma 01K4 we see that \(f\) is quasi-compact. By Schemes, Lemma 01KO we see the morphism \(f\) is quasi-separated. Hence by Schemes, Lemma 01LC the sheaf \(\mathcal{A} = f_*\mathcal{O}_X\) is a quasi-coherent sheaf of \(\mathcal{O}_X\)-algebras. Thus we have the scheme \(g : Y = \underline{\Spec}_S(\mathcal{A}) \to S\) over \(S\). The identity map \(\text{id} : \mathcal{A} = f_*\mathcal{O}_X \to f_*\mathcal{O}_X\) provides, via the definition of the relative spectrum, a morphism \(can : X \to Y\) over \(S\), see Constructions, Lemma 01LY. By assumption, the lemma just cited, and Properties, Lemma 01P9 the restriction \(can|_{f^{-1}(W_j)} : f^{-1}(W_j) \to g^{-1}(W_j)\) is a quasi-compact open immersion. Thus \(can\) is a quasi-compact open immersion. We have shown that (2) implies (4).

Assume (3). Choose any affine open \(U \subset S\). By Constructions, Lemma 01LX we see that the inverse image of \(U\) in the relative spectrum is affine. Hence we conclude that \(f^{-1}(U)\) is quasi-affine (note that quasi-compactness is encoded in (3) as well). Thus (3) implies (1).

Lemma

The composition of quasi-affine morphisms is quasi-affine.

Proof

Let \(f : X \to Y\) and \(g : Y \to Z\) be quasi-affine morphisms. Let \(U \subset Z\) be affine open. Then \(g^{-1}(U)\) is quasi-affine by assumption on \(g\). Let \(j : g^{-1}(U) \to V\) be a quasi-compact open immersion into an affine scheme \(V\). By Lemma 01SM above we see that \(f^{-1}(g^{-1}(U))\) is a quasi-compact open subscheme of the relative spectrum \(\underline{\Spec}_{g^{-1}(U)}(\mathcal{A})\) for some quasi-coherent sheaf of \(\mathcal{O}_{g^{-1}(U)}\)-algebras \(\mathcal{A}\). By Schemes, Lemma 01LC the sheaf \(\mathcal{A}' = j_*\mathcal{A}\) is a quasi-coherent sheaf of \(\mathcal{O}_V\)-algebras with the property that \(j^*\mathcal{A}' = \mathcal{A}\). Hence we get a commutative diagram \[\xymatrix{ f^{-1}(g^{-1}(U)) \ar[r] & \underline{\Spec}_{g^{-1}(U)}(\mathcal{A}) \ar[r] \ar[d] & \underline{\Spec}_V(\mathcal{A}') \ar[d] \\ & g^{-1}(U) \ar[r]^j & V }\] with the square being a fibre square, see Constructions, Lemma 01LX. Note that the upper right corner is an affine scheme. Hence \((g \circ f)^{-1}(U)\) is quasi-affine.

Lemma

The base change of a quasi-affine morphism is quasi-affine.

Proof

Let \(f : X \to S\) be a quasi-affine morphism. By Lemma 01SM above we can find a quasi-coherent sheaf of \(\mathcal{O}_S\)-algebras \(\mathcal{A}\) and a quasi-compact open immersion \(X \to \underline{\Spec}_S(\mathcal{A})\) over \(S\). Let \(g : S' \to S\) be any morphism. Denote \(f' : X_{S'} = S' \times_S X \to S'\) the base change of \(f\). Since the base change of a quasi-compact open immersion is a quasi-compact open immersion we see that \(X_{S'} \to \underline{\Spec}_{S'}(g^*\mathcal{A})\) is a quasi-compact open immersion (we have used Schemes, Lemmas 01K5 and 01JY and Constructions, Lemma 01LX). By Lemma 01SM again we conclude that \(X_{S'} \to S'\) is quasi-affine.

Lemma

A quasi-compact immersion is quasi-affine.

Proof

Let \(X \to S\) be a quasi-compact immersion. We have to show the inverse image of every affine open is quasi-affine. Hence, assuming \(S\) is an affine scheme, we have to show \(X\) is quasi-affine. By Lemma 01RG the morphism \(X \to S\) factors as \(X \to Z \to S\) where \(Z\) is a closed subscheme of \(S\) and \(X \subset Z\) is a quasi-compact open. Since \(S\) is affine Lemma 01QO implies \(Z\) is affine. Hence we win.

Lemma

Let \(S\) be a scheme. Let \(X\) be an affine scheme. A morphism \(f : X \to S\) is quasi-affine if and only if it is quasi-compact. In particular any morphism from an affine scheme to a quasi-separated scheme is quasi-affine.

Proof

Let \(V \subset S\) be an affine open. Then \(f^{-1}(V)\) is an open subscheme of the affine scheme \(X\), hence quasi-affine if and only if it is quasi-compact. This proves the first assertion. The quasi-compactness of any \(f : X \to S\) where \(X\) is affine and \(S\) quasi-separated follows from Schemes, Lemma 03GI applied to \(X \to S \to \Spec(\mathbf{Z})\).

Lemma

Suppose \(g : X \to Y\) is a morphism of schemes over \(S\). If \(X\) is quasi-affine over \(S\) and \(Y\) is quasi-separated over \(S\), then \(g\) is quasi-affine. In particular, any morphism from a quasi-affine scheme to a quasi-separated scheme is quasi-affine.

Proof

The base change \(X \times_S Y \to Y\) is quasi-affine by Lemma 01SO. The morphism \(X \to X \times_S Y\) is a quasi-compact immersion as \(Y \to S\) is quasi-separated, see Schemes, Lemma 01KT. A quasi-compact immersion is quasi-affine by Lemma 02JR and the composition of quasi-affine morphisms is quasi-affine (see Lemma 01SN). Thus we win.

Types of morphisms defined by properties of ring maps

In this section we study what properties of ring maps allow one to define local properties of morphisms of schemes.

Definition

Let \(P\) be a property of ring maps.

We say that \(P\) is local if the following hold:

For any ring map \(R \to A\), and any \(f \in R\) we have \(P(R \to A) \Rightarrow P(R_f \to A_f)\).

For any rings \(R\), \(A\), any \(f \in R\), \(a\in A\), and any ring map \(R_f \to A\) we have \(P(R_f \to A) \Rightarrow P(R \to A_a)\).

For any ring map \(R \to A\), and \(a_i \in A\) such that \((a_1, \ldots, a_n) = A\) then \(\forall i, P(R \to A_{a_i}) \Rightarrow P(R \to A)\).

We say that \(P\) is stable under base change if for any ring maps \(R \to A\), \(R \to R'\) we have \(P(R \to A) \Rightarrow P(R' \to R' \otimes_R A)\).

We say that \(P\) is stable under composition if for any ring maps \(A \to B\), \(B \to C\) we have \(P(A \to B) \wedge P(B \to C) \Rightarrow P(A \to C)\).

Definition

Let \(P\) be a property of ring maps. Let \(f : X \to S\) be a morphism of schemes. We say \(f\) is locally of type \(P\) if for any \(x \in X\) there exists an affine open neighbourhood \(U\) of \(x\) in \(X\) which maps into an affine open \(V \subset S\) such that the induced ring map \(\mathcal{O}_S(V) \to \mathcal{O}_X(U)\) has property \(P\).

This is not a “good” definition unless the property \(P\) is a local property. Even if \(P\) is a local property we will not automatically use this definition to say that a morphism is “locally of type \(P\)” unless we also explicitly state the definition elsewhere.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(P\) be a property of ring maps. Let \(U\) be an affine open of \(X\), and \(V\) an affine open of \(S\) such that \(f(U) \subset V\). If \(f\) is locally of type \(P\) and \(P\) is local, then \(P(\mathcal{O}_S(V) \to \mathcal{O}_X(U))\) holds.

Proof

As \(f\) is locally of type \(P\) for every \(u \in U\) there exists an affine open \(U_u \subset X\) mapping into an affine open \(V_u \subset S\) such that \(P(\mathcal{O}_S(V_u) \to \mathcal{O}_X(U_u))\) holds. Choose an open neighbourhood \(U'_u \subset U \cap U_u\) of \(u\) which is standard affine open in both \(U\) and \(U_u\), see Schemes, Lemma 01IW. By Definition 01SR (1)(b) we see that \(P(\mathcal{O}_S(V_u) \to \mathcal{O}_X(U'_u))\) holds. Hence we may assume that \(U_u \subset U\) is a standard affine open. Choose an open neighbourhood \(V'_u \subset V \cap V_u\) of \(f(u)\) which is standard affine open in both \(V\) and \(V_u\), see Schemes, Lemma 01IW. Then \(U'_u = f^{-1}(V'_u) \cap U_u\) is a standard affine open of \(U_u\) (hence of \(U\)) and we have \(P(\mathcal{O}_S(V'_u) \to \mathcal{O}_X(U'_u))\) by Definition 01SR (1)(a). Hence we may assume both \(U_u \subset U\) and \(V_u \subset V\) are standard affine open. Applying Definition 01SR (1)(b) one more time we conclude that \(P(\mathcal{O}_S(V) \to \mathcal{O}_X(U_u))\) holds. Because \(U\) is quasi-compact we may choose a finite number of points \(u_1, \ldots, u_n \in U\) such that \[U = U_{u_1} \cup \ldots \cup U_{u_n}.\] By Definition 01SR (1)(c) we conclude that \(P(\mathcal{O}_S(V) \to \mathcal{O}_X(U))\) holds.

Lemma

Let \(P\) be a local property of ring maps. Let \(f : X \to S\) be a morphism of schemes. The following are equivalent

The morphism \(f\) is locally of type \(P\).

For every affine opens \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) we have \(P(\mathcal{O}_S(V) \to \mathcal{O}_X(U))\).

There exists an open covering \(S = \bigcup_{j \in J} V_j\) and open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that each of the morphisms \(U_i \to V_j\), \(j\in J, i\in I_j\) is locally of type \(P\).

There exists an affine open covering \(S = \bigcup_{j \in J} V_j\) and affine open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that \(P(\mathcal{O}_S(V_j) \to \mathcal{O}_X(U_i))\) holds, for all \(j\in J, i\in I_j\).

Moreover, if \(f\) is locally of type \(P\) then for any open subschemes \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the restriction \(f|_U : U \to V\) is locally of type \(P\).

Proof

This follows from Lemma 01ST above.

Lemma

Let \(P\) be a property of ring maps. Assume \(P\) is local and stable under composition. The composition of morphisms locally of type \(P\) is locally of type \(P\).

Proof

Let \(f : X \to Y\) and \(g : Y \to Z\) be morphisms locally of type \(P\). Let \(x \in X\). Choose an affine open neighbourhood \(W \subset Z\) of \(g(f(x))\). Choose an affine open neighbourhood \(V \subset g^{-1}(W)\) of \(f(x)\). Choose an affine open neighbourhood \(U \subset f^{-1}(V)\) of \(x\). By Lemma 01SU the ring maps \(\mathcal{O}_Z(W) \to \mathcal{O}_Y(V)\) and \(\mathcal{O}_Y(V) \to \mathcal{O}_X(U)\) satisfy \(P\). Hence \(\mathcal{O}_Z(W) \to \mathcal{O}_X(U)\) satisfies \(P\) as \(P\) is assumed stable under composition.

Lemma

Let \(P\) be a property of ring maps. Assume \(P\) is local and stable under base change. The base change of a morphism locally of type \(P\) is locally of type \(P\).

Proof

Let \(f : X \to S\) be a morphism locally of type \(P\). Let \(S' \to S\) be any morphism. Denote \(f' : X_{S'} = S' \times_S X \to S'\) the base change of \(f\). For every \(s' \in S'\) there exists an open affine neighbourhood \(s' \in V' \subset S'\) which maps into some open affine \(V \subset S\). By Lemma 01SU the open \(f^{-1}(V)\) is a union of affines \(U_i\) such that the ring maps \(\mathcal{O}_S(V) \to \mathcal{O}_X(U_i)\) all satisfy \(P\). By the material in Schemes, Section 01JO we see that \(f^{-1}(U)_{V'} = V' \times_V f^{-1}(V)\) is the union of the affine opens \(V' \times_V U_i\). Since \(\mathcal{O}_{X_{S'}}(V' \times_V U_i) = \mathcal{O}_{S'}(V') \otimes_{\mathcal{O}_S(V)} \mathcal{O}_X(U_i)\) we see that the ring maps \(\mathcal{O}_{S'}(V') \to \mathcal{O}_{X_{S'}}(V' \times_V U_i)\) satisfy \(P\) as \(P\) is assumed stable under base change.

Lemma

The following properties of a ring map \(R \to A\) are local.

(Isomorphism on local rings.) For every prime \(\mathfrak q\) of \(A\) lying over \(\mathfrak p \subset R\) the ring map \(R \to A\) induces an isomorphism \(R_{\mathfrak p} \to A_{\mathfrak q}\).

(Open immersion.) For every prime \(\mathfrak q\) of \(A\) there exists an \(f \in R\), \(\varphi(f) \not \in \mathfrak q\) such that the ring map \(\varphi : R \to A\) induces an isomorphism \(R_f \to A_f\).

(Reduced fibres.) For every prime \(\mathfrak p\) of \(R\) the fibre ring \(A \otimes_R \kappa(\mathfrak p)\) is reduced.

(Fibres of dimension at most \(n\).) For every prime \(\mathfrak p\) of \(R\) the fibre ring \(A \otimes_R \kappa(\mathfrak p)\) has Krull dimension at most \(n\).

(Locally Noetherian on the target.) The ring map \(R \to A\) has the property that \(A\) is Noetherian.

Add more here as needed⁴.

Proof

Omitted.

Lemma

The following properties of ring maps are stable under base change.

(Isomorphism on local rings.) For every prime \(\mathfrak q\) of \(A\) lying over \(\mathfrak p \subset R\) the ring map \(R \to A\) induces an isomorphism \(R_{\mathfrak p} \to A_{\mathfrak q}\).

(Open immersion.) For every prime \(\mathfrak q\) of \(A\) there exists an \(f \in R\), \(\varphi(f) \not \in \mathfrak q\) such that the ring map \(\varphi : R \to A\) induces an isomorphism \(R_f \to A_f\).

Add more here as needed⁵.

Proof

Omitted.

Lemma

The following properties of ring maps are stable under composition.

(Isomorphism on local rings.) For every prime \(\mathfrak q\) of \(A\) lying over \(\mathfrak p \subset R\) the ring map \(R \to A\) induces an isomorphism \(R_{\mathfrak p} \to A_{\mathfrak q}\).

(Open immersion.) For every prime \(\mathfrak q\) of \(A\) there exists an \(f \in R\), \(\varphi(f) \not \in \mathfrak q\) such that the ring map \(\varphi : R \to A\) induces an isomorphism \(R_f \to A_f\).

(Locally Noetherian on the target.) The ring map \(R \to A\) has the property that \(A\) is Noetherian.

Add more here as needed⁶.

Proof

Omitted.

Morphisms of finite type

Recall that a ring map \(R \to A\) is said to be of finite type if \(A\) is isomorphic to a quotient of \(R[x_1, \ldots, x_n]\) as an \(R\)-algebra, see Algebra, Definition 00F3.

Definition

Let \(f : X \to S\) be a morphism of schemes.

We say that \(f\) is of finite type at \(x \in X\) if there exists an affine open neighbourhood \(\Spec(A) = U \subset X\) of \(x\) and an affine open \(\Spec(R) = V \subset S\) with \(f(U) \subset V\) such that the induced ring map \(R \to A\) is of finite type.

We say that \(f\) is locally of finite type if it is of finite type at every point of \(X\).

We say that \(f\) is of finite type if it is locally of finite type and quasi-compact.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent

The morphism \(f\) is locally of finite type.

For all affine opens \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the ring map \(\mathcal{O}_S(V) \to \mathcal{O}_X(U)\) is of finite type.

There exist an open covering \(S = \bigcup_{j \in J} V_j\) and open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that each of the morphisms \(U_i \to V_j\), \(j\in J, i\in I_j\) is locally of finite type.

There exist an affine open covering \(S = \bigcup_{j \in J} V_j\) and affine open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that the ring map \(\mathcal{O}_S(V_j) \to \mathcal{O}_X(U_i)\) is of finite type, for all \(j\in J, i\in I_j\).

Moreover, if \(f\) is locally of finite type then for any open subschemes \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the restriction \(f|_U : U \to V\) is locally of finite type.

Proof

This follows from Lemma 01ST if we show that the property “\(R \to A\) is of finite type” is local. We check conditions (a), (b) and (c) of Definition 01SR. By Algebra, Lemma 05G5 being of finite type is stable under base change and hence we conclude (a) holds. By Algebra, Lemma 00F4 being of finite type is stable under composition and trivially for any ring \(R\) the ring map \(R \to R_f\) is of finite type. We conclude (b) holds. Finally, property (c) is true according to Algebra, Lemma 00EP.

Lemma

The composition of two morphisms which are locally of finite type is locally of finite type. The same is true for morphisms of finite type.

Proof

In the proof of Lemma 01T2 we saw that being of finite type is a local property of ring maps. Hence the first statement of the lemma follows from Lemma 01SV combined with the fact that being of finite type is a property of ring maps that is stable under composition, see Algebra, Lemma 00F4. By the above and the fact that compositions of quasi-compact morphisms are quasi-compact, see Schemes, Lemma 01K6 we see that the composition of morphisms of finite type is of finite type.

Lemma

The base change of a morphism which is locally of finite type is locally of finite type. The same is true for morphisms of finite type.

Proof

In the proof of Lemma 01T2 we saw that being of finite type is a local property of ring maps. Hence the first statement of the lemma follows from Lemma 01SW combined with the fact that being of finite type is a property of ring maps that is stable under base change, see Algebra, Lemma 05G5. By the above and the fact that a base change of a quasi-compact morphism is quasi-compact, see Schemes, Lemma 01K5 we see that the base change of a morphism of finite type is a morphism of finite type.

Lemma

A closed immersion is of finite type. An immersion is locally of finite type.

Proof

This is true because an open immersion is a local isomorphism, and a closed immersion is obviously of finite type.

Lemma

Let \(f : X \to S\) be a morphism. If \(S\) is (locally) Noetherian and \(f\) (locally) of finite type then \(X\) is (locally) Noetherian.

Proof

This follows immediately from the fact that a ring of finite type over a Noetherian ring is Noetherian, see Algebra, Lemma 00FN. (Also: use the fact that the source of a quasi-compact morphism with quasi-compact target is quasi-compact.)

Lemma

Let \(f : X \to S\) be locally of finite type with \(S\) locally Noetherian. Then \(f\) is quasi-separated.

Proof

In fact, it is true that \(X\) is quasi-separated, see Properties, Lemma 01OY and Lemma 01T6 above. Then apply Schemes, Lemma 01KV to conclude that \(f\) is quasi-separated.

Lemma

Let \(X \to Y\) be a morphism of schemes over a base scheme \(S\). If \(X\) is locally of finite type over \(S\), then \(X \to Y\) is locally of finite type.

Proof

Via Lemma 01T2 this translates into the following algebra fact: Given ring maps \(A \to B \to C\) such that \(A \to C\) is of finite type, then \(B \to C\) is of finite type. (See Algebra, Lemma 00F4).

Points of finite type and Jacobson schemes

Let \(S\) be a scheme. A finite type point \(s\) of \(S\) is a point such that the morphism \(\Spec(\kappa(s)) \to S\) is of finite type. The reason for studying this is that finite type points can replace closed points in a certain sense and in certain situations. There are always enough of them for example. Moreover, a scheme is Jacobson if and only if all finite type points are closed points.

Lemma

Let \(S\) be a scheme. Let \(k\) be a field. Let \(f : \Spec(k) \to S\) be a morphism. The following are equivalent:

The morphism \(f\) is of finite type.

The morphism \(f\) is locally of finite type.

There exists an affine open \(U = \Spec(R)\) of \(S\) such that \(f\) corresponds to a finite ring map \(R \to k\).

There exists an affine open \(U = \Spec(R)\) of \(S\) such that the image of \(f\) consists of a closed point \(u\) in \(U\) and the field extension \(k/\kappa(u)\) is finite.

Proof

The equivalence of (1) and (2) is obvious as \(\Spec(k)\) is a singleton and hence any morphism from it is quasi-compact.

Suppose \(f\) is locally of finite type. Choose any affine open \(\Spec(R) = U \subset S\) such that the image of \(f\) is contained in \(U\), and the ring map \(R \to k\) is of finite type. Let \(\mathfrak p \subset R\) be the kernel. Then \(R/\mathfrak p \subset k\) is of finite type. By Algebra, Lemma 00FY there exist a \(\overline{f} \in R/\mathfrak p\) such that \((R/\mathfrak p)_{\overline{f}}\) is a field and \((R/\mathfrak p)_{\overline{f}} \to k\) is a finite field extension. If \(f \in R\) is a lift of \(\overline{f}\), then we see that \(k\) is a finite \(R_f\)-module. Thus (2) \(\Rightarrow\) (3).

Suppose that \(\Spec(R) = U \subset S\) is an affine open such that \(f\) corresponds to a finite ring map \(R \to k\). Then \(f\) is locally of finite type by Lemma 01T2. Thus (3) \(\Rightarrow\) (2).

Suppose \(R \to k\) is finite. The image of \(R \to k\) is a field over which \(k\) is finite by Algebra, Lemma 00GR. Hence the kernel of \(R \to k\) is a maximal ideal. Thus (3) \(\Rightarrow\) (4).

The implication (4) \(\Rightarrow\) (3) is immediate.

Lemma

Let \(S\) be a scheme. Let \(A\) be an Artinian local ring with residue field \(\kappa\). Let \(f : \Spec(A) \to S\) be a morphism of schemes. Then \(f\) is of finite type if and only if the composition \(\Spec(\kappa) \to \Spec(A) \to S\) is of finite type.

Proof

Since the morphism \(\Spec(\kappa) \to \Spec(A)\) is of finite type it is clear that if \(f\) is of finite type so is the composition \(\Spec(\kappa) \to S\) (see Lemma 01T3). For the converse, note that \(\Spec(A) \to S\) maps into some affine open \(U = \Spec(B)\) of \(S\) as \(\Spec(A)\) has only one point. To finish apply Algebra, Lemma 07DT to \(B \to A\).

Recall that given a point \(s\) of a scheme \(S\) there is a canonical morphism \(\Spec(\kappa(s)) \to S\), see Schemes, Section 01J5.

Definition

Let \(S\) be a scheme. Let us say that a point \(s\) of \(S\) is a finite type point if the canonical morphism \(\Spec(\kappa(s)) \to S\) is of finite type. We denote \(S_{\text{ft-pts}}\) the set of finite type points of \(S\).

We can describe the set of finite type points as follows.

Lemma

Let \(S\) be a scheme. We have \[S_{\text{ft-pts}} = \bigcup\nolimits_{U \subset S\text{ open}} U_0\] where \(U_0\) is the set of closed points of \(U\). Here we may let \(U\) range over all opens or over all affine opens of \(S\).

Proof

Immediate from Lemma 01TA.

Lemma

Let \(f : T \to S\) be a morphism of schemes. If \(f\) is locally of finite type, then \(f(T_{\text{ft-pts}}) \subset S_{\text{ft-pts}}\).

Proof

If \(T\) is the spectrum of a field this is Lemma 01TA. In general it follows since the composition of morphisms locally of finite type is locally of finite type (Lemma 01T3).

Lemma

Let \(f : T \to S\) be a morphism of schemes. If \(f\) is locally of finite type and surjective, then \(f(T_{\text{ft-pts}}) = S_{\text{ft-pts}}\).

Proof

We have \(f(T_{\text{ft-pts}}) \subset S_{\text{ft-pts}}\) by Lemma 02J3. Let \(s \in S\) be a finite type point. As \(f\) is surjective the scheme \(T_s = \Spec(\kappa(s)) \times_S T\) is nonempty, therefore has a finite type point \(t \in T_s\) by Lemma 02J2. Now \(T_s \to T\) is a morphism of finite type as a base change of \(s \to S\) (Lemma 01T4). Hence the image of \(t\) in \(T\) is a finite type point by Lemma 02J3 which maps to \(s\) by construction.

Lemma

Let \(S\) be a scheme. For any locally closed subset \(T \subset S\) we have \[T \not = \emptyset \Rightarrow T \cap S_{\text{ft-pts}} \not = \emptyset.\] In particular, for any closed subset \(T \subset S\) we see that \(T \cap S_{\text{ft-pts}}\) is dense in \(T\).

Proof

Note that \(T\) carries a scheme structure (see Schemes, Lemma 01J3) such that \(T \to S\) is a locally closed immersion. Any locally closed immersion is locally of finite type, see Lemma 01T5. Hence by Lemma 02J3 we see \(T_{\text{ft-pts}} \subset S_{\text{ft-pts}}\). Finally, any nonempty affine open of \(T\) has at least one closed point which is a finite type point of \(T\) by Lemma 02J2.

It follows that most of the material from Topology, Section 005T goes through with the set of closed points replaced by the set of points of finite type. In fact, if \(S\) is Jacobson then we recover the closed points as the finite type points.

Lemma

Let \(S\) be a scheme. The following are equivalent:

the scheme \(S\) is Jacobson,

\(S_{\text{ft-pts}}\) is the set of closed points of \(S\),

for all \(T \to S\) locally of finite type closed points map to closed points, and

for all \(T \to S\) locally of finite type closed points \(t \in T\) map to closed points \(s \in S\) with \(\kappa(s) \subset \kappa(t)\) finite.

Proof

We have trivially (4) \(\Rightarrow\) (3) \(\Rightarrow\) (2). Lemma 02J4 shows that (2) implies (1). Hence it suffices to show that (1) implies (4). Suppose that \(T \to S\) is locally of finite type. Choose \(t \in T\) closed and let \(s \in S\) be the image. Choose affine open neighbourhoods \(\Spec(R) = U \subset S\) of \(s\) and \(\Spec(A) = V \subset T\) of \(t\) with \(V\) mapping into \(U\). The induced ring map \(R \to A\) is of finite type (see Lemma 01T2) and \(R\) is Jacobson by Properties, Lemma 01P4. Thus the result follows from Algebra, Proposition 00GB.

Lemma

Let \(S\) be a Jacobson scheme. Any scheme locally of finite type over \(S\) is Jacobson.

Proof

This is clear from Algebra, Proposition 00GB (and Properties, Lemma 01P4 and Lemma 01T2).

Lemma

The following types of schemes are Jacobson.

Any scheme locally of finite type over a field.

Any scheme locally of finite type over \(\mathbf{Z}\).

Any scheme locally of finite type over a \(1\)-dimensional Noetherian domain with infinitely many primes.

A scheme of the form \(\Spec(R) \setminus \{\mathfrak m\}\) where \((R, \mathfrak m)\) is a Noetherian local ring. Also any scheme locally of finite type over it.

Proof

We will use Lemma 02J5 without mention. The spectrum of a field is clearly Jacobson. The spectrum of \(\mathbf{Z}\) is Jacobson, see Algebra, Lemma 00G4. For (3) see Algebra, Lemma 00KX. For (4) see Properties, Lemma 02IM.

Universally catenary schemes

Recall that a topological space \(X\) is called catenary if for every pair of irreducible closed subsets \(T \subset T'\) there exist a maximal chain of irreducible closed subsets \[T = T_0 \subset T_1 \subset \ldots \subset T_e = T'\] and every such chain has the same length. See Topology, Definition 02I1. Recall that a scheme is catenary if its underlying topological space is catenary. See Properties, Definition 02IW.

Definition

Let \(S\) be a scheme. Assume \(S\) is locally Noetherian. We say \(S\) is universally catenary if for every morphism \(X \to S\) locally of finite type the scheme \(X\) is catenary.

This is a “better” notion than catenary as there exist Noetherian schemes which are catenary but not universally catenary. See Examples, Section 02JE. Many schemes are universally catenary, see Lemma 02JB below.

Recall that a ring \(A\) is called catenary if for any pair of prime ideals \(\mathfrak p \subset \mathfrak q\) there exists a maximal chain of primes \[\mathfrak p = \mathfrak p_0 \subset \ldots \subset \mathfrak p_e = \mathfrak q\] and all of these have the same length. See Algebra, Definition 00NI. We have seen the relationship between catenary schemes and catenary rings in Properties, Section 02IV. Recall that a ring \(A\) is called universally catenary if \(A\) is Noetherian and for every finite type ring map \(A \to B\) the ring \(B\) is catenary. See Algebra, Definition 00NL. Many interesting rings which come up in algebraic geometry satisfy this property.

Lemma

Let \(S\) be a locally Noetherian scheme. The following are equivalent

\(S\) is universally catenary,

there exists an open covering of \(S\) all of whose members are universally catenary schemes,

for every affine open \(\Spec(R) = U \subset S\) the ring \(R\) is universally catenary, and

there exists an affine open covering \(S = \bigcup U_i\) such that each \(U_i\) is the spectrum of a universally catenary ring.

Moreover, in this case any scheme locally of finite type over \(S\) is universally catenary as well.

Proof

By Lemma 01T5 an open immersion is locally of finite type. A composition of morphisms locally of finite type is locally of finite type (Lemma 01T3). Thus it is clear that if \(S\) is universally catenary then any open and any scheme locally of finite type over \(S\) is universally catenary as well. This proves the final statement of the lemma and that (1) implies (2).

If \(\Spec(R)\) is a universally catenary scheme, then every scheme \(\Spec(A)\) with \(A\) a finite type \(R\)-algebra is catenary. Hence all these rings \(A\) are catenary by Algebra, Lemma 02IH. Thus \(R\) is universally catenary. Combined with the remarks above we conclude that (1) implies (3), and (2) implies (4). Of course (3) implies (4) trivially.

To finish the proof we show that (4) implies (1). Assume (4) and let \(X \to S\) be a morphism locally of finite type. We can find an affine open covering \(X = \bigcup V_j\) such that each \(V_j \to S\) maps into one of the \(U_i\). By Lemma 01T2 the induced ring map \(\mathcal{O}(U_i) \to \mathcal{O}(V_j)\) is of finite type. Hence \(\mathcal{O}(V_j)\) is catenary. Hence \(X\) is catenary by Properties, Lemma 02IX.

Lemma

Let \(S\) be a locally Noetherian scheme. The following are equivalent:

\(S\) is universally catenary, and

all local rings \(\mathcal{O}_{S, s}\) of \(S\) are universally catenary.

Proof

Assume that all local rings of \(S\) are universally catenary. Let \(f : X \to S\) be locally of finite type. We know that \(X\) is catenary if and only if \(\mathcal{O}_{X, x}\) is catenary for all \(x \in X\). If \(f(x) = s\), then \(\mathcal{O}_{X, x}\) is essentially of finite type over \(\mathcal{O}_{S, s}\). Hence \(\mathcal{O}_{X, x}\) is catenary by the assumption that \(\mathcal{O}_{S, s}\) is universally catenary.

Conversely, assume that \(S\) is universally catenary. Let \(s \in S\). We may replace \(S\) by an affine open neighbourhood of \(s\) by Lemma 02J9. Say \(S = \Spec(R)\) and \(s\) corresponds to the prime ideal \(\mathfrak p\). Any finite type \(R_{\mathfrak p}\)-algebra \(A'\) is of the form \(A_{\mathfrak p}\) for some finite type \(R\)-algebra \(A\). By assumption (and Lemma 02J9 if you like) the ring \(A\) is catenary, and hence \(A'\) (a localization of \(A\)) is catenary. Thus \(R_{\mathfrak p}\) is universally catenary.

Lemma

Let \(S\) be a locally Noetherian scheme. Then \(S\) is universally catenary if and only if the irreducible components of \(S\) are universally catenary.

Proof

Omitted. For the affine case, please see Algebra, Lemma 0AUP.

Lemma

The following types of schemes are universally catenary.

Any scheme locally of finite type over a field.

Any scheme locally of finite type over a Cohen-Macaulay scheme.

Any scheme locally of finite type over \(\mathbf{Z}\).

Any scheme locally of finite type over a \(1\)-dimensional Noetherian domain.

And so on.

Proof

All of these follow from the fact that a Cohen-Macaulay ring is universally catenary, see Algebra, Lemma 00NM. Also, use the last assertion of Lemma 02J9. Some details omitted.

Nagata schemes, reprise

See Properties, Section 033R for the definitions and basic properties of Nagata and universally Japanese schemes.

Lemma

Let \(f : X \to S\) be a morphism. If \(S\) is Nagata and \(f\) locally of finite type then \(X\) is Nagata. If \(S\) is universally Japanese and \(f\) locally of finite type then \(X\) is universally Japanese.

Proof

For “universally Japanese” this follows from Algebra, Lemma 032S. For “Nagata” this follows from Algebra, Proposition 0334.

Lemma

The following types of schemes are Nagata.

Any scheme locally of finite type over a field.

Any scheme locally of finite type over a Noetherian complete local ring.

Any scheme locally of finite type over \(\mathbf{Z}\).

Any scheme locally of finite type over a Dedekind ring of characteristic zero.

And so on.

Proof

By Lemma 035A we only need to show that the rings mentioned above are Nagata rings. For this see Algebra, Proposition 0335.

The singular locus, reprise

We look for a criterion that implies openness of the regular locus for any scheme locally of finite type over the base. Here is the definition.

Definition

Let \(X\) be a locally Noetherian scheme. We say \(X\) is J-2 if for every morphism \(Y \to X\) which is locally of finite type the regular locus \(\text{Reg}(Y)\) is open in \(Y\).

This is the analogue of the corresponding notion for Noetherian rings, see More on Algebra, Definition 07P7.

Lemma

Let \(X\) be a locally Noetherian scheme. The following are equivalent

\(X\) is J-2,

there exists an open covering of \(X\) all of whose members are J-2 schemes,

for every affine open \(\Spec(R) = U \subset X\) the ring \(R\) is J-2, and

there exists an affine open covering \(S = \bigcup U_i\) such that each \(\mathcal{O}(U_i)\) is J-2 for all \(i\).

Moreover, in this case any scheme locally of finite type over \(X\) is J-2 as well.

Proof

By Lemma 01T5 an open immersion is locally of finite type. A composition of morphisms locally of finite type is locally of finite type (Lemma 01T3). Thus it is clear that if \(X\) is J-2 then any open and any scheme locally of finite type over \(X\) is J-2 as well. This proves the final statement of the lemma.

If \(\Spec(R)\) is J-2, then for every finite type \(R\)-algebra \(A\) the regular locus of the scheme \(\Spec(A)\) is open. Hence \(R\) is J-2, by definition (see More on Algebra, Definition 07P7). Combined with the remarks above we conclude that (1) implies (3), and (2) implies (4). Of course (1) \(\Rightarrow\) (2) and (3) \(\Rightarrow\) (4) trivially.

To finish the proof we show that (4) implies (1). Assume (4) and let \(Y \to X\) be a morphism locally of finite type. We can find an affine open covering \(Y = \bigcup V_j\) such that each \(V_j \to X\) maps into one of the \(U_i\). By Lemma 01T2 the induced ring map \(\mathcal{O}(U_i) \to \mathcal{O}(V_j)\) is of finite type. Hence the regular locus of \(V_j = \Spec(\mathcal{O}(V_j))\) is open. Since \(\text{Reg}(Y) \cap V_j = \text{Reg}(V_j)\) we conclude that \(\text{Reg}(Y)\) is open as desired.

Lemma

The following types of schemes are J-2.

Any scheme locally of finite type over a field.

Any scheme locally of finite type over a Noetherian complete local ring.

Any scheme locally of finite type over \(\mathbf{Z}\).

Any scheme locally of finite type over a Noetherian local ring of dimension \(1\).

Any scheme locally of finite type over a Nagata ring of dimension \(1\).

Any scheme locally of finite type over a Dedekind ring of characteristic zero.

And so on.

Proof

By Lemma 07R4 we only need to show that the rings mentioned above are J-2. For this see More on Algebra, Proposition 07PJ.

Excellent schemes

We recall that a ring is quasi-excellent if it is a G-ring and J-2. A ring is excellent if it is quasi-excellent and universally catenary.

Definition

Let \(X\) be a scheme.

We say \(X\) is quasi-excellent if for every \(x \in X\) there exists an affine open neighbourhood \(x \in U \subset X\) such that the ring \(\mathcal{O}_X(U)\) is quasi-excellent (see More on Algebra, Definition 07QT).

We say \(X\) is excellent if for every \(x \in X\) there exists an affine open neighbourhood \(x \in U \subset X\) such that the ring \(\mathcal{O}_X(U)\) is excellent (see More on Algebra, Definition 07QT).

Quasi-excellent schemes are locally Noetherian (G-rings are Noetherian).

Lemma

Let \(X\) be a scheme. The following are equivalent

\(X\) is quasi-excellent, and

\(X\) is a G-scheme and J-2.

Proof

The implication (1) \(\Rightarrow\) (2) follows on combining the definition of quasi-excellent schemes (Definition 0HA7), the definition of quasi-excellent rings (More on Algebra, Definition 07QT), the definition of G-schemes (Properties, Definition 0HA3), and Lemma 07R4. Conversely, if \(X\) is a G-scheme and J-2, then for any \(x \in X\) we can find an affine open neighbourhood \(x \in U \subset X\) such that \(\mathcal{O}_X(U)\) is a G-ring. By Lemma 07R4 the ring \(\mathcal{O}_X(U)\) is also J-2, whence quasi-excellent.

Lemma

A quasi-excellent scheme is Nagata.

Proof

See More on Algebra, Lemma 07QV.

Lemma

Let \(X\) be a scheme. The following are equivalent

\(X\) is excellent, and

\(X\) is quasi-excellent and universally catenary.

Proof

Assume (1). Since excellent rings are quasi-excellent, it is clear that \(X\) is quasi-excellent. Since excellent rings are universally catenary, it also follows that \(X\) is universally catenary by Lemma 02J9. Assume (2). Let \(x \in X\). Since \(X\) is quasi-excellent there exists an affine open neighbourhood \(x \in U \subset X\) such that \(\mathcal{O}_X(U)\) is quasi-excellent. Since \(X\) is universally category, the ring \(\mathcal{O}_X(U)\) is also universally catenary by Lemma 02J9, whence excellent.

Lemma

Let \(X\) be a scheme. The following are equivalent:

The scheme \(X\) is (quasi-)excellent.

For every affine open \(U \subset X\) the ring \(\mathcal{O}_X(U)\) is (quasi-)excellent.

There exists an affine open covering \(X = \bigcup U_i\) such that each \(\mathcal{O}_X(U_i)\) is (quasi-)excellent.

There exists an open covering \(X = \bigcup X_j\) such that each open subscheme \(X_j\) is (quasi-)excellent.

Moreover, if \(X\) is (quasi-)excellent then every open subscheme is (quasi-)excellent.

Proof

By Lemma 0HA8 being quasi-excellent is the same as being a G-scheme and J-2. Thus for the quasi-excellent case, the lemma follows from Lemma 07R4 and Properties, Lemma 0HA5. For the excellent case, use Lemma 0HAA, the quasi-excellent case, and Lemma 02J9.

Lemma

Let \(f : X \to S\) be a morphism. If \(S\) is (quasi-)excellent and \(f\) locally of finite type then \(X\) is (quasi-)excellent.

Proof

See More on Algebra, Lemma 07QU.

Lemma

The following types of schemes are excellent.

Any scheme locally of finite type over a field.

Any scheme locally of finite type over a Noetherian complete local ring.

Any scheme locally of finite type over \(\mathbf{Z}\).

Any scheme locally of finite type over a Dedekind ring of characteristic zero.

Proof

By Lemmas 0HAB and 0HAC we only need to show that the rings mentioned above are excellent. For this see More on Algebra, Proposition 07QW.

Quasi-finite morphisms

A solid treatment of quasi-finite morphisms is the basis of many developments further down the road. It will lead to various versions of Zariski’s Main Theorem, behaviour of dimensions of fibres, descent for étale morphisms, etc, etc. Before reading this section it may be a good idea to take a look at the algebra results in Algebra, Section 02MK.

Recall that a finite type ring map \(R \to A\) is quasi-finite at a prime \(\mathfrak q\) if \(\mathfrak q\) defines an isolated point of its fibre, see Algebra, Definition 00PL.

Definition

Let \(f : X \to S\) be a morphism of schemes.

We say that \(f\) is quasi-finite at a point \(x \in X\) if there exist an affine neighbourhood \(\Spec(A) = U \subset X\) of \(x\) and an affine open \(\Spec(R) = V \subset S\) such that \(f(U) \subset V\), the ring map \(R \to A\) is of finite type, and \(R \to A\) is quasi-finite at the prime of \(A\) corresponding to \(x\) (see above).

We say \(f\) is locally quasi-finite if \(f\) is quasi-finite at every point \(x\) of \(X\).

We say that \(f\) is quasi-finite if \(f\) is of finite type and every point \(x\) is an isolated point of its fibre.

Trivially, a locally quasi-finite morphism is locally of finite type. We will see below that a morphism \(f\) which is locally of finite type is quasi-finite at \(x\) if and only if \(x\) is isolated in its fibre. Moreover, the set of points at which a morphism is quasi-finite is open; we will see this in Section 03GS on Zariski’s Main Theorem.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(x \in X\) be a point. Set \(s = f(x)\). If \(\kappa(x)/\kappa(s)\) is an algebraic field extension, then

\(x\) is a closed point of its fibre, and

if in addition \(s\) is a closed point of \(S\), then \(x\) is a closed point of \(X\).

Proof

The second statement follows from the first by elementary topology. According to Schemes, Lemma 01K1 to prove the first statement we may replace \(X\) by \(X_s\) and \(S\) by \(\Spec(\kappa(s))\). Thus we may assume that \(S = \Spec(k)\) is the spectrum of a field. In this case, let \(\Spec(A) = U \subset X\) be any affine open containing \(x\). The point \(x\) corresponds to a prime ideal \(\mathfrak q \subset A\) such that \(\kappa(\mathfrak q)/k\) is an algebraic field extension. By Algebra, Lemma 00GA we see that \(\mathfrak q\) is a maximal ideal, i.e., \(x \in U\) is a closed point. Since the affine opens form a basis of the topology of \(X\) we conclude that \(\{x\}\) is closed.

The following lemma is a version of the Hilbert Nullstellensatz.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(x \in X\) be a point. Set \(s = f(x)\). Assume \(f\) is locally of finite type. Then \(x\) is a closed point of its fibre if and only if \(\kappa(x)/\kappa(s)\) is a finite field extension.

Proof

If the extension is finite, then \(x\) is a closed point of the fibre by Lemma 01TE above. For the converse, assume that \(x\) is a closed point of its fibre. Choose affine opens \(\Spec(A) = U \subset X\) and \(\Spec(R) = V \subset S\) such that \(f(U) \subset V\). By Lemma 01T2 the ring map \(R \to A\) is of finite type. Let \(\mathfrak q \subset A\), resp. \(\mathfrak p \subset R\) be the prime ideal corresponding to \(x\), resp. \(s\). Consider the fibre ring \(\overline{A} = A \otimes_R \kappa(\mathfrak p)\). Let \(\overline{\mathfrak q}\) be the prime of \(\overline{A}\) corresponding to \(\mathfrak q\). The assumption that \(x\) is a closed point of its fibre implies that \(\overline{\mathfrak q}\) is a maximal ideal of \(\overline{A}\). Since \(\overline{A}\) is an algebra of finite type over the field \(\kappa(\mathfrak p)\) we see by the Hilbert Nullstellensatz, see Algebra, Theorem 00FV, that \(\kappa(\overline{\mathfrak q})\) is a finite extension of \(\kappa(\mathfrak p)\). Since \(\kappa(s) = \kappa(\mathfrak p)\) and \(\kappa(x) = \kappa(\mathfrak q) = \kappa(\overline{\mathfrak q})\) we win.

Lemma

Let \(f : X \to S\) be a morphism of schemes which is locally of finite type. Let \(g : S' \to S\) be any morphism. Denote \(f' : X' \to S'\) the base change. If \(x' \in X'\) maps to a point \(x \in X\) which is closed in \(X_{f(x)}\) then \(x'\) is closed in \(X'_{f'(x')}\).

Proof

The residue field \(\kappa(x')\) is a quotient of \(\kappa(f'(x')) \otimes_{\kappa(f(x))} \kappa(x)\), see Schemes, Lemma 01JT. Hence it is a finite extension of \(\kappa(f'(x'))\) as \(\kappa(x)\) is a finite extension of \(\kappa(f(x))\) by Lemma 01TF. Thus we see that \(x'\) is closed in its fibre by applying that lemma one more time.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(x \in X\) be a point. Set \(s = f(x)\). If \(f\) is quasi-finite at \(x\), then the residue field extension \(\kappa(x)/\kappa(s)\) is finite.

Proof

This is clear from Algebra, Definition 00PL.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(x \in X\) be a point. Set \(s = f(x)\). Let \(X_s\) be the fibre of \(f\) at \(s\). Assume \(f\) is locally of finite type. The following are equivalent:

The morphism \(f\) is quasi-finite at \(x\).

The point \(x\) is isolated in \(X_s\).

The point \(x\) is closed in \(X_s\) and there is no point \(x' \in X_s\), \(x' \not = x\) which specializes to \(x\).

For any pair of affine opens \(\Spec(A) = U \subset X\), \(\Spec(R) = V \subset S\) with \(f(U) \subset V\) and \(x \in U\) corresponding to \(\mathfrak q \subset A\) the ring map \(R \to A\) is quasi-finite at \(\mathfrak q\).

Proof

Assume \(f\) is quasi-finite at \(x\). By assumption there exist opens \(U \subset X\), \(V \subset S\) such that \(f(U) \subset V\), \(x \in U\) and \(x\) an isolated point of \(U_s\). Hence \(\{x\} \subset U_s\) is an open subset. Since \(U_s = U \cap X_s \subset X_s\) is also open we conclude that \(\{x\} \subset X_s\) is an open subset also. Thus we conclude that \(x\) is an isolated point of \(X_s\).

Note that \(X_s\) is a Jacobson scheme by Lemma 02J6 (and Lemma 01T4). If \(x\) is isolated in \(X_s\), i.e., \(\{x\} \subset X_s\) is open, then \(\{x\}\) contains a closed point (by the Jacobson property), hence \(x\) is closed in \(X_s\). It is clear that there is no point \(x' \in X_s\), distinct from \(x\), specializing to \(x\).

Assume that \(x\) is closed in \(X_s\) and that there is no point \(x' \in X_s\), distinct from \(x\), specializing to \(x\). Consider a pair of affine opens \(\Spec(A) = U \subset X\), \(\Spec(R) = V \subset S\) with \(f(U) \subset V\) and \(x \in U\). Let \(\mathfrak q \subset A\) correspond to \(x\) and \(\mathfrak p \subset R\) correspond to \(s\). By Lemma 01T2 the ring map \(R \to A\) is of finite type. Consider the fibre ring \(\overline{A} = A \otimes_R \kappa(\mathfrak p)\). Let \(\overline{\mathfrak q}\) be the prime of \(\overline{A}\) corresponding to \(\mathfrak q\). Since \(\Spec(\overline{A})\) is an open subscheme of the fibre \(X_s\) we see that \(\overline{q}\) is a maximal ideal of \(\overline{A}\) and that there is no point of \(\Spec(\overline{A})\) specializing to \(\overline{\mathfrak q}\). This implies that \(\dim(\overline{A}_{\overline{q}}) = 0\). Hence by Algebra, Definition 00PL we see that \(R \to A\) is quasi-finite at \(\mathfrak q\), i.e., \(X \to S\) is quasi-finite at \(x\) by definition.

At this point we have shown conditions (1) – (3) are all equivalent. It is clear that (4) implies (1). And it is also clear that (2) implies (4) since if \(x\) is an isolated point of \(X_s\) then it is also an isolated point of \(U_s\) for any open \(U\) which contains it.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(s \in S\). Assume that

\(f\) is locally of finite type, and

\(f^{-1}(\{s\})\) is a finite set.

Then \(X_s\) is a finite discrete topological space, and \(f\) is quasi-finite at each point of \(X\) lying over \(s\).

Proof

Suppose \(T\) is a scheme which (a) is locally of finite type over a field \(k\), and (b) has finitely many points. Then Lemma 02J6 shows \(T\) is a Jacobson scheme. A finite Jacobson space is discrete, see Topology, Lemma 07JU. Apply this remark to the fibre \(X_s\) which is locally of finite type over \(\Spec(\kappa(s))\) to see the first statement. Finally, apply Lemma 01TH to see the second.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Assume \(f\) is locally of finite type. Then the following are equivalent

\(f\) is locally quasi-finite,

for every \(s \in S\) the fibre \(X_s\) is a discrete topological space, and

for every morphism \(\Spec(k) \to S\) where \(k\) is a field the base change \(X_k\) has an underlying discrete topological space.

Proof

It is immediate that (3) implies (2). Lemma 01TH shows that (2) is equivalent to (1). Assume (2) and let \(\Spec(k) \to S\) be as in (3). Denote \(s \in S\) the image of \(\Spec(k) \to S\). Then \(X_k\) is the base change of \(X_s\) via \(\Spec(k) \to \Spec(\kappa(s))\). Hence every point of \(X_k\) is closed by Lemma 053M. As \(X_k \to \Spec(k)\) is locally of finite type (by Lemma 01T4), we may apply Lemma 01TH to conclude that every point of \(X_k\) is isolated, i.e., \(X_k\) has a discrete underlying topological space.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Then \(f\) is quasi-finite if and only if \(f\) is locally quasi-finite and quasi-compact.

Proof

Assume \(f\) is quasi-finite. It is quasi-compact by Definition 01T1. Let \(x \in X\). We see that \(f\) is quasi-finite at \(x\) by Lemma 01TH. Hence \(f\) is quasi-compact and locally quasi-finite.

Assume \(f\) is quasi-compact and locally quasi-finite. Then \(f\) is of finite type. Let \(x \in X\) be a point. By Lemma 01TH we see that \(x\) is an isolated point of its fibre. The lemma is proved.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent:

\(f\) is quasi-finite, and

\(f\) is locally of finite type, quasi-compact, and has finite fibres.

Proof

Assume \(f\) is quasi-finite. In particular \(f\) is locally of finite type and quasi-compact (since it is of finite type). Let \(s \in S\). Since every \(x \in X_s\) is isolated in \(X_s\) we see that \(X_s = \bigcup_{x \in X_s} \{x\}\) is an open covering. As \(f\) is quasi-compact, the fibre \(X_s\) is quasi-compact. Hence we see that \(X_s\) is finite.

Conversely, assume \(f\) is locally of finite type, quasi-compact and has finite fibres. Then it is locally quasi-finite by Lemma 02NG. Hence it is quasi-finite by Lemma 01TJ.

Recall that a ring map \(R \to A\) is quasi-finite if it is of finite type and quasi-finite at all primes of \(A\), see Algebra, Definition 00PL.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent

The morphism \(f\) is locally quasi-finite.

For every pair of affine opens \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the ring map \(\mathcal{O}_S(V) \to \mathcal{O}_X(U)\) is quasi-finite.

There exists an open covering \(S = \bigcup_{j \in J} V_j\) and open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that each of the morphisms \(U_i \to V_j\), \(j\in J, i\in I_j\) is locally quasi-finite.

There exists an affine open covering \(S = \bigcup_{j \in J} V_j\) and affine open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that the ring map \(\mathcal{O}_S(V_j) \to \mathcal{O}_X(U_i)\) is quasi-finite, for all \(j\in J, i\in I_j\).

Moreover, if \(f\) is locally quasi-finite then for any open subschemes \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the restriction \(f|_U : U \to V\) is locally quasi-finite.

Proof

For a ring map \(R \to A\) let us define \(P(R \to A)\) to mean “\(R \to A\) is quasi-finite” (see remark above lemma). We claim that \(P\) is a local property of ring maps. We check conditions (a), (b) and (c) of Definition 01SR. In the proof of Lemma 01T2 we have seen that (a), (b) and (c) hold for the property of being “of finite type”. Note that, for a finite type ring map \(R \to A\), the property \(R \to A\) is quasi-finite at \(\mathfrak q\) depends only on the local ring \(A_{\mathfrak q}\) as an algebra over \(R_{\mathfrak p}\) where \(\mathfrak p = R \cap \mathfrak q\) (usual abuse of notation). Using these remarks (a), (b) and (c) of Definition 01SR follow immediately. For example, suppose \(R \to A\) is a ring map such that all of the ring maps \(R \to A_{a_i}\) are quasi-finite for \(a_1, \ldots, a_n \in A\) generating the unit ideal. We conclude that \(R \to A\) is of finite type. Also, for any prime \(\mathfrak q \subset A\) the local ring \(A_{\mathfrak q}\) is isomorphic as an \(R\)-algebra to the local ring \((A_{a_i})_{\mathfrak q_i}\) for some \(i\) and some \(\mathfrak q_i \subset A_{a_i}\). Hence we conclude that \(R \to A\) is quasi-finite at \(\mathfrak q\).

We conclude that Lemma 01ST applies with \(P\) as in the previous paragraph. Hence it suffices to prove that \(f\) is locally quasi-finite is equivalent to \(f\) is locally of type \(P\). Since \(P(R \to A)\) is “\(R \to A\) is quasi-finite” which means \(R \to A\) is quasi-finite at every prime of \(A\), this follows from Lemma 01TH.

Lemma

The composition of two morphisms which are locally quasi-finite is locally quasi-finite. The same is true for quasi-finite morphisms.

Proof

In the proof of Lemma 01TK we saw that \(P =\)“quasi-finite” is a local property of ring maps, and that a morphism of schemes is locally quasi-finite if and only if it is locally of type \(P\) as in Definition 01SS. Hence the first statement of the lemma follows from Lemma 01SV combined with the fact that being quasi-finite is a property of ring maps that is stable under composition, see Algebra, Lemma 00PO. By the above, Lemma 01TJ and the fact that compositions of quasi-compact morphisms are quasi-compact, see Schemes, Lemma 01K6 we see that the composition of quasi-finite morphisms is quasi-finite.

We will see later (Lemma 01TI) that the set \(U\) of the following lemma is open.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(g : S' \to S\) be a morphism of schemes. Denote \(f' : X' \to S'\) the base change of \(f\) by \(g\) and denote \(g' : X' \to X\) the projection. Assume \(X\) is locally of finite type over \(S\).

Let \(U \subset X\) (resp. \(U' \subset X'\)) be the set of points where \(f\) (resp. \(f'\)) is quasi-finite. Then \(U' = U \times_S S' = (g')^{-1}(U)\).

The base change of a locally quasi-finite morphism is locally quasi-finite.

The base change of a quasi-finite morphism is quasi-finite.

Proof

The first and second assertion follow from the corresponding algebra result, see Algebra, Lemma 00PP (combined with the fact that \(f'\) is also locally of finite type by Lemma 01T4). By the above, Lemma 01TJ and the fact that a base change of a quasi-compact morphism is quasi-compact, see Schemes, Lemma 01K5 we see that the base change of a quasi-finite morphism is quasi-finite.

Lemma

Let \(f : X \to S\) be a morphism of schemes of finite type. Let \(s \in S\). There are at most finitely many points of \(X\) lying over \(s\) at which \(f\) is quasi-finite.

Proof

The fibre \(X_s\) is a scheme of finite type over a field, hence Noetherian (Lemma 01T6). Hence the topology on \(X_s\) is Noetherian (Properties, Lemma 01OZ) and can have at most a finite number of isolated points (by elementary topology). Thus our lemma follows from Lemma 01TH.

Lemma

Let \(f : X \to Y\) be a morphism of schemes. If \(f\) is locally of finite type and a monomorphism, then \(f\) is separated and locally quasi-finite.

Proof

A monomorphism is separated by Schemes, Lemma 01L4. A monomorphism is injective, hence we get \(f\) is quasi-finite at every \(x \in X\) for example by Lemma 01TH.

Lemma

Any immersion is locally quasi-finite.

Proof

This is true because an open immersion is a local isomorphism and a closed immersion is clearly quasi-finite.

Lemma

Let \(X \to Y\) be a morphism of schemes over a base scheme \(S\). Let \(x \in X\). If \(X \to S\) is quasi-finite at \(x\), then \(X \to Y\) is quasi-finite at \(x\). If \(X\) is locally quasi-finite over \(S\), then \(X \to Y\) is locally quasi-finite.

Proof

Via Lemma 01TK this translates into the following algebra fact: Given ring maps \(A \to B \to C\) such that \(A \to C\) is quasi-finite, then \(B \to C\) is quasi-finite. This follows from Algebra, Lemma 00PN with \(R = A\), \(S = S' = C\) and \(R' = B\).

Lemma

Let \(f : X \to Y\) and \(g : Y \to S\) be morphisms of schemes. If \(f\) is surjective, \(g \circ f\) locally quasi-finite, and \(g\) locally of finite type, then \(g : Y \to S\) is locally quasi-finite.

Proof

Let \(x \in X\) with images \(y \in Y\) and \(s \in S\). Since \(g \circ f\) is locally quasi-finite by Lemma 01TG the extension \(\kappa(x)/\kappa(s)\) is finite. Hence \(\kappa(y)/\kappa(s)\) is finite. Hence \(y\) is a closed point of \(Y_s\) by Lemma 01TE. Since \(f\) is surjective, we see that every point of \(Y\) is closed in its fibre over \(S\). Thus by Lemma 01TH we conclude that \(g\) is quasi-finite at every point.

Morphisms of finite presentation

Recall that a ring map \(R \to A\) is of finite presentation if \(A\) is isomorphic to \(R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)\) as an \(R\)-algebra for some \(n, m\) and some polynomials \(f_j\), see Algebra, Definition 00F3.

Definition

Let \(f : X \to S\) be a morphism of schemes.

We say that \(f\) is of finite presentation at \(x \in X\) if there exists an affine open neighbourhood \(\Spec(A) = U \subset X\) of \(x\) and affine open \(\Spec(R) = V \subset S\) with \(f(U) \subset V\) such that the induced ring map \(R \to A\) is of finite presentation.

We say that \(f\) is locally of finite presentation if it is of finite presentation at every point of \(X\).

We say that \(f\) is of finite presentation if it is locally of finite presentation, quasi-compact and quasi-separated.

Note that a morphism of finite presentation is not just a quasi-compact morphism which is locally of finite presentation. Later we will characterize morphisms which are locally of finite presentation as those morphisms such that \[\colim \Mor_S(T_i, X) = \Mor_S(\lim T_i, X)\] for any directed system of affine schemes \(T_i\) over \(S\). See Limits, Proposition 01ZC. In Limits, Section 01ZL we show that, if \(S = \lim_i S_i\) is a limit of affine schemes, any scheme \(X\) of finite presentation over \(S\) descends to a scheme \(X_i\) over \(S_i\) for some \(i\).

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent

The morphism \(f\) is locally of finite presentation.

For every affine opens \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the ring map \(\mathcal{O}_S(V) \to \mathcal{O}_X(U)\) is of finite presentation.

There exist an open covering \(S = \bigcup_{j \in J} V_j\) and open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that each of the morphisms \(U_i \to V_j\), \(j\in J, i\in I_j\) is locally of finite presentation.

There exist an affine open covering \(S = \bigcup_{j \in J} V_j\) and affine open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that the ring map \(\mathcal{O}_S(V_j) \to \mathcal{O}_X(U_i)\) is of finite presentation, for all \(j\in J, i\in I_j\).

Moreover, if \(f\) is locally of finite presentation then for any open subschemes \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the restriction \(f|_U : U \to V\) is locally of finite presentation.

Proof

This follows from Lemma 01SU if we show that the property “\(R \to A\) is of finite presentation” is local. We check conditions (a), (b) and (c) of Definition 01SR. By Algebra, Lemma 05G5 being of finite presentation is stable under base change and hence we conclude (a) holds. By Algebra, Lemma 00F4 being of finite presentation is stable under composition and trivially for any ring \(R\) the ring map \(R \to R_f\) is of finite presentation. We conclude (b) holds. Finally, property (c) is true according to Algebra, Lemma 00EP.

Lemma

The composition of two morphisms which are locally of finite presentation is locally of finite presentation. The same is true for morphisms of finite presentation.

Proof

In the proof of Lemma 01TQ we saw that being of finite presentation is a local property of ring maps. Hence the first statement of the lemma follows from Lemma 01SV combined with the fact that being of finite presentation is a property of ring maps that is stable under composition, see Algebra, Lemma 00F4. By the above and the fact that compositions of quasi-compact, quasi-separated morphisms are quasi-compact and quasi-separated, see Schemes, Lemmas 01K6 and 01KU we see that the composition of morphisms of finite presentation is of finite presentation.

Lemma

The base change of a morphism which is locally of finite presentation is locally of finite presentation. The same is true for morphisms of finite presentation.

Proof

In the proof of Lemma 01TQ we saw that being of finite presentation is a local property of ring maps. Hence the first statement of the lemma follows from Lemma 01SV combined with the fact that being of finite presentation is a property of ring maps that is stable under base change, see Algebra, Lemma 05G5. By the above and the fact that a base change of a quasi-compact, quasi-separated morphism is quasi-compact and quasi-separated, see Schemes, Lemmas 01K5 and 01KU we see that the base change of a morphism of finite presentation is a morphism of finite presentation.

Lemma

Any open immersion is locally of finite presentation.

Proof

This is true because an open immersion is a local isomorphism.

Lemma

Any open immersion is of finite presentation if and only if it is quasi-compact.

Proof

We have seen (Lemma 01TT) that an open immersion is locally of finite presentation. We have seen (Schemes, Lemma 01L7) that an immersion is separated and hence quasi-separated. From this and Definition 01TP the lemma follows.

Lemma

A closed immersion \(i : Z \to X\) is of finite presentation if and only if the associated quasi-coherent sheaf of ideals \(\mathcal{I} = \Ker(\mathcal{O}_X \to i_*\mathcal{O}_Z)\) is of finite type (as an \(\mathcal{O}_X\)-module).

Proof

On any affine open \(\Spec(R) \subset X\) we have \(i^{-1}(\Spec(R)) = \Spec(R/I)\) and \(\mathcal{I} = \widetilde{I}\). Moreover, \(\mathcal{I}\) is of finite type if and only if \(I\) is a finite \(R\)-module for every such affine open (see Properties, Lemma 01PB). And \(R/I\) is of finite presentation over \(R\) if and only if \(I\) is a finite \(R\)-module. Hence we win.

Lemma

A morphism which is locally of finite presentation is locally of finite type. A morphism of finite presentation is of finite type.

Proof

Omitted.

Lemma

Let \(f : X \to S\) be a morphism.

If \(S\) is locally Noetherian and \(f\) locally of finite type then \(f\) is locally of finite presentation.

If \(S\) is locally Noetherian and \(f\) of finite type then \(f\) is of finite presentation.

Proof

The first statement follows from the fact that a ring of finite type over a Noetherian ring is of finite presentation, see Algebra, Lemma 00FP. Suppose that \(f\) is of finite type and \(S\) is locally Noetherian. Then \(f\) is quasi-compact and locally of finite presentation by (1). Hence it suffices to prove that \(f\) is quasi-separated. This follows from Lemma 01T7 (and Lemma 01TW).

Lemma

Let \(S\) be a scheme which is quasi-compact and quasi-separated. If \(X\) is of finite presentation over \(S\), then \(X\) is quasi-compact and quasi-separated.

Proof

Omitted.

Lemma

Let \(f : X \to Y\) be a morphism of schemes over \(S\).

If \(X\) is locally of finite presentation over \(S\) and \(Y\) is locally of finite type over \(S\), then \(f\) is locally of finite presentation.

If \(X\) is of finite presentation over \(S\) and \(Y\) is quasi-separated and locally of finite type over \(S\), then \(f\) is of finite presentation.

Proof

Proof of (1). Via Lemma 01TQ this translates into the following algebra fact: Given ring maps \(A \to B \to C\) such that \(A \to C\) is of finite presentation and \(A \to B\) is of finite type, then \(B \to C\) is of finite presentation. See Algebra, Lemma 00F4.

Part (2) follows from (1) and Schemes, Lemmas 01KV and 03GI.

Lemma

Let \(f : X \to Y\) be a morphism of schemes with diagonal \(\Delta : X \to X \times_Y X\). If \(f\) is locally of finite type then \(\Delta\) is locally of finite presentation. If \(f\) is quasi-separated and locally of finite type, then \(\Delta\) is of finite presentation.

Proof

Note that \(\Delta\) is a morphism of schemes over \(X\) (via the second projection \(X \times_Y X \to X\)). Assume \(f\) is locally of finite type. Note that \(X\) is of finite presentation over \(X\) and \(X \times_Y X\) is locally of finite type over \(X\) (by Lemma 01T4). Thus the first statement holds by Lemma 02FV. The second statement follows from the first, the definitions, and the fact that a diagonal morphism is a monomorphism, hence separated (Schemes, Lemma 01L4).

Constructible sets

Constructible and locally constructible sets of schemes have been discussed in Properties, Section 054B. In this section we prove some results concerning images and inverse images of (locally) constructible sets. The main result is Chevalley’s theorem which states that the image of a locally constructible set under a morphism of finite presentation is locally constructible.

Lemma

Let \(f : X \to Y\) be a morphism of schemes. Let \(E \subset Y\) be a subset. If \(E\) is (locally) constructible in \(Y\), then \(f^{-1}(E)\) is (locally) constructible in \(X\).

Proof

To show that the inverse image of every constructible subset is constructible it suffices to show that the inverse image of every retrocompact open \(V\) of \(Y\) is retrocompact in \(X\), see Topology, Lemma 005I. The significance of \(V\) being retrocompact in \(Y\) is just that the open immersion \(V \to Y\) is quasi-compact. Hence the base change \(f^{-1}(V) = X \times_Y V \to X\) is quasi-compact too, see Schemes, Lemma 01K5. Hence we see \(f^{-1}(V)\) is retrocompact in \(X\). Suppose \(E\) is locally constructible in \(Y\). Choose \(x \in X\). Choose an affine neighbourhood \(V\) of \(f(x)\) and an affine neighbourhood \(U \subset X\) of \(x\) such that \(f(U) \subset V\). Thus we think of \(f|_U : U \to V\) as a morphism into \(V\). By Properties, Lemma 054C we see that \(E \cap V\) is constructible in \(V\). By the constructible case we see that \((f|_U)^{-1}(E \cap V)\) is constructible in \(U\). Since \((f|_U)^{-1}(E \cap V) = f^{-1}(E) \cap U\) we win.

Lemma

Let \(f : X \to Y\) be a morphism of schemes. Assume

\(f\) is quasi-compact and locally of finite presentation, and

\(Y\) is quasi-compact and quasi-separated.

Then the image of every constructible subset of \(X\) is constructible in \(Y\).

Proof

By Properties, Lemma 054E it suffices to prove this lemma in case \(Y\) is affine. In this case \(X\) is quasi-compact. Hence we can write \(X = U_1 \cup \ldots \cup U_n\) with each \(U_i\) affine open in \(X\). If \(E \subset X\) is constructible, then each \(E \cap U_i\) is constructible too, see Topology, Lemma 005J. Hence, since \(f(E) = \bigcup f(E \cap U_i)\) and since finite unions of constructible sets are constructible, this reduces us to the case where \(X\) is affine. In this case the result is Algebra, Theorem 00FE.

Theorem

Let \(f : X \to Y\) be a morphism of schemes. Assume \(f\) is quasi-compact and locally of finite presentation. Then the image of every locally constructible subset is locally constructible.

Proof

Let \(E \subset X\) be locally constructible. We have to show that \(f(E)\) is locally constructible too. We will show that \(f(E) \cap V\) is constructible for any affine open \(V \subset Y\). Thus we reduce to the case where \(Y\) is affine. In this case \(X\) is quasi-compact. Hence we can write \(X = U_1 \cup \ldots \cup U_n\) with each \(U_i\) affine open in \(X\). If \(E \subset X\) is locally constructible, then each \(E \cap U_i\) is constructible, see Properties, Lemma 054C. Hence, since \(f(E) = \bigcup f(E \cap U_i)\) and since finite unions of constructible sets are constructible, this reduces us to the case where \(X\) is affine. In this case the result is Algebra, Theorem 00FE.

Lemma

Let \(X\) be a scheme. Let \(x \in X\). Let \(E \subset X\) be a locally constructible subset. If \(\{x' \mid x' \leadsto x\} \subset E\), then \(E\) contains an open neighbourhood of \(x\).

Proof

Assume \(\{x' \mid x' \leadsto x\} \subset E\). We may assume \(X\) is affine. In this case \(E\) is constructible, see Properties, Lemma 054C. In particular, also the complement \(E^c\) is constructible. By Algebra, Lemma 00F8 we can find a morphism of affine schemes \(f : Y \to X\) such that \(E^c = f(Y)\). Let \(Z \subset X\) be the scheme theoretic image of \(f\). By Lemma 02JQ and the assumption \(\{x' \mid x' \leadsto x\} \subset E\) we see that \(x \not \in Z\). Hence \(X \setminus Z \subset E\) is an open neighbourhood of \(x\) contained in \(E\).

Open morphisms

Definition

Let \(f : X \to S\) be a morphism.

We say \(f\) is open if the map on underlying topological spaces is open.

We say \(f\) is universally open if for any morphism of schemes \(S' \to S\) the base change \(f' : X_{S'} \to S'\) is open.

According to Topology, Lemma 0066 generalizations lift along certain types of open maps of topological spaces. In fact generalizations lift along any open morphism of schemes (see Lemma 040F). Also, we will see that generalizations lift along flat morphisms of schemes (Lemma 03HV). This sometimes in turn implies that the morphism is open.

Lemma

Let \(f : X \to S\) be a morphism.

If \(f\) is locally of finite presentation and generalizations lift along \(f\), then \(f\) is open.

If \(f\) is locally of finite presentation and generalizations lift along every base change of \(f\), then \(f\) is universally open.

Proof

It suffices to prove the first assertion. This reduces to the case where both \(X\) and \(S\) are affine. In this case the result follows from Algebra, Lemma 00HW and Proposition 00I1.

See also Lemma 01UA for the case of a morphism flat of finite presentation.

Lemma

A composition of (universally) open morphisms is (universally) open.

Proof

Omitted.

Lemma

Let \(k\) be a field. Let \(X\) be a scheme over \(k\). The structure morphism \(X \to \Spec(k)\) is universally open.

Proof

Let \(S \to \Spec(k)\) be a morphism. We have to show that the base change \(X_S \to S\) is open. The question is local on \(S\) and \(X\), hence we may assume that \(S\) and \(X\) are affine. In this case the result is Algebra, Lemma 037G.

Lemma

Let \(\varphi : X \to Y\) be a morphism of schemes. If \(\varphi\) is open, then \(\varphi\) is generizing (i.e., generalizations lift along \(\varphi\)). If \(\varphi\) is universally open, then \(\varphi\) is universally generizing.

Proof

Assume \(\varphi\) is open. Let \(y' \leadsto y\) be a specialization of points of \(Y\). Let \(x \in X\) with \(\varphi(x) = y\). Choose affine opens \(U \subset X\) and \(V \subset Y\) such that \(\varphi(U) \subset V\) and \(x \in U\). Then also \(y' \in V\). Hence we may replace \(X\) by \(U\) and \(Y\) by \(V\) and assume \(X\), \(Y\) affine. The affine case is Algebra, Lemma 0407 (combined with Algebra, Lemma 00HW).

Lemma

Let \(f : X \to Y\) be a morphism of schemes. Let \(g : Y' \to Y\) be open and surjective such that the base change \(f' : X' \to Y'\) is quasi-compact. Then \(f\) is quasi-compact.

Proof

Let \(V \subset Y\) be a quasi-compact open. As \(g\) is open and surjective we can find a quasi-compact open \(W' \subset W\) such that \(g(W') = V\). By assumption \((f')^{-1}(W')\) is quasi-compact. The image of \((f')^{-1}(W')\) in \(X\) is equal to \(f^{-1}(V)\), see Lemma 0495. Hence \(f^{-1}(V)\) is quasi-compact as the image of a quasi-compact space, see Topology, Lemma 04Z9. Thus \(f\) is quasi-compact.

Submersive morphisms

Definition

Let \(f : X \to Y\) be a morphism of schemes.

We say \(f\) is submersive⁷ if the continuous map of underlying topological spaces is submersive, see Topology, Definition 0406.

We say \(f\) is universally submersive if for every morphism of schemes \(Y' \to Y\) the base change \(Y' \times_Y X \to Y'\) is submersive.

We note that a submersive morphism is in particular surjective.

Lemma

The base change of a universally submersive morphism of schemes by any morphism of schemes is universally submersive.

Proof

This is immediate from the definition.

Lemma

The composition of a pair of (universally) submersive morphisms of schemes is (universally) submersive.

Proof

Omitted.

Flat morphisms

Flatness is one of the most important technical tools in algebraic geometry. In this section we introduce this notion. We intentionally limit the discussion to straightforward observations, apart from Lemma 01UA. A very important class of results, namely criteria for flatness, are discussed in Algebra, Sections 00MD, 051E, 00R3, and More on Morphisms, Section 039A. There is a chapter dedicated to advanced material on flat morphisms of schemes, namely More on Flatness, Section 057N.

Recall that a module \(M\) over a ring \(R\) is flat if the functor \(-\otimes_R M : \text{Mod}_R \to \text{Mod}_R\) is exact. A ring map \(R \to A\) is said to be flat if \(A\) is flat as an \(R\)-module. See Algebra, Definition 00HB.

Definition

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{F}\) be a quasi-coherent sheaf of \(\mathcal{O}_X\)-modules.

We say \(f\) is flat at a point \(x \in X\) if the local ring \(\mathcal{O}_{X, x}\) is flat over the local ring \(\mathcal{O}_{S, f(x)}\).

We say that \(\mathcal{F}\) is flat over \(S\) at a point \(x \in X\) if the stalk \(\mathcal{F}_x\) is a flat \(\mathcal{O}_{S, f(x)}\)-module.

We say \(f\) is flat if \(f\) is flat at every point of \(X\).

We say that \(\mathcal{F}\) is flat over \(S\) if \(\mathcal{F}\) is flat over \(S\) at every point \(x\) of \(X\).

Thus we see that \(f\) is flat if and only if the structure sheaf \(\mathcal{O}_X\) is flat over \(S\).

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{F}\) be a quasi-coherent sheaf of \(\mathcal{O}_X\)-modules. The following are equivalent

The sheaf \(\mathcal{F}\) is flat over \(S\).

For every affine opens \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the \(\mathcal{O}_S(V)\)-module \(\mathcal{F}(U)\) is flat.

There exists an open covering \(S = \bigcup_{j \in J} V_j\) and open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that each of the modules \(\mathcal{F}|_{U_i}\) is flat over \(V_j\), for all \(j\in J, i\in I_j\).

There exists an affine open covering \(S = \bigcup_{j \in J} V_j\) and affine open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that \(\mathcal{F}(U_i)\) is a flat \(\mathcal{O}_S(V_j)\)-module, for all \(j\in J, i\in I_j\).

Moreover, if \(\mathcal{F}\) is flat over \(S\) then for any open subschemes \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the restriction \(\mathcal{F}|_U\) is flat over \(V\).

Proof

Let \(R \to A\) be a ring map. Let \(M\) be an \(A\)-module. If \(M\) is \(R\)-flat, then for all primes \(\mathfrak q\) the module \(M_{\mathfrak q}\) is flat over \(R_{\mathfrak p}\) with \(\mathfrak p\) the prime of \(R\) lying under \(\mathfrak q\). Conversely, if \(M_{\mathfrak q}\) is flat over \(R_{\mathfrak p}\) for all primes \(\mathfrak q\) of \(A\), then \(M\) is flat over \(R\). See Algebra, Lemma 00HT. This equivalence easily implies the statements of the lemma.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent

The morphism \(f\) is flat.

For every affine opens \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the ring map \(\mathcal{O}_S(V) \to \mathcal{O}_X(U)\) is flat.

There exists an open covering \(S = \bigcup_{j \in J} V_j\) and open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that each of the morphisms \(U_i \to V_j\), \(j\in J, i\in I_j\) is flat.

There exists an affine open covering \(S = \bigcup_{j \in J} V_j\) and affine open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that \(\mathcal{O}_S(V_j) \to \mathcal{O}_X(U_i)\) is flat, for all \(j\in J, i\in I_j\).

Moreover, if \(f\) is flat then for any open subschemes \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the restriction \(f|_U : U \to V\) is flat.

Proof

This is a special case of Lemma 01U4 above.

Lemma

Let \(f : X \to Y\) be an affine morphism of schemes over a base scheme \(S\). Let \(\mathcal{F}\) be a quasi-coherent \(\mathcal{O}_X\)-module. Then \(\mathcal{F}\) is flat over \(S\) if and only if \(f_*\mathcal{F}\) is flat over \(S\).

Proof

By Lemma 01U4 and the fact that \(f\) is an affine morphism, this reduces us to the affine case. Say \(X \to Y \to S\) corresponds to the ring maps \(C \leftarrow B \leftarrow A\). Let \(N\) be the \(C\)-module corresponding to \(\mathcal{F}\). Recall that \(f_*\mathcal{F}\) corresponds to \(N\) viewed as a \(B\)-module, see Schemes, Lemma 01I9. Thus the result is clear.

Lemma

Let \(X \to Y \to Z\) be morphisms of schemes. Let \(\mathcal{F}\) be a quasi-coherent \(\mathcal{O}_X\)-module. Let \(x \in X\) with image \(y\) in \(Y\). If \(\mathcal{F}\) is flat over \(Y\) at \(x\), and \(Y\) is flat over \(Z\) at \(y\), then \(\mathcal{F}\) is flat over \(Z\) at \(x\).

Proof

See Algebra, Lemma 00HC.

Lemma

The composition of flat morphisms is flat.

Proof

This is a special case of Lemma 01U6.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{F}\) be a quasi-coherent sheaf of \(\mathcal{O}_X\)-modules. Let \(g : S' \to S\) be a morphism of schemes. Denote \(g' : X' = X_{S'} \to X\) the projection. Let \(x' \in X'\) be a point with image \(x = g'(x') \in X\). If \(\mathcal{F}\) is flat over \(S\) at \(x\), then \((g')^*\mathcal{F}\) is flat over \(S'\) at \(x'\). In particular, if \(\mathcal{F}\) is flat over \(S\), then \((g')^*\mathcal{F}\) is flat over \(S'\).

Proof

See Algebra, Lemma 00HI.

Lemma

The base change of a flat morphism is flat.

Proof

This is a special case of Lemma 01U8.

Lemma

Let \(f : X \to S\) be a flat morphism of schemes. Then generalizations lift along \(f\), see Topology, Definition 0063.

Proof

See Algebra, Section 00HU.

Lemma

A flat morphism locally of finite presentation is universally open.

Proof

This follows from Lemmas 03HV and Lemma 01U1 above. We can also argue directly as follows.

Let \(f : X \to S\) be flat and locally of finite presentation. By Lemmas 01U9 and 01TS any base change of \(f\) is flat and locally of finite presentation. Hence it suffices to show \(f\) is open. To show \(f\) is open it suffices to show that we may cover \(X\) by open affines \(X = \bigcup U_i\) such that \(U_i \to S\) is open. We may cover \(X\) by affine opens \(U_i \subset X\) such that each \(U_i\) maps into an affine open \(V_i \subset S\) and such that the induced ring map \(\mathcal{O}_S(V_i) \to \mathcal{O}_X(U_i)\) is flat and of finite presentation (Lemmas 01U5 and 01TQ). Then \(U_i \to V_i\) is open by Algebra, Proposition 00I1 and the proof is complete.

Lemma

Let \(f : X \to Y\) be a morphism of schemes. Let \(\mathcal{F}\) be a quasi-coherent \(\mathcal{O}_X\)-module. Assume \(f\) locally finite presentation, \(\mathcal{F}\) of finite type, \(X = \text{Supp}(\mathcal{F})\), and \(\mathcal{F}\) flat over \(Y\). Then \(f\) is universally open.

Proof

By Lemmas 01U8, 01TS, and 056J the assumptions are preserved under base change. By Lemma 01U1 it suffices to show that generalizations lift along \(f\). This follows from Algebra, Lemma 080T.

Lemma

Let \(f : X \to Y\) be a quasi-compact, surjective, flat morphism. A subset \(T \subset Y\) is open (resp. closed) if and only \(f^{-1}(T)\) is open (resp. closed). In other words, \(f\) is a submersive morphism.

Proof

The question is local on \(Y\), hence we may assume that \(Y\) is affine. In this case \(X\) is quasi-compact as \(f\) is quasi-compact. Write \(X = X_1 \cup \ldots \cup X_n\) as a finite union of affine opens. Then \(f' : X' = X_1 \amalg \ldots \amalg X_n \to Y\) is a surjective flat morphism of affine schemes. Note that for \(T \subset Y\) we have \((f')^{-1}(T) = f^{-1}(T) \cap X_1 \amalg \ldots \amalg f^{-1}(T) \cap X_n\). Hence, \(f^{-1}(T)\) is open if and only if \((f')^{-1}(T)\) is open. Thus we may assume both \(X\) and \(Y\) are affine.

Let \(f : \Spec(B) \to \Spec(A)\) be a surjective morphism of affine schemes corresponding to a flat ring map \(A \to B\). Suppose that \(f^{-1}(T)\) is closed, say \(f^{-1}(T) = V(J)\) for \(J \subset B\) an ideal. Then \(T = f(f^{-1}(T)) = f(V(J))\) is the image of \(\Spec(B/J) \to \Spec(A)\) (here we use that \(f\) is surjective). On the other hand, generalizations lift along \(f\) (Lemma 03HV). Hence by Topology, Lemma 0065 we see that \(Y \setminus T = f(X \setminus f^{-1}(T))\) is stable under generalization. Hence \(T\) is stable under specialization (Topology, Lemma 0062). Thus \(T\) is closed by Algebra, Lemma 00HY.

Lemma

Let \(h : X \to Y\) be a morphism of schemes over \(S\). Let \(\mathcal{G}\) be a quasi-coherent sheaf on \(Y\). Let \(x \in X\) with \(y = h(x) \in Y\). If \(h\) is flat at \(x\), then \[\mathcal{G}\text{ flat over }S\text{ at }y \Leftrightarrow h^*\mathcal{G}\text{ flat over }S\text{ at }x.\] In particular: If \(h\) is surjective and flat, then \(\mathcal{G}\) is flat over \(S\), if and only if \(h^*\mathcal{G}\) is flat over \(S\). If \(h\) is surjective and flat, and \(X\) is flat over \(S\), then \(Y\) is flat over \(S\).

Proof

You can prove this by applying Algebra, Lemma 0584. Here is a direct proof. Let \(s \in S\) be the image of \(y\). Consider the local ring maps \(\mathcal{O}_{S, s} \to \mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}\). By assumption the ring map \(\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}\) is faithfully flat, see Algebra, Lemma 00HR. Let \(N = \mathcal{G}_y\). Note that \(h^*\mathcal{G}_x = N \otimes_{\mathcal{O}_{Y, y}} \mathcal{O}_{X, x}\), see Sheaves, Lemma 0098. Let \(M' \to M\) be an injection of \(\mathcal{O}_{S, s}\)-modules. By the faithful flatness mentioned above we have \[\begin{align*} \Ker( M' \otimes_{\mathcal{O}_{S, s}} N \to M \otimes_{\mathcal{O}_{S, s}} N) \otimes_{\mathcal{O}_{Y, y}} \mathcal{O}_{X, x} \\ = \Ker( M' \otimes_{\mathcal{O}_{S, s}} N \otimes_{\mathcal{O}_{Y, y}} \mathcal{O}_{X, x} \to M \otimes_{\mathcal{O}_{S, s}} N \otimes_{\mathcal{O}_{Y, y}} \mathcal{O}_{X, x}) \end{align*}\] Hence the equivalence of the lemma follows from the second characterization of flatness in Algebra, Lemma 00HD.

Lemma

Let \(f : Y \to X\) be a morphism of schemes. Let \(\mathcal{F}\) be a finite type quasi-coherent \(\mathcal{O}_X\)-module with scheme theoretic support \(Z \subset X\). If \(f\) is flat, then \(f^{-1}(Z)\) is the scheme theoretic support of \(f^*\mathcal{F}\).

Proof

Using the characterization of scheme theoretic support on affines as given in Lemma 05JU we reduce to Algebra, Lemma 07T8.

Lemma

Let \(f : X \to Y\) be a flat morphism of schemes. Let \(V \subset Y\) be a retrocompact open which is scheme theoretically dense. Then \(f^{-1}V\) is scheme theoretically dense in \(X\).

Proof

We will use the characterization of Lemma 01RE. We have to show that for any open \(U \subset X\) the map \(\mathcal{O}_X(U) \to \mathcal{O}_X(U \cap f^{-1}V)\) is injective. It suffices to prove this when \(U\) is an affine open which maps into an affine open \(W \subset Y\). Say \(W = \Spec(A)\) and \(U = \Spec(B)\). Then \(V \cap W = D(f_1) \cup \ldots \cup D(f_n)\) for some \(f_i \in A\), see Algebra, Lemma 00F6. Thus we have to show that \(B \to B_{f_1} \times \ldots \times B_{f_n}\) is injective. We are given that \(A \to A_{f_1} \times \ldots \times A_{f_n}\) is injective and that \(A \to B\) is flat. Since \(B_{f_i} = A_{f_i} \otimes_A B\) we win.

Lemma

Let \(f : X \to Y\) be a flat morphism of schemes. Let \(g : V \to Y\) be a quasi-compact morphism of schemes. Let \(Z \subset Y\) be the scheme theoretic image of \(g\) and let \(Z' \subset X\) be the scheme theoretic image of the base change \(V \times_Y X \to X\). Then \(Z' = f^{-1}Z\).

Proof

Recall that \(Z\) is cut out by \(\mathcal{I} = \Ker(\mathcal{O}_Y \to g_*\mathcal{O}_V)\) and \(Z'\) is cut out by \(\mathcal{I}' = \Ker(\mathcal{O}_X \to (V \times_Y X \to X)_*\mathcal{O}_{V \times_Y X})\), see Lemma 01R8. Hence the question is local on \(X\) and \(Y\) and we may assume \(X\) and \(Y\) affine. Note that we may replace \(V\) by \(\coprod V_i\) where \(V = V_1 \cup \ldots \cup V_n\) is a finite affine open covering. Hence we may assume \(g\) is affine. In this case \((V \times_Y X \to X)_*\mathcal{O}_{V \times_Y X}\) is the pullback of \(g_*\mathcal{O}_V\) by \(f\). Since \(f\) is flat we conclude that \(f^*\mathcal{I} = \mathcal{I}'\) and the lemma holds.

Flat closed immersions

Connected components of schemes are not always open. But they do always have a canonical scheme structure. We explain this in this section.

Lemma

Let \(X\) be a scheme. The rule which associates to a closed subscheme of \(X\) its underlying closed subset defines a bijection \[\left\{ \begin{matrix} \text{closed subschemes }Z \subset X \\ \text{such that }Z \to X\text{ is flat} \end{matrix} \right\} \leftrightarrow \left\{ \begin{matrix} \text{closed subsets }Z \subset X \\ \text{closed under generalizations} \end{matrix} \right\}\] If \(Z \subset X\) is such a closed subscheme, every morphism of schemes \(g : Y \to X\) with \(g(Y) \subset Z\) set theoretically factors (scheme theoretically) through \(Z\).

Proof

The affine case of the bijection is Algebra, Lemma 04PU. For general schemes \(X\) the bijection follows by covering \(X\) by affines and glueing. Details omitted. For the final assertion, observe that the projection \(Z \times_{X, g} Y \to Y\) is a flat (Lemma 01U9) closed immersion which is bijective on underlying topological spaces and hence must be an isomorphism by the bijection established in the first part of the proof.

Lemma

A flat closed immersion of finite presentation is the open immersion of an open and closed subscheme.

Proof

The affine case is Algebra, Lemma 05KK. In general the lemma follows by covering \(X\) by affines. Details omitted.

Note that a connected component \(T\) of a scheme \(X\) is a closed subset stable under generalization. Hence the following definition makes sense.

Definition

Let \(X\) be a scheme. Let \(T \subset X\) be a connected component. The canonical scheme structure on \(T\) is the unique scheme structure on \(T\) such that the closed immersion \(T \to X\) is flat, see Lemma 04PW.

It turns out that we can determine when every finite flat \(\mathcal{O}_X\)-module is finite locally free using the previous lemma.

Lemma

Let \(X\) be a scheme. The following are equivalent

every finite flat quasi-coherent \(\mathcal{O}_X\)-module is finite locally free, and

every closed subset \(Z \subset X\) which is closed under generalizations is open.

Proof

In the affine case this is Algebra, Lemma 052U. The scheme case does not follow directly from the affine case, so we simply repeat the arguments.

Assume (1). Consider a closed immersion \(i : Z \to X\) such that \(i\) is flat. Then \(i_*\mathcal{O}_Z\) is quasi-coherent and flat, hence finite locally free by (1). Thus \(Z = \text{Supp}(i_*\mathcal{O}_Z)\) is also open and we see that (2) holds. Hence the implication (1) \(\Rightarrow\) (2) follows from the characterization of flat closed immersions in Lemma 04PW.

For the converse assume that \(X\) satisfies (2). Let \(\mathcal{F}\) be a finite flat quasi-coherent \(\mathcal{O}_X\)-module. The support \(Z = \text{Supp}(\mathcal{F})\) of \(\mathcal{F}\) is closed, see Modules, Lemma 01BA. On the other hand, if \(x \leadsto x'\) is a specialization, then by Algebra, Lemma 00NZ the module \(\mathcal{F}_{x'}\) is free over \(\mathcal{O}_{X, x'}\), and \[\mathcal{F}_x = \mathcal{F}_{x'} \otimes_{\mathcal{O}_{X, x'}} \mathcal{O}_{X, x}.\] Hence \(x' \in \text{Supp}(\mathcal{F}) \Rightarrow x \in \text{Supp}(\mathcal{F})\), in other words, the support is closed under generalization. As \(X\) satisfies (2) we see that the support of \(\mathcal{F}\) is open and closed. The modules \(\wedge^i(\mathcal{F})\), \(i = 1, 2, 3, \ldots\) are finite flat quasi-coherent \(\mathcal{O}_X\)-modules also, see Modules, Section 01CF. Note that \(\text{Supp}(\wedge^{i + 1}(\mathcal{F})) \subset \text{Supp}(\wedge^i(\mathcal{F}))\). Thus we see that there exists a decomposition \[X = U_0 \amalg U_1 \amalg U_2 \amalg \ldots\] by open and closed subsets such that the support of \(\wedge^i(\mathcal{F})\) is \(U_i \cup U_{i + 1} \cup \ldots\) for all \(i\). Let \(x\) be a point of \(X\), and say \(x \in U_r\). Note that \(\wedge^i(\mathcal{F})_x \otimes \kappa(x) = \wedge^i(\mathcal{F}_x \otimes \kappa(x))\). Hence, \(x \in U_r\) implies that \(\mathcal{F}_x \otimes \kappa(x)\) is a vector space of dimension \(r\). By Nakayama’s lemma, see Algebra, Lemma 00DV we can choose an affine open neighbourhood \(U \subset U_r \subset X\) of \(x\) and sections \(s_1, \ldots, s_r \in \mathcal{F}(U)\) such that the induced map \[\mathcal{O}_U^{\oplus r} \longrightarrow \mathcal{F}|_U, \quad (f_1, \ldots, f_r) \longmapsto \sum f_i s_i\] is surjective. This means that \(\wedge^r(\mathcal{F}|_U)\) is a finite flat quasi-coherent \(\mathcal{O}_U\)-module whose support is all of \(U\). By the above it is generated by a single element, namely \(s_1 \wedge \ldots \wedge s_r\). Hence \(\wedge^r(\mathcal{F}|_U) \cong \mathcal{O}_U/\mathcal{I}\) for some quasi-coherent sheaf of ideals \(\mathcal{I}\) such that \(\mathcal{O}_U/\mathcal{I}\) is flat over \(\mathcal{O}_U\) and such that \(V(\mathcal{I}) = U\). It follows that \(\mathcal{I} = 0\) by applying Lemma 04PW. Thus \(s_1 \wedge \ldots \wedge s_r\) is a basis for \(\wedge^r(\mathcal{F}|_U)\) and it follows that the displayed map is injective as well as surjective. This proves that \(\mathcal{F}\) is finite locally free as desired.

Generic flatness

A scheme of finite type over an integral base is flat over a dense open of the base. In Algebra, Section 051Q we proved a Noetherian version, a version for morphisms of finite presentation, and a general version. We only state and prove the general version here. However, it turns out that this will be superseded by Proposition 052B which shows the result holds if we only assume the base is reduced.

Proposition

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{F}\) be a quasi-coherent sheaf of \(\mathcal{O}_X\)-modules. Assume

\(S\) is integral,

\(f\) is of finite type, and

\(\mathcal{F}\) is a finite type \(\mathcal{O}_X\)-module.

Then there exists an open dense subscheme \(U \subset S\) such that \(X_U \to U\) is flat and of finite presentation and such that \(\mathcal{F}|_{X_U}\) is flat over \(U\) and of finite presentation over \(\mathcal{O}_{X_U}\).

Proof

As \(S\) is integral it is irreducible (see Properties, Lemma 01ON) and any nonempty open is dense. Hence we may replace \(S\) by an affine open of \(S\) and assume that \(S = \Spec(A)\) is affine. As \(S\) is integral we see that \(A\) is a domain. As \(f\) is of finite type, it is quasi-compact, so \(X\) is quasi-compact. Hence we can find a finite affine open cover \(X = \bigcup_{i = 1, \ldots, n} X_i\). Write \(X_i = \Spec(B_i)\). Then \(B_i\) is a finite type \(A\)-algebra, see Lemma 01T2. Moreover there are finite type \(B_i\)-modules \(M_i\) such that \(\mathcal{F}|_{X_i}\) is the quasi-coherent sheaf associated to the \(B_i\)-module \(M_i\), see Properties, Lemma 01PB. Next, for each pair of indices \(i, j\) choose an ideal \(I_{ij} \subset B_i\) such that \(X_i \setminus X_i \cap X_j = V(I_{ij})\) inside \(X_i = \Spec(B_i)\). Set \(M_{ij} = B_i/I_{ij}\) and think of it as a \(B_i\)-module. Then \(V(I_{ij}) = \text{Supp}(M_{ij})\) and \(M_{ij}\) is a finite \(B_i\)-module.

At this point we apply Algebra, Lemma 051T the pairs \((A \to B_i, M_{ij})\) and to the pairs \((A \to B_i, M_i)\). Thus we obtain nonzero \(f_{ij}, f_i \in A\) such that (a) \(A_{f_{ij}} \to B_{i, f_{ij}}\) is flat and of finite presentation and \(M_{ij, f_{ij}}\) is flat over \(A_{f_{ij}}\) and of finite presentation over \(B_{i, f_{ij}}\), and (b) \(B_{i, f_i}\) is flat and of finite presentation over \(A_f\) and \(M_{i, f_i}\) is flat and of finite presentation over \(B_{i, f_i}\). Set \(f = (\prod f_i) (\prod f_{ij})\). We claim that taking \(U = D(f)\) works.

To prove our claim we may replace \(A\) by \(A_f\), i.e., perform the base change by \(U = \Spec(A_f) \to S\). After this base change we see that each of \(A \to B_i\) is flat and of finite presentation and that \(M_i\), \(M_{ij}\) are flat over \(A\) and of finite presentation over \(B_i\). This already proves that \(X \to S\) is quasi-compact, locally of finite presentation, flat, and that \(\mathcal{F}\) is flat over \(S\) and of finite presentation over \(\mathcal{O}_X\), see Lemma 01TQ and Properties, Lemma 01PC. Since \(M_{ij}\) is of finite presentation over \(B_i\) we see that \(X_i \cap X_j = X_i \setminus \text{Supp}(M_{ij})\) is a quasi-compact open of \(X_i\), see Algebra, Lemma 051B. Hence we see that \(X \to S\) is quasi-separated by Schemes, Lemma 01KO. This proves the proposition.

It actually turns out that there is also a version of generic flatness over an arbitrary reduced base. Here it is.

Proposition

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{F}\) be a quasi-coherent sheaf of \(\mathcal{O}_X\)-modules. Assume

\(S\) is reduced,

\(f\) is of finite type, and

\(\mathcal{F}\) is a finite type \(\mathcal{O}_X\)-module.

Then there exists an open dense subscheme \(U \subset S\) such that \(X_U \to U\) is flat and of finite presentation and such that \(\mathcal{F}|_{X_U}\) is flat over \(U\) and of finite presentation over \(\mathcal{O}_{X_U}\).

Proof

For the impatient reader: This proof is a repeat of the proof of Proposition 052A using Algebra, Lemma 051Z instead of Algebra, Lemma 051T.

Since being flat and being of finite presentation is local on the base, see Lemmas 01U4 and 01TQ, we may work affine locally on \(S\). Thus we may assume that \(S = \Spec(A)\), where \(A\) is a reduced ring (see Properties, Lemma 01OL). As \(f\) is of finite type, it is quasi-compact, so \(X\) is quasi-compact. Hence we can find a finite affine open cover \(X = \bigcup_{i = 1, \ldots, n} X_i\). Write \(X_i = \Spec(B_i)\). Then \(B_i\) is a finite type \(A\)-algebra, see Lemma 01T2. Moreover there are finite type \(B_i\)-modules \(M_i\) such that \(\mathcal{F}|_{X_i}\) is the quasi-coherent sheaf associated to the \(B_i\)-module \(M_i\), see Properties, Lemma 01PB. Next, for each pair of indices \(i, j\) choose an ideal \(I_{ij} \subset B_i\) such that \(X_i \setminus X_i \cap X_j = V(I_{ij})\) inside \(X_i = \Spec(B_i)\). Set \(M_{ij} = B_i/I_{ij}\) and think of it as a \(B_i\)-module. Then \(V(I_{ij}) = \text{Supp}(M_{ij})\) and \(M_{ij}\) is a finite \(B_i\)-module.

At this point we apply Algebra, Lemma 051Z to the pairs \((A \to B_i, M_{ij})\) and to the pairs \((A \to B_i, M_i)\). Thus we obtain dense opens \(U(A \to B_i, M_{ij}) \subset S\) and dense opens \(U(A \to B_i, M_i) \subset S\) with notation as in Algebra, Equation (051V). Since a finite intersection of dense opens is dense open, we see that \[U = \bigcap\nolimits_{i, j} U(A \to B_i, M_{ij}) \quad\cap\quad \bigcap\nolimits_i U(A \to B_i, M_i)\] is open and dense in \(S\). We claim that \(U\) is the desired open.

Pick \(u \in U\). By definition of the loci \(U(A \to B_i, M_{ij})\) and \(U(A \to B, M_i)\) there exist \(f_{ij}, f_i \in A\) such that (a) \(u \in D(f_i)\) and \(u \in D(f_{ij})\), (b) \(A_{f_{ij}} \to B_{i, f_{ij}}\) is flat and of finite presentation and \(M_{ij, f_{ij}}\) is flat over \(A_{f_{ij}}\) and of finite presentation over \(B_{i, f_{ij}}\), and (c) \(B_{i, f_i}\) is flat and of finite presentation over \(A_{f_i}\) and \(M_{i, f_i}\) is of finite presentation over \(B_{i, f_i}\) and flat over \(A_{f_i}\). Set \(f = (\prod f_i) (\prod f_{ij})\). Now it suffices to prove that \(X \to S\) is flat and of finite presentation over \(D(f)\) and that \(\mathcal{F}\) restricted to \(X_{D(f)}\) is flat over \(D(f)\) and of finite presentation over the structure sheaf of \(X_{D(f)}\).

Hence we may replace \(A\) by \(A_f\), i.e., perform the base change by \(\Spec(A_f) \to S\). After this base change we see that each of \(A \to B_i\) is flat and of finite presentation and that \(M_i\), \(M_{ij}\) are flat over \(A\) and of finite presentation over \(B_i\). This already proves that \(X \to S\) is quasi-compact, locally of finite presentation, flat, and that \(\mathcal{F}\) is flat over \(S\) and of finite presentation over \(\mathcal{O}_X\), see Lemma 01TQ and Properties, Lemma 01PC. Since \(M_{ij}\) is of finite presentation over \(B_i\) we see that \(X_i \cap X_j = X_i \setminus \text{Supp}(M_{ij})\) is a quasi-compact open of \(X_i\), see Algebra, Lemma 051B. Hence we see that \(X \to S\) is quasi-separated by Schemes, Lemma 01KO. This proves the proposition.

Remark

The results above are a first step towards more refined flattening techniques for morphisms of schemes. The article [GruRay] by Raynaud and Gruson contains many wonderful results in this direction.

Morphisms and dimensions of fibres

Let \(X\) be a topological space, and \(x \in X\). Recall that we have defined \(\dim_x(X)\) as the minimum of the dimensions of the open neighbourhoods of \(x\) in \(X\). See Topology, Definition 0055.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(x \in X\) and set \(s = f(x)\). Assume \(f\) is locally of finite type. Then \[\dim_x(X_s) = \dim(\mathcal{O}_{X_s, x}) + \text{trdeg}_{\kappa(s)}(\kappa(x)).\]

Proof

This immediately reduces to the case \(S = s\), and \(X\) affine. In this case the result follows from Algebra, Lemma 00P1.

Lemma

Let \(f : X \to Y\) and \(g : Y \to S\) be morphisms of schemes. Let \(x \in X\) and set \(y = f(x)\), \(s = g(y)\). Assume \(f\) and \(g\) locally of finite type. Then \[\dim_x(X_s) \leq \dim_x(X_y) + \dim_y(Y_s).\] Moreover, equality holds if \(\mathcal{O}_{X_s, x}\) is flat over \(\mathcal{O}_{Y_s, y}\), which holds for example if \(\mathcal{O}_{X, x}\) is flat over \(\mathcal{O}_{Y, y}\).

Proof

Note that \(\text{trdeg}_{\kappa(s)}(\kappa(x)) = \text{trdeg}_{\kappa(y)}(\kappa(x)) + \text{trdeg}_{\kappa(s)}(\kappa(y))\). Thus by Lemma 02FX the statement is equivalent to \[\dim(\mathcal{O}_{X_s, x}) \leq \dim(\mathcal{O}_{X_y, x}) + \dim(\mathcal{O}_{Y_s, y}).\] For this see Algebra, Lemma 00OM. For the flat case see Algebra, Lemma 00ON.

Lemma

Let \[\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^f \\ S' \ar[r]^g & S }\] be a fibre product diagram of schemes. Assume \(f\) locally of finite type. Suppose that \(x' \in X'\), \(x = g'(x')\), \(s' = f'(x')\) and \(s = g(s') = f(x)\). Then

\(\dim_x(X_s) = \dim_{x'}(X'_{s'})\),

if \(F\) is the fibre of the morphism \(X'_{s'} \to X_s\) over \(x\), then \[\dim(\mathcal{O}_{F, x'}) = \dim(\mathcal{O}_{X'_{s'}, x'}) - \dim(\mathcal{O}_{X_s, x}) = \text{trdeg}_{\kappa(s)}(\kappa(x)) - \text{trdeg}_{\kappa(s')}(\kappa(x'))\] In particular \(\dim(\mathcal{O}_{X'_{s'}, x'}) \geq \dim(\mathcal{O}_{X_s, x})\) and \(\text{trdeg}_{\kappa(s)}(\kappa(x)) \geq \text{trdeg}_{\kappa(s')}(\kappa(x'))\).

given \(s', s, x\) there exists a choice of \(x'\) such that \(\dim(\mathcal{O}_{X'_{s'}, x'}) = \dim(\mathcal{O}_{X_s, x})\) and \(\text{trdeg}_{\kappa(s)}(\kappa(x)) = \text{trdeg}_{\kappa(s')}(\kappa(x'))\).

Proof

Part (1) follows immediately from Algebra, Lemma 00P4. Parts (2) and (3) from Algebra, Lemma 0CWE.

The following lemma follows from a nontrivial algebraic result. Namely, the algebraic version of Zariski’s main theorem.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(n \geq 0\). Assume \(f\) is locally of finite type. The set \[U_n = \{x \in X \mid \dim_x X_{f(x)} \leq n\}\] is open in \(X\).

Proof

This is immediate from Algebra, Lemma 00QH

Lemma

Let \(f : X \to Y\) be a morphism of finite type with \(Y\) quasi-compact. Then the dimension of the fibres of \(f\) is bounded.

Proof

By Lemma 02FZ the set \(U_n \subset X\) of points where the dimension of the fibre is \(\leq n\) is open. Since \(f\) is of finite type, every point is contained in some \(U_n\) (because the dimension of a finite type algebra over a field is finite). Since \(Y\) is quasi-compact and \(f\) is of finite type, we see that \(X\) is quasi-compact. Hence \(X = U_n\) for some \(n\).

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(n \geq 0\). Assume \(f\) is locally of finite presentation. The open \[U_n = \{x \in X \mid \dim_x X_{f(x)} \leq n\}\] of Lemma 02FZ is retrocompact in \(X\). (See Topology, Definition 005A.)

Proof

The topological space \(X\) has a basis for its topology consisting of affine opens \(U \subset X\) such that the induced morphism \(f|_U : U \to S\) factors through an affine open \(V \subset S\). Hence it is enough to show that \(U \cap U_n\) is quasi-compact for such a \(U\). Note that \(U_n \cap U\) is the same as the open \(\{x \in U \mid \dim_x U_{f(x)} \leq n\}\). This reduces us to the case where \(X\) and \(S\) are affine. In this case the lemma follows from Algebra, Lemma 00QJ (and Lemma 01TQ).

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(x \leadsto x'\) be a nontrivial specialization of points in \(X\) lying over the same point \(s \in S\). Assume \(f\) is locally of finite type. Then

\(\dim_x(X_s) \leq \dim_{x'}(X_s)\),

\(\dim(\mathcal{O}_{X_s, x}) < \dim(\mathcal{O}_{X_s, x'})\), and

\(\text{trdeg}_{\kappa(s)}(\kappa(x)) > \text{trdeg}_{\kappa(s)}(\kappa(x'))\).

Proof

Part (1) follows from the fact that any open of \(X_s\) containing \(x'\) also contains \(x\). Part (2) follows since \(\mathcal{O}_{X_s, x}\) is a localization of \(\mathcal{O}_{X_s, x'}\) at a prime ideal, hence any chain of prime ideals in \(\mathcal{O}_{X_s, x}\) is part of a strictly longer chain of primes in \(\mathcal{O}_{X_s, x'}\). The last inequality follows from Algebra, Lemma 06RP.

Morphisms of given relative dimension

In order to be able to speak comfortably about morphisms of a given relative dimension we introduce the following notion.

Definition

Let \(f : X \to S\) be a morphism of schemes. Assume \(f\) is locally of finite type.

We say \(f\) is of relative dimension \(\leq d\) at \(x\) if \(\dim_x(X_{f(x)}) \leq d\).

We say \(f\) is of relative dimension \(\leq d\) if \(\dim_x(X_{f(x)}) \leq d\) for all \(x \in X\).

We say \(f\) is of relative dimension \(d\) if all nonempty fibres \(X_s\) are equidimensional of dimension \(d\).

This is not a particularly well behaved notion, but it works well in a number of situations.

Lemma

Let \(f : X \to S\) be a morphism of schemes which is locally of finite type. If \(f\) has relative dimension \(d\), then so does any base change of \(f\). Same for relative dimension \(\leq d\).

Proof

This is immediate from Lemma 02FY.

Lemma

Let \(f : X \to Y\), \(g : Y \to Z\) be locally of finite type. If \(f\) has relative dimension \(\leq d\) and \(g\) has relative dimension \(\leq e\) then \(g \circ f\) has relative dimension \(\leq d + e\). If

\(f\) has relative dimension \(d\),

\(g\) has relative dimension \(e\), and

\(f\) is flat,

then \(g \circ f\) has relative dimension \(d + e\).

Proof

This is immediate from Lemma 02JS.

In general it is not possible to decompose a morphism into its pieces where the relative dimension is a given one. However, it is possible if the morphism has Cohen-Macaulay fibres and is flat of finite presentation.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Assume that

\(f\) is flat,

\(f\) is locally of finite presentation, and

for all \(s \in S\) the fibre \(X_s\) is Cohen-Macaulay (Properties, Definition 02IO)

Then there exist open and closed subschemes \(X_d \subset X\) such that \(X = \coprod_{d \geq 0} X_d\) and \(f|_{X_d} : X_d \to S\) has relative dimension \(d\).

Proof

This is immediate from Algebra, Lemma 00RL.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Assume \(f\) is locally of finite type. Let \(x \in X\) with \(s = f(x)\). Then \(f\) is quasi-finite at \(x\) if and only if \(\dim_x(X_s) = 0\). In particular, \(f\) is locally quasi-finite if and only if \(f\) has relative dimension \(0\).

Proof

First proof. If \(f\) is quasi-finite at \(x\) then \(\kappa(x)\) is a finite extension of \(\kappa(s)\) (by Lemma 01TG) and \(x\) is isolated in \(X_s\) (by Lemma 01TH), hence \(\dim_x(X_s) = 0\) by Lemma 02FX. Conversely, if \(\dim_x(X_s) = 0\) then by Lemma 02FX we see \(\kappa(s) \subset \kappa(x)\) is algebraic and there are no other points of \(X_s\) specializing to \(x\). Hence \(x\) is closed in its fibre by Lemma 01TE and by Lemma 01TH (3) we conclude that \(f\) is quasi-finite at \(x\).

Second proof. The fibre \(X_s\) is a scheme locally of finite type over a field, hence locally Noetherian (Lemma 01T6). The result now follows from Lemma 01TH and Properties, Lemma 0H7C.

Lemma

Let \(f : X \to Y\) be a morphism of locally Noetherian schemes which is flat, locally of finite type and of relative dimension \(d\). For every point \(x\) in \(X\) with image \(y\) in \(Y\) we have \(\dim_x(X) = \dim_y(Y) + d\).

Proof

After shrinking \(X\) and \(Y\) to open neighborhoods of \(x\) and \(y\), we can assume that \(\dim(X) = \dim_x(X)\) and \(\dim(Y) = \dim_y(Y)\), by definition of the dimension of a scheme at a point (Properties, Definition 04MT). The morphism \(f\) is open by Lemmas 01TX and 01UA. Hence we can shrink \(Y\) to arrange that \(f\) is surjective. It remains to show that \(\dim(X) = \dim(Y) + d\).

Let \(a\) be a point in \(X\) with image \(b\) in \(Y\). By Algebra, Lemma 00ON, \[\dim(\mathcal{O}_{X,a}) = \dim(\mathcal{O}_{Y,b}) + \dim(\mathcal{O}_{X_b, a}).\] Taking the supremum over all points \(a\) in \(X\), it follows that \(\dim(X) = \dim(Y) + d\), as we want, see Properties, Lemma 04MU.

Syntomic morphisms

An algebra \(A\) over a field \(k\) is called a global complete intersection over \(k\) if \(A \cong k[x_1, \ldots, x_n]/(f_1, \ldots, f_c)\) and \(\dim(A) = n - c\). An algebra \(A\) over a field \(k\) is called a local complete intersection if \(\Spec(A)\) can be covered by standard opens each of which are global complete intersections over \(k\). See Algebra, Section 00S8. Recall that a ring map \(R \to A\) is syntomic if it is of finite presentation, flat with local complete intersection rings as fibres, see Algebra, Definition 00SL.

Definition

Let \(f : X \to S\) be a morphism of schemes.

We say that \(f\) is syntomic at \(x \in X\) if there exists an affine open neighbourhood \(\Spec(A) = U \subset X\) of \(x\) and affine open \(\Spec(R) = V \subset S\) with \(f(U) \subset V\) such that the induced ring map \(R \to A\) is syntomic.

We say that \(f\) is syntomic if it is syntomic at every point of \(X\).

If \(S = \Spec(k)\) and \(f\) is syntomic, then we say that \(X\) is a local complete intersection over \(k\).

A morphism of affine schemes \(f : X \to S\) is called standard syntomic if there exists a global relative complete intersection \(R \to R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)\) (see Algebra, Definition 00SP) such that \(X \to S\) is isomorphic to \[\Spec(R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)) \to \Spec(R).\]

In the literature a syntomic morphism is sometimes referred to as a flat local complete intersection morphism. It turns out this is a convenient class of morphisms. For example one can define a syntomic topology using these, which is finer than the smooth and étale topologies, but has many of the same formal properties.

A global relative complete intersection (which we used to define standard syntomic ring maps) is in particular flat. In More on Morphisms, Section 068E we will consider morphisms \(X \to S\) which locally are of the form \[\Spec(R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)) \to \Spec(R).\] for some Koszul-regular sequence \(f_1, \ldots, f_r\) in \(R[x_1, \ldots, x_n]\). Such a morphism will be called a local complete intersection morphism. Once we have this definition in place it will be the case that a morphism is syntomic if and only if it is a flat, local complete intersection morphism.

Note that there is no separation or quasi-compactness hypotheses in the definition of a syntomic morphism. Hence the question of being syntomic is local in nature on the source. Here is the precise result.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent

The morphism \(f\) is syntomic.

For every affine opens \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the ring map \(\mathcal{O}_S(V) \to \mathcal{O}_X(U)\) is syntomic.

There exists an open covering \(S = \bigcup_{j \in J} V_j\) and open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that each of the morphisms \(U_i \to V_j\), \(j\in J, i\in I_j\) is syntomic.

There exists an affine open covering \(S = \bigcup_{j \in J} V_j\) and affine open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that the ring map \(\mathcal{O}_S(V_j) \to \mathcal{O}_X(U_i)\) is syntomic, for all \(j\in J, i\in I_j\).

Moreover, if \(f\) is syntomic then for any open subschemes \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the restriction \(f|_U : U \to V\) is syntomic.

Proof

This follows from Lemma 01ST if we show that the property “\(R \to A\) is syntomic” is local. We check conditions (a), (b) and (c) of Definition 01SR. By Algebra, Lemma 00SN being syntomic is stable under base change and hence we conclude (a) holds. By Algebra, Lemma 00SZ being syntomic is stable under composition and trivially for any ring \(R\) the ring map \(R \to R_f\) is syntomic. We conclude (b) holds. Finally, property (c) is true according to Algebra, Lemma 00SO.

Lemma

The composition of two morphisms which are syntomic is syntomic.

Proof

In the proof of Lemma 01UD we saw that being syntomic is a local property of ring maps. Hence the first statement of the lemma follows from Lemma 01SV combined with the fact that being syntomic is a property of ring maps that is stable under composition, see Algebra, Lemma 00SZ.

Lemma

The base change of a morphism which is syntomic is syntomic.

Proof

In the proof of Lemma 01UD we saw that being syntomic is a local property of ring maps. Hence the lemma follows from Lemma 01SV combined with the fact that being syntomic is a property of ring maps that is stable under base change, see Algebra, Lemma 00SN.

Lemma

Any open immersion is syntomic.

Proof

This is true because an open immersion is a local isomorphism.

Lemma

A syntomic morphism is locally of finite presentation.

Proof

True because a syntomic ring map is of finite presentation by definition.

Lemma

A syntomic morphism is flat.

Proof

True because a syntomic ring map is flat by definition.

Lemma

A syntomic morphism is universally open.

Proof

Combine Lemmas 01UK, 01UL, and 01UA.

Let \(k\) be a field. Let \(A\) be a local \(k\)-algebra essentially of finite type over \(k\). Recall that \(A\) is called a complete intersection over \(k\) if we can write \(A \cong R/(f_1, \ldots, f_c)\) where \(R\) is a regular local ring essentially of finite type over \(k\), and \(f_1, \ldots, f_c\) is a regular sequence in \(R\), see Algebra, Definition 00SD.

Lemma

Let \(k\) be a field. Let \(X\) be a scheme locally of finite type over \(k\). The following are equivalent:

\(X\) is a local complete intersection over \(k\),

for every \(x \in X\) there exists an affine open \(U = \Spec(R) \subset X\) neighbourhood of \(x\) such that \(R \cong k[x_1, \ldots, x_n]/(f_1, \ldots, f_c)\) is a global complete intersection over \(k\), and

for every \(x \in X\) the local ring \(\mathcal{O}_{X, x}\) is a complete intersection over \(k\).

Proof

The corresponding algebra results can be found in Algebra, Lemmas 00SG and 00SH.

The following lemma says locally any syntomic morphism is standard syntomic. Hence we can use standard syntomic morphisms as a local model for a syntomic morphism. Moreover, it says that a flat morphism of finite presentation is syntomic if and only if the fibres are local complete intersection schemes.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(x \in X\) be a point with image \(s = f(x)\). Let \(V \subset S\) be an affine open neighbourhood of \(s\). The following are equivalent

The morphism \(f\) is syntomic at \(x\).

There exist an affine open \(U \subset X\) with \(x \in U\) and \(f(U) \subset V\) such that \(f|_U : U \to V\) is standard syntomic.

The morphism \(f\) is of finite presentation at \(x\), the local ring map \(\mathcal{O}_{S, s} \to \mathcal{O}_{X, x}\) is flat and \(\mathcal{O}_{X, x}/\mathfrak m_s \mathcal{O}_{X, x}\) is a complete intersection over \(\kappa(s)\) (see Algebra, Definition 00SD).

Proof

Follows from the definitions and Algebra, Lemma 00SY.

Lemma

Let \(f : X \to S\) be a morphism of schemes. If \(f\) is flat, locally of finite presentation, and all fibres \(X_s\) are local complete intersections, then \(f\) is syntomic.

Proof

Clear from Lemmas 01UG and 01UE and the isomorphisms of local rings \(\mathcal{O}_{X, x}/\mathfrak m_s \mathcal{O}_{X, x} \cong \mathcal{O}_{X_s, x}\).

Lemma

Let \(f : X \to S\) be a morphism of schemes. Assume \(f\) locally of finite type. Formation of the set \[T = \{x \in X \mid \mathcal{O}_{X_{f(x)}, x} \text{ is a complete intersection over }\kappa(f(x))\}\] commutes with arbitrary base change: For any morphism \(g : S' \to S\), consider the base change \(f' : X' \to S'\) of \(f\) and the projection \(g' : X' \to X\). Then the corresponding set \(T'\) for the morphism \(f'\) is equal to \(T' = (g')^{-1}(T)\). In particular, if \(f\) is assumed flat, and locally of finite presentation then the same holds for the open set of points where \(f\) is syntomic.

Proof

Let \(s' \in S'\) be a point, and let \(s = g(s')\). Then we have \[X'_{s'} = \Spec(\kappa(s')) \times_{\Spec(\kappa(s))} X_s\] In other words the fibres of the base change are the base changes of the fibres. Hence the first part is equivalent to Algebra, Lemma 00SI. The second part follows from the first because in that case \(T\) is the set of points where \(f\) is syntomic according to Lemma 01UE.

Lemma

Let \(R\) be a ring. Let \(R \to A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)\) be a relative global complete intersection. Set \(S = \Spec(R)\) and \(X = \Spec(A)\). Consider the morphism \(f : X \to S\) associated to the ring map \(R \to A\). The function \(x \mapsto \dim_x(X_{f(x)})\) is constant with value \(n - c\).

Proof

By Algebra, Definition 00SP \(R \to A\) being a relative global complete intersection means all nonzero fibre rings have dimension \(n - c\). Thus for a prime \(\mathfrak p\) of \(R\) the fibre ring \(\kappa(\mathfrak p)[x_1, \ldots, x_n]/(\overline{f}_1, \ldots, \overline{f}_c)\) is either zero or a global complete intersection ring of dimension \(n - c\). By the discussion following Algebra, Definition 00S9 this implies it is equidimensional of dimension \(n - c\). Whence the lemma.

Lemma

Let \(f : X \to S\) be a syntomic morphism. The function \(x \mapsto \dim_x(X_{f(x)})\) is locally constant on \(X\).

Proof

By Lemma 01UE the morphism \(f\) locally looks like a standard syntomic morphism of affines. Hence the result follows from Lemma 02K0.

Lemma 02K1 says that the following definition makes sense.

Definition

Let \(d \geq 0\) be an integer. We say a morphism of schemes \(f : X \to S\) is syntomic of relative dimension \(d\) if \(f\) is syntomic and the function \(\dim_x(X_{f(x)}) = d\) for all \(x \in X\).

In other words, \(f\) is syntomic and the nonempty fibres are equidimensional of dimension \(d\).

Lemma

Let \[\xymatrix{ X \ar[rr]_f \ar[rd]_p & & Y \ar[dl]^q \\ & S }\] be a commutative diagram of morphisms of schemes. Assume that

\(f\) is surjective and syntomic,

\(p\) is syntomic, and

\(q\) is locally of finite presentation⁸.

Then \(q\) is syntomic.

Proof

By Lemma 02JZ we see that \(q\) is flat. Hence it suffices to show that the fibres of \(Y \to S\) are local complete intersections, see Lemma 01UF. Let \(s \in S\). Consider the morphism \(X_s \to Y_s\). This is a base change of the morphism \(X \to Y\) and hence surjective, and syntomic (Lemma 01UI). For the same reason \(X_s\) is syntomic over \(\kappa(s)\). Moreover, \(Y_s\) is locally of finite type over \(\kappa(s)\) (Lemma 01T4). In this way we reduce to the case where \(S\) is the spectrum of a field.

Assume \(S = \Spec(k)\). Let \(y \in Y\). Choose an affine open \(\Spec(A) \subset Y\) neighbourhood of \(y\). Let \(\Spec(B) \subset X\) be an affine open such that \(f(\Spec(B)) \subset \Spec(A)\), containing a point \(x \in X\) such that \(f(x) = y\). Choose a surjection \(k[x_1, \ldots, x_n] \to A\) with kernel \(I\). Choose a surjection \(A[y_1, \ldots, y_m] \to B\), which gives rise in turn to a surjection \(k[x_i, y_j] \to B\) with kernel \(J\). Let \(\mathfrak q \subset k[x_i, y_j]\) be the prime corresponding to \(y \in \Spec(B)\) and let \(\mathfrak p \subset k[x_i]\) the prime corresponding to \(x \in \Spec(A)\). Since \(x\) maps to \(y\) we have \(\mathfrak p = \mathfrak q \cap k[x_i]\). Consider the following commutative diagram of local rings: \[\xymatrix{ \mathcal{O}_{X, x} \ar@{=}[r] & B_{\mathfrak q} & k[x_1, \ldots, x_n, y_1, \ldots, y_m]_{\mathfrak q} \ar[l] \\ \mathcal{O}_{Y, y} \ar@{=}[r] \ar[u] & A_{\mathfrak p} \ar[u] & k[x_1, \ldots, x_n]_{\mathfrak p} \ar[l] \ar[u] }\] We claim that the hypotheses of Algebra, Lemma 02JP are satisfied. Conditions (1) and (2) are trivial. Condition (4) follows as \(X \to Y\) is flat. Condition (3) follows as the rings \(\mathcal{O}_{Y, y}\) and \(\mathcal{O}_{X_y, x} = \mathcal{O}_{X, x}/\mathfrak m_y\mathcal{O}_{X, x}\) are complete intersection rings by our assumptions that \(f\) and \(p\) are syntomic, see Lemma 01UE. The output of Algebra, Lemma 02JP is exactly that \(\mathcal{O}_{Y, y}\) is a complete intersection ring! Hence by Lemma 01UE again we see that \(Y\) is syntomic over \(k\) at \(y\) as desired.

Conormal sheaf of an immersion

Let \(i : Z \to X\) be a closed immersion. Let \(\mathcal{I} \subset \mathcal{O}_X\) be the corresponding quasi-coherent sheaf of ideals. Consider the short exact sequence \[0 \to \mathcal{I}^2 \to \mathcal{I} \to \mathcal{I}/\mathcal{I}^2 \to 0\] of quasi-coherent sheaves on \(X\). Since the sheaf \(\mathcal{I}/\mathcal{I}^2\) is annihilated by \(\mathcal{I}\) it corresponds to a sheaf on \(Z\) by Lemma 01QY. This quasi-coherent \(\mathcal{O}_Z\)-module is called the conormal sheaf of \(Z\) in \(X\) and is often simply denoted \(\mathcal{I}/\mathcal{I}^2\) by the abuse of notation mentioned in Section 01QX.

In case \(i : Z \to X\) is a (locally closed) immersion we define the conormal sheaf of \(i\) as the conormal sheaf of the closed immersion \(i : Z \to X \setminus \partial Z\), where \(\partial Z = \overline{Z} \setminus Z\). It is often denoted \(\mathcal{I}/\mathcal{I}^2\) where \(\mathcal{I}\) is the ideal sheaf of the closed immersion \(i : Z \to X \setminus \partial Z\).

Definition

Let \(i : Z \to X\) be an immersion. The conormal sheaf \(\mathcal{C}_{Z/X}\) of \(Z\) in \(X\) or the conormal sheaf of \(i\) is the quasi-coherent \(\mathcal{O}_Z\)-module \(\mathcal{I}/\mathcal{I}^2\) described above.

In [EGA, IV Definition 16.1.2] this sheaf is denoted \(\mathcal{N}_{Z/X}\). We will not follow this convention since we would like to reserve the notation \(\mathcal{N}_{Z/X}\) for the normal sheaf of the immersion. It is defined as \[\mathcal{N}_{Z/X} = \SheafHom_{\mathcal{O}_Z}(\mathcal{C}_{Z/X}, \mathcal{O}_Z) = \SheafHom_{\mathcal{O}_Z}(\mathcal{I}/\mathcal{I}^2, \mathcal{O}_Z)\] provided the conormal sheaf is of finite presentation (otherwise the normal sheaf may not even be quasi-coherent). We will come back to the normal sheaf later (insert future reference here).

Lemma

Let \(i : Z \to X\) be an immersion. The conormal sheaf of \(i\) has the following properties:

Let \(U \subset X\) be any open subscheme such that \(i\) factors as \(Z \xrightarrow{i'} U \to X\) where \(i'\) is a closed immersion. Let \(\mathcal{I} = \Ker((i')^\sharp) \subset \mathcal{O}_U\). Then \[\mathcal{C}_{Z/X} = (i')^*\mathcal{I}\quad\text{and}\quad i'_*\mathcal{C}_{Z/X} = \mathcal{I}/\mathcal{I}^2\]

For any affine open \(\Spec(R) = U \subset X\) such that \(Z \cap U = \Spec(R/I)\) there is a canonical isomorphism \(\Gamma(Z \cap U, \mathcal{C}_{Z/X}) = I/I^2\).

Proof

Mostly clear from the definitions. Note that given a ring \(R\) and an ideal \(I\) of \(R\) we have \(I/I^2 = I \otimes_R R/I\). Details omitted.

Lemma

Let \[\xymatrix{ Z \ar[r]_i \ar[d]_f & X \ar[d]^g \\ Z' \ar[r]^{i'} & X' }\] be a commutative diagram in the category of schemes. Assume \(i\), \(i'\) immersions. There is a canonical map of \(\mathcal{O}_Z\)-modules \[f^*\mathcal{C}_{Z'/X'} \longrightarrow \mathcal{C}_{Z/X}\] characterized by the following property: For every pair of affine opens \((\Spec(R) = U \subset X, \Spec(R') = U' \subset X')\) with \(g(U) \subset U'\) such that \(Z \cap U = \Spec(R/I)\) and \(Z' \cap U' = \Spec(R'/I')\) the induced map \[\Gamma(Z' \cap U', \mathcal{C}_{Z'/X'}) = I'/I'^2 \longrightarrow I/I^2 = \Gamma(Z \cap U, \mathcal{C}_{Z/X})\] is the one induced by the ring map \(g^\sharp : R' \to R\) which has the property \(g^\sharp(I') \subset I\).

Proof

Let \(\partial Z' = \overline{Z'} \setminus Z'\) and \(\partial Z = \overline{Z} \setminus Z\). These are closed subsets of \(X'\) and of \(X\). Replacing \(X'\) by \(X' \setminus \partial Z'\) and \(X\) by \(X \setminus \big(g^{-1}(\partial Z') \cup \partial Z\big)\) we see that we may assume that \(i\) and \(i'\) are closed immersions.

The fact that \(g \circ i\) factors through \(i'\) implies that \(g^*\mathcal{I}'\) maps into \(\mathcal{I}\) under the canonical map \(g^*\mathcal{I}' \to \mathcal{O}_X\), see Schemes, Lemmas 01HP and 01HQ. Hence we get an induced map of quasi-coherent sheaves \(g^*(\mathcal{I}'/(\mathcal{I}')^2) \to \mathcal{I}/\mathcal{I}^2\). Pulling back by \(i\) gives \(i^*g^*(\mathcal{I}'/(\mathcal{I}')^2) \to i^*(\mathcal{I}/\mathcal{I}^2)\). Note that \(i^*(\mathcal{I}/\mathcal{I}^2) = \mathcal{C}_{Z/X}\). On the other hand, \(i^*g^*(\mathcal{I}'/(\mathcal{I}')^2) = f^*(i')^*(\mathcal{I}'/(\mathcal{I}')^2) = f^*\mathcal{C}_{Z'/X'}\). This gives the desired map.

Checking that the map is locally described as the given map \(I'/(I')^2 \to I/I^2\) is a matter of unwinding the definitions and is omitted. Another observation is that given any \(x \in i(Z)\) there do exist affine open neighbourhoods \(U\), \(U'\) with \(f(U) \subset U'\) and \(Z \cap U\) as well as \(U' \cap Z'\) closed such that \(x \in U\). Proof omitted. Hence the requirement of the lemma indeed characterizes the map (and could have been used to define it).

Lemma

Let \[\xymatrix{ Z \ar[r]_i \ar[d]_f & X \ar[d]^g \\ Z' \ar[r]^{i'} & X' }\] be a fibre product diagram in the category of schemes with \(i\), \(i'\) immersions. Then the canonical map \(f^*\mathcal{C}_{Z'/X'} \to \mathcal{C}_{Z/X}\) of Lemma 01R4 is surjective. If \(g\) is flat, then it is an isomorphism.

Proof

Let \(R' \to R\) be a ring map, and \(I' \subset R'\) an ideal. Set \(I = I'R\). Then \(I'/(I')^2 \otimes_{R'} R \to I/I^2\) is surjective. If \(R' \to R\) is flat, then \(I = I' \otimes_{R'} R\) and \(I^2 = (I')^2 \otimes_{R'} R\) and we see the map is an isomorphism.

Lemma

Let \(Z \to Y \to X\) be immersions of schemes. Then there is a canonical exact sequence \[i^*\mathcal{C}_{Y/X} \to \mathcal{C}_{Z/X} \to \mathcal{C}_{Z/Y} \to 0\] where the maps come from Lemma 01R4 and \(i : Z \to Y\) is the first morphism.

Proof

Via Lemma 01R4 this translates into the following algebra fact. Suppose that \(C \to B \to A\) are surjective ring maps. Let \(I = \Ker(B \to A)\), \(J = \Ker(C \to A)\) and \(K = \Ker(C \to B)\). Then there is an exact sequence \[K/K^2 \otimes_B A \to J/J^2 \to I/I^2 \to 0.\] This follows immediately from the observation that \(I = J/K\).

Sheaf of differentials of a morphism

We suggest the reader take a look at the corresponding section in the chapter on commutative algebra (Algebra, Section 00RM) and the corresponding section in the chapter on sheaves of modules (Modules, Section 08RL).

Definition

Let \(f : X \to S\) be a morphism of schemes. The sheaf of differentials \(\Omega_{X/S}\) of \(X\) over \(S\) is the sheaf of differentials of \(f\) viewed as a morphism of ringed spaces (Modules, Definition 08RT) equipped with its universal \(S\)-derivation \[\text{d}_{X/S} : \mathcal{O}_X \longrightarrow \Omega_{X/S}.\]

It turns out that \(\Omega_{X/S}\) is a quasi-coherent \(\mathcal{O}_X\)-module for example as it is isomorphic to the conormal sheaf of the diagonal morphism \(\Delta : X \to X \times_S X\) (Lemma 08S2). We have defined the module of differentials of \(X\) over \(S\) using a universal property, namely as the receptacle of the universal derivation. If you have any other construction of the sheaf of relative differentials which satisfies this universal property then, by the Yoneda lemma, it will be canonically isomorphic to the one defined above. For convenience we restate the universal property here.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The map \[\Hom_{\mathcal{O}_X}(\Omega_{X/S}, \mathcal{F}) \longrightarrow \text{Der}_S(\mathcal{O}_X, \mathcal{F}),\quad \alpha \longmapsto \alpha \circ \text{d}_{X/S}\] is an isomorphism of functors \(\textit{Mod}(\mathcal{O}_X) \to \textit{Sets}\).

Proof

This is just a restatement of the definition.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(U \subset X\), \(V \subset S\) be open subschemes such that \(f(U) \subset V\). Then there is a unique isomorphism \(\Omega_{X/S}|_U = \Omega_{U/V}\) of \(\mathcal{O}_U\)-modules such that \(\text{d}_{X/S}|_U = \text{d}_{U/V}\).

Proof

This is a special case of Modules, Lemma 08RQ if we use the canonical identification \(f^{-1}\mathcal{O}_S|_U = (f|_U)^{-1}\mathcal{O}_V\).

From now on we will use these canonical identifications and simply write \(\Omega_{U/S}\) or \(\Omega_{U/V}\) for the restriction of \(\Omega_{X/S}\) to \(U\).

Lemma

Let \(R \to A\) be a ring map. Let \(\mathcal{F}\) be a sheaf of \(\mathcal{O}_X\)-modules on \(X = \Spec(A)\). Set \(S = \Spec(R)\). The rule which associates to an \(S\)-derivation on \(\mathcal{F}\) its action on global sections defines a bijection between the set of \(S\)-derivations of \(\mathcal{F}\) and the set of \(R\)-derivations on \(M = \Gamma(X, \mathcal{F})\).

Proof

Let \(D : A \to M\) be an \(R\)-derivation. We have to show there exists a unique \(S\)-derivation on \(\mathcal{F}\) which gives rise to \(D\) on global sections. Let \(U = D(f) \subset X\) be a standard affine open. Any element of \(\Gamma(U, \mathcal{O}_X)\) is of the form \(a/f^n\) for some \(a \in A\) and \(n \geq 0\). By the Leibniz rule we have \[D(a)|_U = a/f^n D(f^n)|_U + f^n D(a/f^n)\] in \(\Gamma(U, \mathcal{F})\). Since \(f\) acts invertibly on \(\Gamma(U, \mathcal{F})\) this completely determines the value of \(D(a/f^n) \in \Gamma(U, \mathcal{F})\). This proves uniqueness. Existence follows by simply defining \[D(a/f^n) := (1/f^n) D(a)|_U - a/f^{2n} D(f^n)|_U\] and proving this has all the desired properties (on the basis of standard opens of \(X\)). Details omitted.

Lemma

Let \(f : X \to S\) be a morphism of schemes. For any pair of affine opens \(\Spec(A) = U \subset X\), \(\Spec(R) = V \subset S\) with \(f(U) \subset V\) there is a unique isomorphism \[\Gamma(U, \Omega_{X/S}) = \Omega_{A/R}.\] compatible with \(\text{d}_{X/S}\) and \(\text{d} : A \to \Omega_{A/R}\).

Proof

By Lemma 01US we may replace \(X\) and \(S\) by \(U\) and \(V\). Thus we may assume \(X = \Spec(A)\) and \(S = \Spec(R)\) and we have to show the lemma with \(U = X\) and \(V = S\). Consider the \(A\)-module \(M = \Gamma(X, \Omega_{X/S})\) together with the \(R\)-derivation \(\text{d}_{X/S} : A \to M\). Let \(N\) be another \(A\)-module and denote \(\widetilde{N}\) the quasi-coherent \(\mathcal{O}_X\)-module associated to \(N\), see Schemes, Section 01I6. Precomposing by \(\text{d}_{X/S} : A \to M\) we get an arrow \[\alpha : \Hom_A(M, N) \longrightarrow \text{Der}_R(A, N)\] Using Lemmas 01UR and 01UO we get identifications \[\Hom_{\mathcal{O}_X}(\Omega_{X/S}, \widetilde{N}) = \text{Der}_S(\mathcal{O}_X, \widetilde{N}) = \text{Der}_R(A, N)\] Taking global sections determines an arrow \(\Hom_{\mathcal{O}_X}(\Omega_{X/S}, \widetilde{N}) \to \Hom_R(M, N)\). Combining this arrow and the identifications above we get an arrow \[\beta : \text{Der}_R(A, N) \longrightarrow \Hom_R(M, N)\] Checking what happens on global sections, we find that \(\alpha\) and \(\beta\) are each others inverse. Hence we see that \(\text{d}_{X/S} : A \to M\) satisfies the same universal property as \(\text{d} : A \to \Omega_{A/R}\), see Algebra, Lemma 00RO. Thus the Yoneda lemma (Categories, Lemma 001P) implies there is a unique isomorphism of \(A\)-modules \(M \cong \Omega_{A/R}\) compatible with derivations.

Remark

The lemma above gives a second way of constructing the module of differentials. Namely, let \(f : X \to S\) be a morphism of schemes. Consider the collection of all affine opens \(U \subset X\) which map into an affine open of \(S\). These form a basis for the topology on \(X\). Thus it suffices to define \(\Gamma(U, \Omega_{X/S})\) for such \(U\). We simply set \(\Gamma(U, \Omega_{X/S}) = \Omega_{A/R}\) if \(A\), \(R\) are as in Lemma 01UT above. This works, but it takes somewhat more algebraic preliminaries to construct the restriction mappings and to verify the sheaf condition with this ansatz.

The following lemma gives yet another way to define the sheaf of differentials and it in particular shows that \(\Omega_{X/S}\) is quasi-coherent if \(X\) and \(S\) are schemes.

Lemma

Let \(f : X \to S\) be a morphism of schemes. There is a canonical isomorphism between \(\Omega_{X/S}\) and the conormal sheaf of the diagonal morphism \(\Delta_{X/S} : X \longrightarrow X \times_S X\).

Proof

We first establish the existence of a couple of “global” sheaves and global maps of sheaves, and further down we describe the constructions over some affine opens.

Recall that \(\Delta = \Delta_{X/S} : X \to X \times_S X\) is an immersion, see Schemes, Lemma 01KJ. Let \(\mathcal{J}\) be the ideal sheaf of the immersion which lives over some open subscheme \(W\) of \(X \times_S X\) such that \(\Delta(X) \subset W\) is closed. Let us take the one that was found in the proof of Schemes, Lemma 01KJ. Note that the sheaf of rings \(\mathcal{O}_W/\mathcal{J}^2\) is supported on \(\Delta(X)\). Moreover it sits in a short exact sequence of sheaves \[0 \to \mathcal{J}/\mathcal{J}^2 \to \mathcal{O}_W/\mathcal{J}^2 \to \Delta_*\mathcal{O}_X \to 0.\] Using \(\Delta^{-1}\) we can think of this as a surjection of sheaves of \(f^{-1}\mathcal{O}_S\)-algebras with kernel the conormal sheaf of \(\Delta\) (see Definition 01R2 and Lemma 01R3). \[0 \to \mathcal{C}_{X/X \times_S X} \to \Delta^{-1}(\mathcal{O}_W/\mathcal{J}^2) \to \mathcal{O}_X \to 0\] This places us in the situation of Modules, Lemma 01UP. The projection morphisms \(p_i : X \times_S X \to X\), \(i = 1, 2\) induce maps of sheaves of rings \((p_i)^\sharp : (p_i)^{-1}\mathcal{O}_X \to \mathcal{O}_{X \times_S X}\). We may restrict to \(W\) and quotient by \(\mathcal{J}^2\) to get \((p_i)^{-1}\mathcal{O}_X \to \mathcal{O}_W/\mathcal{J}^2\). Since \(\Delta^{-1}p_i^{-1}\mathcal{O}_X = \mathcal{O}_X\) we get maps \[s_i : \mathcal{O}_X \to \Delta^{-1}(\mathcal{O}_W/\mathcal{J}^2).\] Both \(s_1\) and \(s_2\) are sections to the map \(\Delta^{-1}(\mathcal{O}_W/\mathcal{J}^2) \to \mathcal{O}_X\), as in Modules, Lemma 01UP. Thus we get an \(S\)-derivation \(\text{d} = s_2 - s_1 : \mathcal{O}_X \to \mathcal{C}_{X/X \times_S X}\). By the universal property of the module of differentials we find a unique \(\mathcal{O}_X\)-linear map \[\Omega_{X/S} \longrightarrow \mathcal{C}_{X/X \times_S X},\quad f\text{d}g \longmapsto fs_2(g) - fs_1(g)\] To see the map is an isomorphism, let us work this out over suitable affine opens. We can cover \(X\) by affine opens \(\Spec(A) = U \subset X\) whose image is contained in an affine open \(\Spec(R) = V \subset S\). According to the proof of Schemes, Lemma 01KJ \(U \times_V U \subset X \times_S X\) is an affine open contained in the open \(W\) mentioned above. Also \(U \times_V U = \Spec(A \otimes_R A)\). The sheaf \(\mathcal{J}\) corresponds to the ideal \(J = \Ker(A \otimes_R A \to A)\). The short exact sequence to the short exact sequence of \(A \otimes_R A\)-modules \[0 \to J/J^2 \to (A \otimes_R A)/J^2 \to A \to 0\] The sections \(s_i\) correspond to the ring maps \[A \longrightarrow (A \otimes_R A)/J^2,\quad s_1 : a \mapsto a \otimes 1,\quad s_2 : a \mapsto 1 \otimes a.\] By Lemma 01R3 we have \(\Gamma(U, \mathcal{C}_{X/X \times_S X}) = J/J^2\) and by Lemma 01UT we have \(\Gamma(U, \Omega_{X/S}) = \Omega_{A/R}\). The map above is the map \(a \text{d}b \mapsto a \otimes b - ab \otimes 1\) which is shown to be an isomorphism in Algebra, Lemma 00RW.

Lemma

Let \[\xymatrix{ X' \ar[d] \ar[r]_f & X \ar[d] \\ S' \ar[r] & S }\] be a commutative diagram of schemes. The canonical map \(\mathcal{O}_X \to f_*\mathcal{O}_{X'}\) composed with the map \(f_*\text{d}_{X'/S'} : f_*\mathcal{O}_{X'} \to f_*\Omega_{X'/S'}\) is a \(S\)-derivation. Hence we obtain a canonical map of \(\mathcal{O}_X\)-modules \(\Omega_{X/S} \to f_*\Omega_{X'/S'}\), and by adjointness of \(f_*\) and \(f^*\) a canonical \(\mathcal{O}_{X'}\)-module homomorphism \[c_f : f^*\Omega_{X/S} \longrightarrow \Omega_{X'/S'}.\] It is uniquely characterized by the property that \(f^*\text{d}_{X/S}(h)\) maps to \(\text{d}_{X'/S'}(f^* h)\) for any local section \(h\) of \(\mathcal{O}_X\).

Proof

This is a special case of Modules, Lemma 08RU. In the case of schemes we can also use the functoriality of the conormal sheaves (see Lemma 01R4) and Lemma 08S2 to define \(c_f\). Or we can use the characterization in the last line of the lemma to glue maps defined on affine patches (see Algebra, Equation (00RQ)).

Lemma

Let \(f : X \to Y\), \(g : Y \to S\) be morphisms of schemes. Then there is a canonical exact sequence \[f^*\Omega_{Y/S} \to \Omega_{X/S} \to \Omega_{X/Y} \to 0\] where the maps come from applications of Lemma 01UV.

Proof

This is the sheafified version of Algebra, Lemma 00RS. Alternatively, there is a general version for morphisms of ringed spaces, see Modules, Lemma 0H79.

Lemma

Let \(X \to S\) be a morphism of schemes. Let \(g : S' \to S\) be a morphism of schemes. Let \(X' = X_{S'}\) be the base change of \(X\). Denote \(g' : X' \to X\) the projection. Then the map \[(g')^*\Omega_{X/S} \to \Omega_{X'/S'}\] of Lemma 01UV is an isomorphism.

Proof

This is the sheafified version of Algebra, Lemma 00RV.

Lemma

Let \(f : X \to S\) and \(g : Y \to S\) be morphisms of schemes with the same target. Let \(p : X \times_S Y \to X\) and \(q : X \times_S Y \to Y\) be the projection morphisms. The maps from Lemma 01UV \[p^*\Omega_{X/S} \oplus q^*\Omega_{Y/S} \longrightarrow \Omega_{X \times_S Y/S}\] give an isomorphism.

Proof

By Lemma 01V0 the composition \(p^*\Omega_{X/S} \to \Omega_{X \times_S Y/S} \to \Omega_{X \times_S Y/Y}\) is an isomorphism, and similarly for \(q\). Moreover, the cokernel of \(p^*\Omega_{X/S} \to \Omega_{X \times_S Y/S}\) is \(\Omega_{X \times_S Y/X}\) by Lemma 01UX. The result follows.

Lemma

Let \(f : X \to S\) be a morphism of schemes. If \(f\) is locally of finite type, then \(\Omega_{X/S}\) is a finite type \(\mathcal{O}_X\)-module.

Proof

Immediate from Algebra, Lemma 00RZ, Lemma 01UT, Lemma 01T2, and Properties, Lemma 01PB.

Lemma

Let \(f : X \to S\) be a morphism of schemes. If \(f\) is locally of finite presentation, then \(\Omega_{X/S}\) is an \(\mathcal{O}_X\)-module of finite presentation.

Proof

Immediate from Algebra, Lemma 00RY, Lemma 01UT, Lemma 01TQ, and Properties, Lemma 01PC.

Lemma

If \(X \to S\) is an immersion, or more generally a monomorphism, then \(\Omega_{X/S}\) is zero.

Proof

This is true because \(\Delta_{X/S}\) is an isomorphism in this case and hence has trivial conormal sheaf. Hence \(\Omega_{X/S} = 0\) by Lemma 08S2. The algebraic version is Algebra, Lemma 00RP.

Lemma

Let \(i : Z \to X\) be an immersion of schemes over \(S\). There is a canonical exact sequence \[\mathcal{C}_{Z/X} \to i^*\Omega_{X/S} \to \Omega_{Z/S} \to 0\] where the first arrow is induced by \(\text{d}_{X/S}\) and the second arrow comes from Lemma 01UV.

Proof

This is the sheafified version of Algebra, Lemma 00RU. However we should make sure we can define the first arrow globally. Hence we explain the meaning of “induced by \(\text{d}_{X/S}\)” here. Namely, we may assume that \(i\) is a closed immersion by shrinking \(X\). Let \(\mathcal{I} \subset \mathcal{O}_X\) be the sheaf of ideals corresponding to \(Z \subset X\). Then \(\text{d}_{X/S} : \mathcal{I} \to \Omega_{X/S}\) maps the subsheaf \(\mathcal{I}^2 \subset \mathcal{I}\) to \(\mathcal{I}\Omega_{X/S}\). Hence it induces a map \(\mathcal{I}/\mathcal{I}^2 \to \Omega_{X/S}/\mathcal{I}\Omega_{X/S}\) which is \(\mathcal{O}_X/\mathcal{I}\)-linear. By Lemma 01QY this corresponds to a map \(\mathcal{C}_{Z/X} \to i^*\Omega_{X/S}\) as desired.

Lemma

Let \(i : Z \to X\) be an immersion of schemes over \(S\), and assume \(i\) (locally) has a left inverse. Then the canonical sequence \[0 \to \mathcal{C}_{Z/X} \to i^*\Omega_{X/S} \to \Omega_{Z/S} \to 0\] of Lemma 01UZ is (locally) split exact. In particular, if \(s : S \to X\) is a section of the structure morphism \(X \to S\) then the map \(\mathcal{C}_{S/X} \to s^*\Omega_{X/S}\) induced by \(\text{d}_{X/S}\) is an isomorphism.

Proof

It follows from Algebra, Lemma 02HP that the result holds when \(X, Z, S\) are affine. In general we conclude the sequence is locally split exact. If \(g : X \to Z\) is a left inverse of \(i\), then \(i^*c_g\) is a right inverse of the map \(i^*\Omega_{X/S} \to \Omega_{Z/S}\) by Modules, Lemma 01UW. This implies the sequence is split by Homology, Lemma 010G. Finally, if \(s\) is a section, then it is an immersion \(s : Z = S \to X\) over \(S\) (see Schemes, Lemma 01KT) and in that case \(\Omega_{Z/S} = 0\).

Remark

Let \(X \to S\) be a morphism of schemes. According to Lemma 01V1 we have \[\Omega_{X \times_S X/S} = \text{pr}_1^*\Omega_{X/S} \oplus \text{pr}_2^*\Omega_{X/S}\] On the other hand, the diagonal morphism \(\Delta : X \to X \times_S X\) is an immersion, which locally has a left inverse. Hence by Lemma 0474 we obtain a canonical short exact sequence \[0 \to \mathcal{C}_{X/X \times_S X} \to \Omega_{X/S} \oplus \Omega_{X/S} \to \Omega_{X/S} \to 0\] Note that the right arrow is \((1, 1)\) which is indeed a split surjection. On the other hand, by Lemma 08S2 we have an identification \(\Omega_{X/S} = \mathcal{C}_{X/X \times_S X}\). Because we chose \(\text{d}_{X/S}(f) = s_2(f) - s_1(f)\) in this identification it turns out that the left arrow is the map \((-1, 1)\)⁹.

Lemma

Let \[\xymatrix{ Z \ar[r]_i \ar[rd]_j & X \ar[d] \\ & Y }\] be a commutative diagram of schemes where \(i\) and \(j\) are immersions. Then there is a canonical exact sequence \[\mathcal{C}_{Z/Y} \to \mathcal{C}_{Z/X} \to i^*\Omega_{X/Y} \to 0\] where the first arrow comes from Lemma 01R4 and the second from Lemma 01UZ.

Proof

The algebraic version of this is Algebra, Lemma 065V.

Finite order differential operators

We suggest the reader take a look at the corresponding section in the chapter on commutative algebra (Algebra, Section 09CH) and the corresponding section in the chapter on sheaves of modules (Modules, Section 0G3P).

Lemma

Let \(R \to A\) be a ring map. Denote \(f : X \to S\) the corresponding morphism of affine schemes. Let \(\mathcal{F}\) and \(\mathcal{G}\) be \(\mathcal{O}_X\)-modules. If \(\mathcal{F}\) is quasi-coherent then the map \[\text{Diff}^k_{X/S}(\mathcal{F}, \mathcal{G}) \to \text{Diff}^k_{A/R}(\Gamma(X, \mathcal{F}), \Gamma(X, \mathcal{G}))\] sending a differential operator to its action on global sections is bijective.

Proof

Write \(\mathcal{F} = \widetilde{M}\) for some \(A\)-module \(M\). Set \(N = \Gamma(X, \mathcal{G})\). Let \(D : M \to N\) be a differential operator of order \(k\). We have to show there exists a unique differential operator \(\mathcal{F} \to \mathcal{G}\) of order \(k\) which gives rise to \(D\) on global sections. Let \(U = D(f) \subset X\) be a standard affine open. Then \(\mathcal{F}(U) = M_f\) is the localization. By Algebra, Lemma 0G36 the differential operator \(D\) extends to a unique differential operator \[D_f : \mathcal{F}(U) = \widetilde{M}(U) = M_f \to N_f = \widetilde{N}(U)\] The uniqueness shows that these maps \(D_f\) glue to give a map of sheaves \(\widetilde{M} \to \widetilde{N}\) on the basis of all standard opens of \(X\). Hence we get a unique map of sheaves \(\widetilde{D} : \widetilde{M} \to \widetilde{N}\) agreeing with these maps by the material in Sheaves, Section 009H. Since \(\widetilde{D}\) is given by differential operators of order \(k\) on the standard opens, we find that \(\widetilde{D}\) is a differential operator of order \(k\) (small detail omitted). Finally, we can post-compose with the canonical \(\mathcal{O}_X\)-module map \(c : \widetilde{N} \to \mathcal{G}\) (Schemes, Lemma 01I7) to get \(c \circ \widetilde{D} : \mathcal{F} \to \mathcal{G}\) which is a differential operator of order \(k\) by Modules, Lemma 0G3R. This proves existence. We omit the proof of uniqueness.

Lemma

Let \(a : X \to S\) and \(b : Y \to S\) be morphisms of schemes. Let \(\mathcal{F}\) and \(\mathcal{F}'\) be quasi-coherent \(\mathcal{O}_X\)-modules. Let \(D : \mathcal{F} \to \mathcal{F}'\) be a differential operator of order \(k\) on \(X/S\). Let \(\mathcal{G}\) be a quasi-coherent \(\mathcal{O}_Y\)-module. Then there is a unique differential operator \[D' : \text{pr}_1^*\mathcal{F} \otimes_{\mathcal{O}_{X \times_S Y}} \text{pr}_2^*\mathcal{G} \longrightarrow \text{pr}_1^*\mathcal{F}' \otimes_{\mathcal{O}_{X \times_S Y}} \text{pr}_2^*\mathcal{G}\] of order \(k\) on \(X \times_S Y / Y\) such that \(D'(s \otimes t) = D(s) \otimes t\) for local sections \(s\) of \(\mathcal{F}\) and \(t\) of \(\mathcal{G}\).

Proof

In case \(X\), \(Y\), and \(S\) are affine, this follows, via Lemma 0G44, from the corresponding algebra result, see Algebra, Lemma 0G37. In general, one uses coverings by affines (for example as in Schemes, Lemma 01JS) to construct \(D'\) globally. Details omitted.

Remark

Let \(a : X \to S\) and \(b : Y \to S\) be morphisms of schemes. Denote \(p : X \times_S Y \to X\) and \(q : X \times_S Y \to Y\) the projections. In this remark, given an \(\mathcal{O}_X\)-module \(\mathcal{F}\) and an \(\mathcal{O}_Y\)-module \(\mathcal{G}\) let us set \[\mathcal{F} \boxtimes \mathcal{G} = p^*\mathcal{F} \otimes_{\mathcal{O}_{X \times_S Y}} q^*\mathcal{G}\] Denote \(\mathcal{A}_{X/S}\) the additive category whose objects are quasi-coherent \(\mathcal{O}_X\)-modules and whose morphisms are differential operators of finite order on \(X/S\). Similarly for \(\mathcal{A}_{Y/S}\) and \(\mathcal{A}_{X \times_S Y/S}\). The construction of Lemma 0G45 determines a functor \[\boxtimes : \mathcal{A}_{X/S} \times \mathcal{A}_{Y/S} \longrightarrow \mathcal{A}_{X \times_S Y/S}, \quad (\mathcal{F}, \mathcal{G}) \longmapsto \mathcal{F} \boxtimes \mathcal{G}\] which is bilinear on morphisms. If \(X = \Spec(A)\), \(Y = \Spec(B)\), and \(S = \Spec(R)\), then via the identification of quasi-coherent sheaves with modules this functor is given by \((M, N) \mapsto M \otimes_R N\) on objects and sends the morphism \((D, D') : (M, N) \to (M', N')\) to \(D \otimes D' : M \otimes_R N \to M' \otimes_R N'\).

Smooth morphisms

Let \(f : X \to Y\) be a continuous map of topological spaces. Consider the following condition: For every \(x \in X\) there exist open neighbourhoods \(x \in U \subset X\) and \(f(x) \in V \subset Y\), and an integer \(d\) such that \(f(U) \subset V\) and such that we obtain a commutative diagram \[\xymatrix{ X \ar[d] & U \ar[l] \ar[d] \ar[r]_-\pi & V \times \mathbf{R}^d \ar[ld] \\ Y & V \ar[l] }\] where \(\pi\) is a homeomorphism onto an open subset. Smooth morphisms of schemes are the analogue of these maps in the category of schemes. See Lemma 01V7 and Lemma 054L.

Contrary to expectations (perhaps) the notion of a smooth ring map is not defined solely in terms of the module of differentials. Namely, recall that \(R \to A\) is a smooth ring map if \(A\) is of finite presentation over \(R\) and if the naive cotangent complex of \(A\) over \(R\) is quasi-isomorphic to a projective module placed in degree \(0\), see Algebra, Definition 00T2.

Definition

Let \(f : X \to S\) be a morphism of schemes.

We say that \(f\) is smooth at \(x \in X\) if there exist an affine open neighbourhood \(\Spec(A) = U \subset X\) of \(x\) and affine open \(\Spec(R) = V \subset S\) with \(f(U) \subset V\) such that the induced ring map \(R \to A\) is smooth.

We say that \(f\) is smooth if it is smooth at every point of \(X\).

A morphism of affine schemes \(f : X \to S\) is called standard smooth if there exists a standard smooth ring map \(R \to R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)\) (see Algebra, Definition 00T6) such that \(X \to S\) is isomorphic to \[\Spec(R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)) \to \Spec(R).\]

A pleasing feature of this definition is that the set of points where a morphism is smooth is automatically open.

Note that there is no separation or quasi-compactness hypotheses in the definition. Hence the question of being smooth is local in nature on the source. Here is the precise result.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent

The morphism \(f\) is smooth.

For every affine opens \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the ring map \(\mathcal{O}_S(V) \to \mathcal{O}_X(U)\) is smooth.

There exists an open covering \(S = \bigcup_{j \in J} V_j\) and open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that each of the morphisms \(U_i \to V_j\), \(j\in J, i\in I_j\) is smooth.

There exists an affine open covering \(S = \bigcup_{j \in J} V_j\) and affine open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that the ring map \(\mathcal{O}_S(V_j) \to \mathcal{O}_X(U_i)\) is smooth, for all \(j\in J, i\in I_j\).

Moreover, if \(f\) is smooth then for any open subschemes \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the restriction \(f|_U : U \to V\) is smooth.

Proof

This follows from Lemma 01ST if we show that the property “\(R \to A\) is smooth” is local. We check conditions (a), (b) and (c) of Definition 01SR. By Algebra, Lemma 00T4 being smooth is stable under base change and hence we conclude (a) holds. By Algebra, Lemma 00TD being smooth is stable under composition and for any ring \(R\) the ring map \(R \to R_f\) is (standard) smooth. We conclude (b) holds. Finally, property (c) is true according to Algebra, Lemma 00TC.

The following lemma characterizes a smooth morphism as a flat, finitely presented morphism with smooth fibres. Note that schemes smooth over a field are discussed in more detail in Varieties, Section 04QM.

Lemma

Let \(f : X \to S\) be a morphism of schemes. If \(f\) is flat, locally of finite presentation, and all fibres \(X_s\) are smooth, then \(f\) is smooth.

Proof

Follows from Algebra, Lemma 00TF.

Lemma

The composition of two morphisms which are smooth is smooth.

Proof

In the proof of Lemma 01V6 we saw that being smooth is a local property of ring maps. Hence the first statement of the lemma follows from Lemma 01SV combined with the fact that being smooth is a property of ring maps that is stable under composition, see Algebra, Lemma 00TD.

Lemma

The base change of a morphism which is smooth is smooth.

Proof

In the proof of Lemma 01V6 we saw that being smooth is a local property of ring maps. Hence the lemma follows from Lemma 01SV combined with the fact that being smooth is a property of ring maps that is stable under base change, see Algebra, Lemma 00T4.

Lemma

Any open immersion is smooth.

Proof

This is true because an open immersion is a local isomorphism.

Lemma

A smooth morphism is syntomic.

Proof

See Algebra, Lemma 00TA.

Lemma

A smooth morphism is locally of finite presentation.

Proof

True because a smooth ring map is of finite presentation by definition.

Lemma

A smooth morphism is flat.

Proof

Combine Lemmas 01UL and 01VD.

Lemma

A smooth morphism is universally open.

Proof

Combine Lemmas 01VF, 01VE, and 01UA. Or alternatively, combine Lemmas 01VD, 056F.

The following lemma says locally any smooth morphism is standard smooth. Hence we can use standard smooth morphisms as a local model for a smooth morphism.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(x \in X\) be a point. Let \(V \subset S\) be an affine open neighbourhood of \(f(x)\). The following are equivalent

The morphism \(f\) is smooth at \(x\).

There exists an affine open \(U \subset X\), with \(x \in U\) and \(f(U) \subset V\) such that the induced morphism \(f|_U : U \to V\) is standard smooth.

Proof

Follows from the definitions and Algebra, Lemmas 00T7 and 00TA.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Assume \(f\) is smooth. Then the module of differentials \(\Omega_{X/S}\) of \(X\) over \(S\) is finite locally free and \[\text{rank}_x(\Omega_{X/S}) = \dim_x(X_{f(x)})\] for every \(x \in X\).

Proof

The statement is local on \(X\) and \(S\). By Lemma 01V7 above we may assume that \(f\) is a standard smooth morphism of affines. In this case the result follows from Algebra, Lemma 00T7 (and the definition of a relative global complete intersection, see Algebra, Definition 00SP).

Lemma 02G1 says that the following definition makes sense.

Definition

Let \(d \geq 0\) be an integer. We say a morphism of schemes \(f : X \to S\) is smooth of relative dimension \(d\) if \(f\) is smooth and \(\Omega_{X/S}\) is finite locally free of constant rank \(d\).

In other words, \(f\) is smooth and the nonempty fibres are equidimensional of dimension \(d\). By Lemma 01V9 below this is also the same as requiring: (a) \(f\) is locally of finite presentation, (b) \(f\) is flat, (c) all nonempty fibres equidimensional of dimension \(d\), and (d) \(\Omega_{X/S}\) finite locally free of rank \(d\). It is not enough to simply assume that \(f\) is flat, of finite presentation, and \(\Omega_{X/S}\) is finite locally free of rank \(d\). A counter example is given by \(\Spec(\mathbf{F}_p[t]) \to \Spec(\mathbf{F}_p[t^p])\).

Here is a differential criterion of smoothness at a point. There are many variants of this result all of which may be useful at some point. We will just add them here as needed.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(x \in X\). Set \(s = f(x)\). Assume \(f\) is locally of finite presentation. The following are equivalent:

The morphism \(f\) is smooth at \(x\).

The local ring map \(\mathcal{O}_{S, s} \to \mathcal{O}_{X, x}\) is flat and \(X_s \to \Spec(\kappa(s))\) is smooth at \(x\).

The local ring map \(\mathcal{O}_{S, s} \to \mathcal{O}_{X, x}\) is flat and the \(\mathcal{O}_{X, x}\)-module \(\Omega_{X/S, x}\) can be generated by at most \(\dim_x(X_{f(x)})\) elements.

The local ring map \(\mathcal{O}_{S, s} \to \mathcal{O}_{X, x}\) is flat and the \(\kappa(x)\)-vector space \[\Omega_{X_s/s, x} \otimes_{\mathcal{O}_{X_s, x}} \kappa(x) = \Omega_{X/S, x} \otimes_{\mathcal{O}_{X, x}} \kappa(x)\] can be generated by at most \(\dim_x(X_{f(x)})\) elements.

There exist affine opens \(U \subset X\), and \(V \subset S\) such that \(x \in U\), \(f(U) \subset V\) and the induced morphism \(f|_U : U \to V\) is standard smooth.

There exist affine opens \(\Spec(A) = U \subset X\) and \(\Spec(R) = V \subset S\) with \(x \in U\) corresponding to \(\mathfrak q \subset A\), and \(f(U) \subset V\) such that there exists a presentation \[A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)\] with \[g = \det \left( \begin{matrix} \partial f_1/\partial x_1 & \partial f_2/\partial x_1 & \ldots & \partial f_c/\partial x_1 \\ \partial f_1/\partial x_2 & \partial f_2/\partial x_2 & \ldots & \partial f_c/\partial x_2 \\ \ldots & \ldots & \ldots & \ldots \\ \partial f_1/\partial x_c & \partial f_2/\partial x_c & \ldots & \partial f_c/\partial x_c \end{matrix} \right)\] mapping to an element of \(A\) not in \(\mathfrak q\).

Proof

Note that if \(f\) is smooth at \(x\), then we see from Lemma 01V7 that (5) holds, and (6) is a slightly weakened version of (5). Moreover, \(f\) smooth implies that the ring map \(\mathcal{O}_{S, s} \to \mathcal{O}_{X, x}\) is flat (see Lemma 01VF) and that \(\Omega_{X/S}\) is finite locally free of rank equal to \(\dim_x(X_s)\) (see Lemma 02G1). Thus (1) implies (3) and (4). By Lemma 01VB we also see that (1) implies (2).

By Lemma 01V0 the module of differentials \(\Omega_{X_s/s}\) of the fibre \(X_s\) over \(\kappa(s)\) is the pullback of the module of differentials \(\Omega_{X/S}\) of \(X\) over \(S\). Hence the displayed equality in part (4) of the lemma. By Lemma 01V2 these modules are of finite type. Hence the minimal number of generators of the modules \(\Omega_{X/S, x}\) and \(\Omega_{X_s/s, x}\) is the same and equal to the dimension of this \(\kappa(x)\)-vector space by Nakayama’s Lemma (Algebra, Lemma 00DV). This in particular shows that (3) and (4) are equivalent.

Algebra, Lemma 00TF shows that (2) implies (1). Algebra, Lemma 00TT shows that (3) and (4) imply (2). Finally, (6) implies (5) see for example Algebra, Example 00T8 and (5) implies (1) by Algebra, Lemma 00T7.

Lemma

Let \[\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^f \\ S' \ar[r]^g & S }\] be a cartesian diagram of schemes. Let \(W \subset X\), resp. \(W' \subset X'\) be the open subscheme of points where \(f\), resp. \(f'\) is smooth. Then \(W' = (g')^{-1}(W)\) if

\(f\) is flat and locally of finite presentation, or

\(f\) is locally of finite presentation and \(g\) is flat.

Proof

Assume first that \(f\) locally of finite type. Consider the set \[T = \{x \in X \mid X_{f(x)}\text{ is smooth over }\kappa(f(x))\text{ at }x\}\] and the corresponding set \(T' \subset X'\) for \(f'\). Then we claim \(T' = (g')^{-1}(T)\). Namely, let \(s' \in S'\) be a point, and let \(s = g(s')\). Then we have \[X'_{s'} = \Spec(\kappa(s')) \times_{\Spec(\kappa(s))} X_s\] In other words the fibres of the base change are the base changes of the fibres. Hence the claim is equivalent to Algebra, Lemma 02UQ.

Thus case (1) follows because in case (1) \(T\) is the (open) set of points where \(f\) is smooth by Lemma 01V9.

In case (2) let \(x' \in W'\). Then \(g'\) is flat at \(x'\) (Lemma 01U8) and \(g \circ f'\) is flat at \(x'\) (Lemma 01U6). It follows that \(f\) is flat at \(x = g'(x')\) by Lemma 02JZ. On the other hand, since \(x' \in T'\) (Lemma 01VB) we see that \(x \in T\). Hence \(f\) is smooth at \(x\) by Lemma 01V9.

Here is a lemma that actually uses the vanishing of \(H^{-1}\) of the naive cotangent complex for a smooth ring map.

Lemma

Let \(f : X \to Y\), \(g : Y \to S\) be morphisms of schemes. Assume \(f\) is smooth. Then \[0 \to f^*\Omega_{Y/S} \to \Omega_{X/S} \to \Omega_{X/Y} \to 0\] (see Lemma 01UX) is short exact.

Proof

The algebraic version of this lemma is the following: Given ring maps \(A \to B \to C\) with \(B \to C\) smooth, then the sequence \[0 \to C \otimes_B \Omega_{B/A} \to \Omega_{C/A} \to \Omega_{C/B} \to 0\] of Algebra, Lemma 00RS is exact. This is Algebra, Lemma 04B2.

Lemma

Let \(i : Z \to X\) be an immersion of schemes over \(S\). Assume that \(Z\) is smooth over \(S\). Then the canonical exact sequence \[0 \to \mathcal{C}_{Z/X} \to i^*\Omega_{X/S} \to \Omega_{Z/S} \to 0\] of Lemma 01UZ is short exact.

Proof

The algebraic version of this lemma is the following: Given ring maps \(A \to B \to C\) with \(A \to C\) smooth and \(B \to C\) surjective with kernel \(J\), then the sequence \[0 \to J/J^2 \to C \otimes_B \Omega_{B/A} \to \Omega_{C/A} \to 0\] of Algebra, Lemma 00RU is exact. This is Algebra, Lemma 06A8.

Lemma

Let \[\xymatrix{ Z \ar[r]_i \ar[rd]_j & X \ar[d] \\ & Y }\] be a commutative diagram of schemes where \(i\) and \(j\) are immersions and \(X \to Y\) is smooth. Then the canonical exact sequence \[0 \to \mathcal{C}_{Z/Y} \to \mathcal{C}_{Z/X} \to i^*\Omega_{X/Y} \to 0\] of Lemma 067L is exact.

Proof

The algebraic version of this lemma is the following: Given ring maps \(A \to B \to C\) with \(A \to C\) surjective and \(A \to B\) smooth, then the sequence \[0 \to I/I^2 \to J/J^2 \to C \otimes_B \Omega_{B/A} \to 0\] of Algebra, Lemma 065V is exact. This is Algebra, Lemma 06A9.

Lemma

Let \[\xymatrix{ X \ar[rr]_f \ar[rd]_p & & Y \ar[dl]^q \\ & S }\] be a commutative diagram of morphisms of schemes. Assume that

\(f\) is surjective, and smooth,

\(p\) is smooth, and

\(q\) is locally of finite presentation¹⁰.

Then \(q\) is smooth.

Proof

By Lemma 02JZ we see that \(q\) is flat. Pick a point \(y \in Y\). Pick a point \(x \in X\) mapping to \(y\). Suppose \(f\) has relative dimension \(a\) at \(x\) and \(p\) has relative dimension \(b\) at \(x\). By Lemma 02G1 this means that \(\Omega_{X/S, x}\) is free of rank \(b\) and \(\Omega_{X/Y, x}\) is free of rank \(a\). By the short exact sequence of Lemma 02K4 this means that \((f^*\Omega_{Y/S})_x\) is free of rank \(b - a\). By Nakayama’s Lemma this implies that \(\Omega_{Y/S, y}\) can be generated by \(b - a\) elements. Also, by Lemma 02JS we see that \(\dim_y(Y_s) = b - a\). Hence we conclude that \(Y \to S\) is smooth at \(y\) by Lemma 01V9 part (2).

In the situation of the following lemma the image of \(\sigma\) is locally on \(X\) cut out by a regular sequence, see Divisors, Lemma 067R.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(\sigma : S \to X\) be a section of \(f\). Let \(s \in S\) be a point such that \(f\) is smooth at \(x = \sigma(s)\). Then there exist affine open neighbourhoods \(\Spec(A) = U \subset S\) of \(s\) and \(\Spec(B) = V \subset X\) of \(x\) such that

\(f(V) \subset U\) and \(\sigma(U) \subset V\),

with \(I = \Ker(\sigma^\# : B \to A)\) the module \(I/I^2\) is a free \(A\)-module, and

\(B^\wedge \cong A[[x_1, \ldots, x_d]]\) as \(A\)-algebras where \(B^\wedge\) denotes the completion of \(B\) with respect to \(I\).

Proof

Pick an affine open \(U \subset S\) containing \(s\) Pick an affine open \(V \subset f^{-1}(U)\) containing \(x\). Pick an affine open \(U' \subset \sigma^{-1}(V)\) containing \(s\). Note that \(V' = f^{-1}(U') \cap V\) is affine as it is equal to the fibre product \(V' = U' \times_U V\). Then \(U'\) and \(V'\) satisfy (1). Write \(U' = \Spec(A')\) and \(V' = \Spec(B')\). By Algebra, Lemma 05D5 the module \(I'/(I')^2\) is finite locally free as a \(A'\)-module. Hence after replacing \(U'\) by a smaller affine open \(U'' \subset U'\) and \(V'\) by \(V'' = V' \cap f^{-1}(U'')\) we obtain the situation where \(I''/(I'')^2\) is free, i.e., (2) holds. In this case (3) holds also by Algebra, Lemma 05D5.

The dimension of a scheme \(X\) at a point \(x\) (Properties, Definition 04MT) is just the dimension of \(X\) at \(x\) as a topological space, see Topology, Definition 0055. This is not the dimension of the local ring \(\mathcal{O}_{X,x}\), in general.

Lemma

Let \(f : X \to Y\) be a smooth morphism of locally Noetherian schemes. For every point \(x\) in \(X\) with image \(y\) in \(Y\), \[\dim_x(X) = \dim_y(Y) + \dim_x(X_y),\] where \(X_y\) denotes the fiber over \(y\).

Proof

After replacing \(X\) by an open neighborhood of \(x\), there is a natural number \(d\) such that all fibers of \(X \to Y\) have dimension \(d\) at every point, see Lemma 02G1. Then \(f\) is flat (Lemma 01VF), locally of finite type (Lemma 01VE), and of relative dimension \(d\). Hence the result follows from Lemma 0AFE.

Unramified morphisms

We briefly discuss unramified morphisms before the (perhaps) more interesting class of étale morphisms. Recall that a ring map \(R \to A\) is unramified if it is of finite type and \(\Omega_{A/R} = 0\) (this is the definition of [Henselian]). A ring map \(R \to A\) is called G-unramified if it is of finite presentation and \(\Omega_{A/R} = 0\) (this is the definition of [EGA]). See Algebra, Definition 00UT.

Definition

Let \(f : X \to S\) be a morphism of schemes.

We say that \(f\) is unramified at \(x \in X\) if there exists an affine open neighbourhood \(\Spec(A) = U \subset X\) of \(x\) and affine open \(\Spec(R) = V \subset S\) with \(f(U) \subset V\) such that the induced ring map \(R \to A\) is unramified.

We say that \(f\) is G-unramified at \(x \in X\) if there exists an affine open neighbourhood \(\Spec(A) = U \subset X\) of \(x\) and affine open \(\Spec(R) = V \subset S\) with \(f(U) \subset V\) such that the induced ring map \(R \to A\) is G-unramified.

We say that \(f\) is unramified if it is unramified at every point of \(X\).

We say that \(f\) is G-unramified if it is G-unramified at every point of \(X\).

Note that a G-unramified morphism is unramified. Hence any result for unramified morphisms implies the corresponding result for G-unramified morphisms. Moreover, if \(S\) is locally Noetherian then there is no difference between G-unramified and unramified morphisms, see Lemma 04EV. A pleasing feature of this definition is that the set of points where a morphism is unramified (resp. G-unramified) is automatically open.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Then

\(f\) is unramified if and only if \(f\) is locally of finite type and \(\Omega_{X/S} = 0\), and

\(f\) is G-unramified if and only if \(f\) is locally of finite presentation and \(\Omega_{X/S} = 0\).

Proof

By definition a ring map \(R \to A\) is unramified (resp. G-unramified) if and only if it is of finite type (resp. finite presentation) and \(\Omega_{A/R} = 0\). Hence the lemma follows directly from the definitions and Lemma 01UT.

Note that there is no separation or quasi-compactness hypotheses in the definition. Hence the question of being unramified is local in nature on the source. Here is the precise result.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent

The morphism \(f\) is unramified (resp. G-unramified).

For every affine open \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the ring map \(\mathcal{O}_S(V) \to \mathcal{O}_X(U)\) is unramified (resp. G-unramified).

There exists an open covering \(S = \bigcup_{j \in J} V_j\) and open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that each of the morphisms \(U_i \to V_j\), \(j\in J, i\in I_j\) is unramified (resp. G-unramified).

There exists an affine open covering \(S = \bigcup_{j \in J} V_j\) and affine open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that the ring map \(\mathcal{O}_S(V_j) \to \mathcal{O}_X(U_i)\) is unramified (resp. G-unramified), for all \(j\in J, i\in I_j\).

Moreover, if \(f\) is unramified (resp. G-unramified) then for any open subschemes \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the restriction \(f|_U : U \to V\) is unramified (resp. G-unramified).

Proof

This follows from Lemma 01ST if we show that the property “\(R \to A\) is unramified” is local. We check conditions (a), (b) and (c) of Definition 01SR. These properties are proved in Algebra, Lemma 00UV.

Lemma

The composition of two morphisms which are unramified is unramified. The same holds for G-unramified morphisms.

Proof

The proof of Lemma 02G6 shows that being unramified (resp. G-unramified) is a local property of ring maps. Hence the first statement of the lemma follows from Lemma 01SV combined with the fact that being unramified (resp. G-unramified) is a property of ring maps that is stable under composition, see Algebra, Lemma 00UV.

Lemma

The base change of a morphism which is unramified is unramified. The same holds for G-unramified morphisms.

Proof

The proof of Lemma 02G6 shows that being unramified (resp. G-unramified) is a local property of ring maps. Hence the lemma follows from Lemma 01SW combined with the fact that being unramified (resp. G-unramified) is a property of ring maps that is stable under base change, see Algebra, Lemma 00UV.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Assume \(S\) is locally Noetherian. Then \(f\) is unramified if and only if \(f\) is G-unramified.

Proof

Follows from the definitions and Lemma 01TX.

Lemma

Any open immersion is G-unramified.

Proof

This is true because an open immersion is a local isomorphism.

Lemma

A closed immersion \(i : Z \to X\) is unramified. It is G-unramified if and only if the associated quasi-coherent sheaf of ideals \(\mathcal{I} = \Ker(\mathcal{O}_X \to i_*\mathcal{O}_Z)\) is of finite type (as an \(\mathcal{O}_X\)-module).

Proof

Follows from Lemma 01TV and Algebra, Lemma 00UV.

Lemma

An unramified morphism is locally of finite type. A G-unramified morphism is locally of finite presentation.

Proof

An unramified ring map is of finite type by definition. A G-unramified ring map is of finite presentation by definition.

Lemma

Let \(f : X \to S\) be a morphism of schemes. If \(f\) is unramified at \(x\) then \(f\) is quasi-finite at \(x\). In particular, an unramified morphism is locally quasi-finite.

Proof

See Algebra, Lemma 02UR.

Lemma

Fibres of unramified morphisms.

Let \(X\) be a scheme over a field \(k\). The structure morphism \(X \to \Spec(k)\) is unramified if and only if \(X\) is a disjoint union of spectra of finite separable field extensions of \(k\).

If \(f : X \to S\) is an unramified morphism then for every \(s \in S\) the fibre \(X_s\) is a disjoint union of spectra of finite separable field extensions of \(\kappa(s)\).

Proof

Part (2) follows from part (1) and Lemma 02GA. Let us prove part (1). We first use Algebra, Lemma 02FM. This lemma implies that if \(X\) is a disjoint union of spectra of finite separable field extensions of \(k\) then \(X \to \Spec(k)\) is unramified. Conversely, suppose that \(X \to \Spec(k)\) is unramified. By Algebra, Lemma 00UW for every \(x \in X\) the residue field extension \(\kappa(x)/k\) is finite separable. Since \(X \to \Spec(k)\) is locally quasi-finite (Lemma 02V5) we see that all points of \(X\) are isolated closed points, see Lemma 01TH. Thus \(X\) is a discrete space, in particular the disjoint union of the spectra of its local rings. By Algebra, Lemma 00UW again these local rings are fields, and we win.

The following lemma characterizes an unramified morphisms as morphisms locally of finite type with unramified fibres.

Lemma

Let \(f : X \to S\) be a morphism of schemes.

If \(f\) is unramified then for any \(x \in X\) the field extension \(\kappa(x)/\kappa(f(x))\) is finite separable.

If \(f\) is locally of finite type, and for every \(s \in S\) the fibre \(X_s\) is a disjoint union of spectra of finite separable field extensions of \(\kappa(s)\) then \(f\) is unramified.

If \(f\) is locally of finite presentation, and for every \(s \in S\) the fibre \(X_s\) is a disjoint union of spectra of finite separable field extensions of \(\kappa(s)\) then \(f\) is G-unramified.

Proof

Follows from Algebra, Lemmas 00UW and 02FM.

Here is a characterization of unramified morphisms in terms of the diagonal morphism.

Lemma

Let \(f : X \to S\) be a morphism.

If \(f\) is unramified, then the diagonal morphism \(\Delta : X \to X \times_S X\) is an open immersion.

If \(f\) is locally of finite type and \(\Delta\) is an open immersion, then \(f\) is unramified.

If \(f\) is locally of finite presentation and \(\Delta\) is an open immersion, then \(f\) is G-unramified.

Proof

The first statement follows from Algebra, Lemma 02FL. The second statement from the fact that \(\Omega_{X/S}\) is the conormal sheaf of the diagonal morphism (Lemma 08S2) and hence clearly zero if \(\Delta\) is an open immersion.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(x \in X\). Set \(s = f(x)\). Assume \(f\) is locally of finite type (resp. locally of finite presentation). The following are equivalent:

The morphism \(f\) is unramified (resp. G-unramified) at \(x\).

The fibre \(X_s\) is unramified over \(\kappa(s)\) at \(x\).

The \(\mathcal{O}_{X, x}\)-module \(\Omega_{X/S, x}\) is zero.

The \(\mathcal{O}_{X_s, x}\)-module \(\Omega_{X_s/s, x}\) is zero.

The \(\kappa(x)\)-vector space \[\Omega_{X_s/s, x} \otimes_{\mathcal{O}_{X_s, x}} \kappa(x) = \Omega_{X/S, x} \otimes_{\mathcal{O}_{X, x}} \kappa(x)\] is zero.

We have \(\mathfrak m_s\mathcal{O}_{X, x} = \mathfrak m_x\) and the field extension \(\kappa(x)/\kappa(s)\) is finite separable.

Proof

Note that if \(f\) is unramified at \(x\), then we see that \(\Omega_{X/S} = 0\) in a neighbourhood of \(x\) by the definitions and the results on modules of differentials in Section 01UM. Hence (1) implies (3) and the vanishing of the right hand vector space in (5). It also implies (2) because by Lemma 01V0 the module of differentials \(\Omega_{X_s/s}\) of the fibre \(X_s\) over \(\kappa(s)\) is the pullback of the module of differentials \(\Omega_{X/S}\) of \(X\) over \(S\). This fact on modules of differentials also implies the displayed equality of vector spaces in part (4). By Lemma 01V2 the modules \(\Omega_{X/S, x}\) and \(\Omega_{X_s/s, x}\) are of finite type. Hence the modules \(\Omega_{X/S, x}\) and \(\Omega_{X_s/s, x}\) are zero if and only if the corresponding \(\kappa(x)\)-vector space in (4) is zero by Nakayama’s Lemma (Algebra, Lemma 00DV). This in particular shows that (3), (4) and (5) are equivalent. The support of \(\Omega_{X/S}\) is closed in \(X\), see Modules, Lemma 01BA. Assumption (3) implies that \(x\) is not in the support. Hence \(\Omega_{X/S}\) is zero in a neighbourhood of \(x\), which implies (1). The equivalence of (1) and (3) applied to \(X_s \to s\) implies the equivalence of (2) and (4). At this point we have seen that (1) – (5) are equivalent.

Alternatively you can use Algebra, Lemma 00UV to see the equivalence of (1) – (5) more directly.

The equivalence of (1) and (6) follows from Lemma 02G8. It also follows more directly from Algebra, Lemmas 00UW and 02FM.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Assume \(f\) locally of finite type. Formation of the open set \[\begin{align*} T & = \{x \in X \mid X_{f(x)}\text{ is unramified over }\kappa(f(x))\text{ at }x\} \\ & = \{x \in X \mid X\text{ is unramified over }S\text{ at }x\} \end{align*}\] commutes with arbitrary base change: For any morphism \(g : S' \to S\), consider the base change \(f' : X' \to S'\) of \(f\) and the projection \(g' : X' \to X\). Then the corresponding set \(T'\) for the morphism \(f'\) is equal to \(T' = (g')^{-1}(T)\). If \(f\) is assumed locally of finite presentation then the same holds for the open set of points where \(f\) is G-unramified.

Proof

Let \(s' \in S'\) be a point, and let \(s = g(s')\). Then we have \[X'_{s'} = \Spec(\kappa(s')) \times_{\Spec(\kappa(s))} X_s\] In other words the fibres of the base change are the base changes of the fibres. In particular \[\Omega_{X_s/s, x} \otimes_{\mathcal{O}_{X_s, x}} \kappa(x') = \Omega_{X'_{s'}/s', x'} \otimes_{\mathcal{O}_{X'_{s'}, x'}} \kappa(x')\] see Lemma 01V0. Whence \(x' \in T'\) if and only if \(x \in T\) by Lemma 02GF. The second part follows from the first because in that case \(T\) is the (open) set of points where \(f\) is G-unramified according to Lemma 02GF.

Lemma

Let \(f : X \to Y\) be a morphism of schemes over \(S\).

If \(X\) is unramified over \(S\), then \(f\) is unramified.

If \(X\) is G-unramified over \(S\) and \(Y\) is locally of finite type over \(S\), then \(f\) is G-unramified.

Proof

Assume that \(X\) is unramified over \(S\). By Lemma 01T8 we see that \(f\) is locally of finite type. By assumption we have \(\Omega_{X/S} = 0\). Hence \(\Omega_{X/Y} = 0\) by Lemma 01UX. Thus \(f\) is unramified. If \(X\) is G-unramified over \(S\) and \(Y\) is locally of finite type over \(S\), then by Lemma 02FV we see that \(f\) is locally of finite presentation and we conclude that \(f\) is G-unramified.

Lemma

Let \(S\) be a scheme. Let \(X\), \(Y\) be schemes over \(S\). Let \(f, g : X \to Y\) be morphisms over \(S\). Let \(x \in X\). Assume that

the structure morphism \(Y \to S\) is unramified,

\(f(x) = g(x)\) in \(Y\), say \(y = f(x) = g(x)\), and

the induced maps \(f^\sharp, g^\sharp : \kappa(y) \to \kappa(x)\) are equal.

Then there exists an open neighbourhood of \(x\) in \(X\) on which \(f\) and \(g\) are equal.

Proof

Consider the morphism \((f, g) : X \to Y \times_S Y\). By assumption (1) and Lemma 02GE the inverse image of \(\Delta_{Y/S}(Y)\) is open in \(X\). And assumptions (2) and (3) imply that \(x\) is in this open subset.

Étale morphisms

The Zariski topology of a scheme is a very coarse topology. This is particularly clear when looking at varieties over \(\mathbf{C}\). It turns out that declaring an étale morphism to be the analogue of a local isomorphism in topology introduces a much finer topology. On varieties over \(\mathbf{C}\) this topology gives rise to the “correct” Betti numbers when computing cohomology with finite coefficients. Another observable is that if \(f : X \to Y\) is an étale morphism of varieties over \(\mathbf{C}\), and if \(x\) is a closed point of \(X\), then \(f\) induces an isomorphism \(\mathcal{O}^{\wedge}_{Y, f(x)} \to \mathcal{O}^{\wedge}_{X, x}\) of complete local rings.

In this section we start our study of these matters. In fact we deliberately restrict our discussion to a minimum since we will discuss more interesting results elsewhere. Recall that a ring map \(R \to A\) is said to be étale if it is smooth and \(\Omega_{A/R} = 0\), see Algebra, Definition 00U1.

Definition

Let \(f : X \to S\) be a morphism of schemes.

We say that \(f\) is étale at \(x \in X\) if there exists an affine open neighbourhood \(\Spec(A) = U \subset X\) of \(x\) and affine open \(\Spec(R) = V \subset S\) with \(f(U) \subset V\) such that the induced ring map \(R \to A\) is étale.

We say that \(f\) is étale if it is étale at every point of \(X\).

A morphism of affine schemes \(f : X \to S\) is called standard étale if \(X \to S\) is isomorphic to \[\Spec(R[x]_h/(g)) \to \Spec(R)\] where \(R \to R[x]_h/(g)\) is a standard étale ring map, see Algebra, Definition 00UB, i.e., \(g\) is monic and \(g'\) invertible in \(R[x]_h/(g)\).

A morphism is étale if and only if it is smooth of relative dimension \(0\) (see Definition 02G2). A pleasing feature of the definition is that the set of points where a morphism is étale is automatically open.

Note that there is no separation or quasi-compactness hypotheses in the definition. Hence the question of being étale is local in nature on the source. Here is the precise result.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent

The morphism \(f\) is étale.

For every affine opens \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the ring map \(\mathcal{O}_S(V) \to \mathcal{O}_X(U)\) is étale.

There exists an open covering \(S = \bigcup_{j \in J} V_j\) and open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that each of the morphisms \(U_i \to V_j\), \(j\in J, i\in I_j\) is étale.

There exists an affine open covering \(S = \bigcup_{j \in J} V_j\) and affine open coverings \(f^{-1}(V_j) = \bigcup_{i \in I_j} U_i\) such that the ring map \(\mathcal{O}_S(V_j) \to \mathcal{O}_X(U_i)\) is étale, for all \(j\in J, i\in I_j\).

Moreover, if \(f\) is étale then for any open subschemes \(U \subset X\), \(V \subset S\) with \(f(U) \subset V\) the restriction \(f|_U : U \to V\) is étale.

Proof

This follows from Lemma 01ST if we show that the property “\(R \to A\) is étale” is local. We check conditions (a), (b) and (c) of Definition 01SR. These all follow from Algebra, Lemma 00U2.

Lemma

The composition of two morphisms which are étale is étale.

Proof

In the proof of Lemma 02GJ we saw that being étale is a local property of ring maps. Hence the first statement of the lemma follows from Lemma 01SV combined with the fact that being étale is a property of ring maps that is stable under composition, see Algebra, Lemma 00U2.

Lemma

The base change of a morphism which is étale is étale.

Proof

In the proof of Lemma 02GJ we saw that being étale is a local property of ring maps. Hence the lemma follows from Lemma 01SV combined with the fact that being étale is a property of ring maps that is stable under base change, see Algebra, Lemma 00U2.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(x \in X\). Then \(f\) is étale at \(x\) if and only if \(f\) is smooth and unramified at \(x\).

Proof

This follows immediately from the definitions.

Lemma

An étale morphism is locally quasi-finite.

Proof

By Lemma 02GK an étale morphism is unramified. By Lemma 02V5 an unramified morphism is locally quasi-finite.

Lemma

Fibres of étale morphisms.

Let \(X\) be a scheme over a field \(k\). The structure morphism \(X \to \Spec(k)\) is étale if and only if \(X\) is a disjoint union of spectra of finite separable field extensions of \(k\).

If \(f : X \to S\) is an étale morphism, then for every \(s \in S\) the fibre \(X_s\) is a disjoint union of spectra of finite separable field extensions of \(\kappa(s)\).

Proof

You can deduce this from Lemma 02G7 via Lemma 02GK above. Here is a direct proof.

We will use Algebra, Lemma 00U3. Hence it is clear that if \(X\) is a disjoint union of spectra of finite separable field extensions of \(k\) then \(X \to \Spec(k)\) is étale. Conversely, suppose that \(X \to \Spec(k)\) is étale. Then for any affine open \(U \subset X\) we see that \(U\) is a finite disjoint union of spectra of finite separable field extensions of \(k\). Hence all points of \(X\) are closed points (see Lemma 01TE for example). Thus \(X\) is a discrete space and we win.

The following lemma characterizes an étale morphism as a flat, finitely presented morphism with “étale fibres”.

Lemma

Let \(f : X \to S\) be a morphism of schemes. If \(f\) is flat, locally of finite presentation, and for every \(s \in S\) the fibre \(X_s\) is a disjoint union of spectra of finite separable field extensions of \(\kappa(s)\), then \(f\) is étale.

Proof

You can deduce this from Algebra, Lemma 00U6. Here is another proof.

By Lemma 02GL a fibre \(X_s\) is étale and hence smooth over \(s\). By Lemma 01V8 we see that \(X \to S\) is smooth. By Lemma 02G8 we see that \(f\) is unramified. We conclude by Lemma 02GK.

Lemma

Any open immersion is étale.

Proof

This is true because an open immersion is a local isomorphism.

Lemma

An étale morphism is syntomic.

Proof

See Algebra, Lemma 00TA and use that an étale morphism is the same as a smooth morphism of relative dimension \(0\).

Lemma

An étale morphism is locally of finite presentation.

Proof

True because an étale ring map is of finite presentation by definition.

Lemma

An étale morphism is flat.

Proof

Combine Lemmas 01UL and 02GQ.

Lemma

An étale morphism is open.

Proof

Combine Lemmas 02GS, 02GR, and 01UA.

The following lemma says locally any étale morphism is standard étale. This is actually kind of a tricky result to prove in complete generality. The tricky parts are hidden in the chapter on commutative algebra. Hence a standard étale morphism is a local model for a general étale morphism.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(x \in X\) be a point. Let \(V \subset S\) be an affine open neighbourhood of \(f(x)\). The following are equivalent

The morphism \(f\) is étale at \(x\).

There exist an affine open \(U \subset X\) with \(x \in U\) and \(f(U) \subset V\) such that the induced morphism \(f|_U : U \to V\) is standard étale (see Definition 02GI).

Proof

Follows from the definitions and Algebra, Proposition 00UE.

Here is a differential criterion of étaleness at a point. There are many variants of this result all of which may be useful at some point. We will just add them here as needed.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(x \in X\). Set \(s = f(x)\). Assume \(f\) is locally of finite presentation. The following are equivalent:

The morphism \(f\) is étale at \(x\).

The local ring map \(\mathcal{O}_{S, s} \to \mathcal{O}_{X, x}\) is flat and \(X_s \to \Spec(\kappa(s))\) is étale at \(x\).

The local ring map \(\mathcal{O}_{S, s} \to \mathcal{O}_{X, x}\) is flat and \(X_s \to \Spec(\kappa(s))\) is unramified at \(x\).

The local ring map \(\mathcal{O}_{S, s} \to \mathcal{O}_{X, x}\) is flat and the \(\mathcal{O}_{X, x}\)-module \(\Omega_{X/S, x}\) is zero.

The local ring map \(\mathcal{O}_{S, s} \to \mathcal{O}_{X, x}\) is flat and the \(\kappa(x)\)-vector space \[\Omega_{X_s/s, x} \otimes_{\mathcal{O}_{X_s, x}} \kappa(x) = \Omega_{X/S, x} \otimes_{\mathcal{O}_{X, x}} \kappa(x)\] is zero.

The local ring map \(\mathcal{O}_{S, s} \to \mathcal{O}_{X, x}\) is flat, we have \(\mathfrak m_s\mathcal{O}_{X, x} = \mathfrak m_x\) and the field extension \(\kappa(x)/\kappa(s)\) is finite separable.

There exist affine opens \(U \subset X\), and \(V \subset S\) such that \(x \in U\), \(f(U) \subset V\) and the induced morphism \(f|_U : U \to V\) is standard smooth of relative dimension \(0\).

There exist affine opens \(\Spec(A) = U \subset X\) and \(\Spec(R) = V \subset S\) with \(x \in U\) corresponding to \(\mathfrak q \subset A\), and \(f(U) \subset V\) such that there exists a presentation \[A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_n)\] with \[g = \det \left( \begin{matrix} \partial f_1/\partial x_1 & \partial f_2/\partial x_1 & \ldots & \partial f_n/\partial x_1 \\ \partial f_1/\partial x_2 & \partial f_2/\partial x_2 & \ldots & \partial f_n/\partial x_2 \\ \ldots & \ldots & \ldots & \ldots \\ \partial f_1/\partial x_n & \partial f_2/\partial x_n & \ldots & \partial f_n/\partial x_n \end{matrix} \right)\] mapping to an element of \(A\) not in \(\mathfrak q\).

There exist affine opens \(U \subset X\), and \(V \subset S\) such that \(x \in U\), \(f(U) \subset V\) and the induced morphism \(f|_U : U \to V\) is standard étale.

There exist affine opens \(\Spec(A) = U \subset X\) and \(\Spec(R) = V \subset S\) with \(x \in U\) corresponding to \(\mathfrak q \subset A\), and \(f(U) \subset V\) such that there exists a presentation \[A = R[x]_Q/(P) = R[x, 1/Q]/(P)\] with \(P, Q \in R[x]\), \(P\) monic and \(P' = \text{d}P/\text{d}x\) mapping to an element of \(A\) not in \(\mathfrak q\).

Proof

Use Lemma 02GT and the definitions to see that (1) implies all of the other conditions. For each of the conditions (2) – (10) combine Lemmas 01V9 and 02GF to see that (1) holds by showing \(f\) is both smooth and unramified at \(x\) and applying Lemma 02GK. Some details omitted.

Lemma

A morphism is étale at a point if and only if it is flat and G-unramified at that point. A morphism is étale if and only if it is flat and G-unramified.

Proof

This is clear from Lemmas 02GU and 02GF.

Lemma

Let \[\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^f \\ S' \ar[r]^g & S }\] be a cartesian diagram of schemes. Let \(W \subset X\), resp. \(W' \subset X'\) be the open subscheme of points where \(f\), resp. \(f'\) is étale. Then \(W' = (g')^{-1}(W)\) if

\(f\) is flat and locally of finite presentation, or

\(f\) is locally of finite presentation and \(g\) is flat.

Proof

Assume first that \(f\) locally of finite type. Consider the set \[T = \{x \in X \mid f\text{ is unramified at }x\}\] and the corresponding set \(T' \subset X'\) for \(f'\). Then \(T' = (g')^{-1}(T)\) by Lemma 0475.

Thus case (1) follows because in case (1) \(T\) is the (open) set of points where \(f\) is étale by Lemma 02GV.

In case (2) let \(x' \in W'\). Then \(g'\) is flat at \(x'\) (Lemma 01U8) and \(g \circ f'\) is flat at \(x'\) (Lemma 01U6). It follows that \(f\) is flat at \(x = g'(x')\) by Lemma 02JZ. On the other hand, since \(x' \in T'\) (Lemma 01VB) we see that \(x \in T\). Hence \(f\) is étale at \(x\) by Lemma 02GU.

Lemma

Let \(f : X \to Y\) be a morphism of schemes over \(S\). If \(X\) is étale over \(S\) and \(Y\) is unramified over \(S\), then \(f\) is étale.

Proof

Consider the factorization \(X \to X \times_S Y \to Y\), where the first arrow is given by \(\text{id}_X\) and \(f\) and the second arrow is the projection. We claim both arrows are étale and hence \(f\) is étale by Lemma 02GN. Namely, the projection is étale as it is the base change of \(X \to S\), see Lemma 02GO. The first arrow is the base change of the diagonal morphism \(Y \to Y \times_S Y\) because the square \[\xymatrix{ X \ar[d] \ar[r] & X \times_S Y \ar[d] \\ Y \ar[r] & Y \times_S Y }\] is cartesian. The diagonal \(Y \to Y \times_S Y\) is an open immersion by Lemma 02GE. The base change of an open immersion is an open immersion (Schemes, Lemma 01JY) and an open immersion is étale (Lemma 02GP).

Lemma

Let \(f : X \to Y\) be a morphism of schemes over \(S\). If \(X\) and \(Y\) are étale over \(S\), then \(f\) is étale.

Proof

Since \(Y \to S\) is unramified by Lemma 02GK we may apply Lemma 0HB1. See also Algebra, Lemma 00U7.

Lemma

Let \[\xymatrix{ X \ar[rr]_f \ar[rd]_p & & Y \ar[dl]^q \\ & S }\] be a commutative diagram of morphisms of schemes. Assume that

\(f\) is surjective, and étale,

\(p\) is étale, and

\(q\) is locally of finite presentation¹¹.

Then \(q\) is étale.

Proof

By Lemma 02K5 we see that \(q\) is smooth. Thus we only need to see that \(q\) has relative dimension \(0\). This follows from Lemma 02JS and the fact that \(f\) and \(p\) have relative dimension \(0\).

A final characterization of smooth morphisms is that a smooth morphism \(f : X \to S\) is locally the composition of an étale morphism by a projection \(\mathbf{A}_S^d \to S\).

Lemma

Let \(\varphi : X \to Y\) be a morphism of schemes. Let \(x \in X\). Let \(V \subset Y\) be an affine open neighbourhood of \(\varphi(x)\). If \(\varphi\) is smooth at \(x\), then there exists an integer \(d \geq 0\) and an affine open \(U \subset X\) with \(x \in U\) and \(\varphi(U) \subset V\) such that there exists a commutative diagram \[\xymatrix{ X \ar[d] & U \ar[l] \ar[d] \ar[r]_-\pi & \mathbf{A}^d_V \ar[ld] \\ Y & V \ar[l] }\] where \(\pi\) is étale.

Proof

By Lemma 01V7 we can find an affine open \(U\) as in the lemma such that \(\varphi|_U : U \to V\) is standard smooth. Write \(U = \Spec(A)\) and \(V = \Spec(R)\) so that we can write \[A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)\] with \[g = \det \left( \begin{matrix} \partial f_1/\partial x_1 & \partial f_2/\partial x_1 & \ldots & \partial f_c/\partial x_1 \\ \partial f_1/\partial x_2 & \partial f_2/\partial x_2 & \ldots & \partial f_c/\partial x_2 \\ \ldots & \ldots & \ldots & \ldots \\ \partial f_1/\partial x_c & \partial f_2/\partial x_c & \ldots & \partial f_c/\partial x_c \end{matrix} \right)\] mapping to an invertible element of \(A\). Then it is clear that \(R[x_{c + 1}, \ldots, x_n] \to A\) is standard smooth of relative dimension \(0\). Hence it is smooth of relative dimension \(0\). In other words the ring map \(R[x_{c + 1}, \ldots, x_n] \to A\) is étale. As \(\mathbf{A}^{n - c}_V = \Spec(R[x_{c + 1}, \ldots, x_n])\) the lemma with \(d = n - c\).

Relatively ample sheaves

Let \(X\) be a scheme and \(\mathcal{L}\) an invertible sheaf on \(X\). Then \(\mathcal{L}\) is ample on \(X\) if \(X\) is quasi-compact and every point of \(X\) is contained in an affine open of the form \(X_s\), where \(s \in \Gamma(X, \mathcal{L}^{\otimes n})\) and \(n \geq 1\), see Properties, Definition 01PS. We turn this into a relative notion as follows.

Definition

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. We say \(\mathcal{L}\) is relatively ample, or \(f\)-relatively ample, or ample on \(X/S\), or \(f\)-ample if \(f : X \to S\) is quasi-compact, and if for every affine open \(V \subset S\) the restriction of \(\mathcal{L}\) to the open subscheme \(f^{-1}(V)\) of \(X\) is ample.

We note that the existence of a relatively ample sheaf on \(X\) does not force the morphism \(X \to S\) to be of finite type.

Lemma

Let \(X \to S\) be a morphism of schemes. Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. Let \(n \geq 1\). Then \(\mathcal{L}\) is \(f\)-ample if and only if \(\mathcal{L}^{\otimes n}\) is \(f\)-ample.

Proof

This follows from Properties, Lemma 01PT.

Lemma

Let \(f : X \to S\) be a morphism of schemes. If there exists an \(f\)-ample invertible sheaf, then \(f\) is separated.

Proof

Being separated is local on the base (see Schemes, Lemma 01KP for example; it also follows easily from the definition). Hence we may assume \(S\) is affine and \(X\) has an ample invertible sheaf. In this case the result follows from Properties, Lemma 09MP.

There are many ways to characterize relatively ample invertible sheaves, analogous to the equivalent conditions in Properties, Proposition 01Q3. We will add these here as needed.

Lemma

Let \(f : X \to S\) be a quasi-compact morphism of schemes. Let \(\mathcal{L}\) be an invertible sheaf on \(X\). The following are equivalent:

The invertible sheaf \(\mathcal{L}\) is \(f\)-ample.

There exists an open covering \(S = \bigcup V_i\) such that each \(\mathcal{L}|_{f^{-1}(V_i)}\) is ample relative to \(f^{-1}(V_i) \to V_i\).

There exists an affine open covering \(S = \bigcup V_i\) such that each \(\mathcal{L}|_{f^{-1}(V_i)}\) is ample.

There exists a quasi-coherent graded \(\mathcal{O}_S\)-algebra \(\mathcal{A}\) and a map of graded \(\mathcal{O}_X\)-algebras \(\psi : f^*\mathcal{A} \to \bigoplus_{d \geq 0} \mathcal{L}^{\otimes d}\) such that \(U(\psi) = X\) and \[r_{\mathcal{L}, \psi} : X \longrightarrow \underline{\text{Proj}}_S(\mathcal{A})\] is an open immersion (see Constructions, Lemma 01O9 for notation).

The morphism \(f\) is quasi-separated and part (4) above holds with \(\mathcal{A} = f_*(\bigoplus_{d \geq 0} \mathcal{L}^{\otimes d})\) and \(\psi\) the adjunction mapping.

Same as (4) but just requiring \(r_{\mathcal{L}, \psi}\) to be an immersion.

Proof

It is immediate from the definition that (1) implies (2) and (2) implies (3). It is clear that (5) implies (4).

Assume (3) holds for the affine open covering \(S = \bigcup V_i\). We are going to show (5) holds. Since each \(f^{-1}(V_i)\) has an ample invertible sheaf we see that \(f^{-1}(V_i)\) is separated (Properties, Lemma 09MP). Hence \(f\) is separated. By Schemes, Lemma 01LC we see that \(\mathcal{A} = f_*(\bigoplus_{d \geq 0} \mathcal{L}^{\otimes d})\) is a quasi-coherent graded \(\mathcal{O}_S\)-algebra. Denote \(\psi : f^*\mathcal{A} \to \bigoplus_{d \geq 0} \mathcal{L}^{\otimes d}\) the adjunction mapping. The description of the open \(U(\psi)\) in Constructions, Section 01O8 and the definition of ampleness of \(\mathcal{L}|_{f^{-1}(V_i)}\) show that \(U(\psi) = X\). Moreover, Constructions, Lemma 01O9 part (3) shows that the restriction of \(r_{\mathcal{L}, \psi}\) to \(f^{-1}(V_i)\) is the same as the morphism from Properties, Lemma 01PZ which is an open immersion according to Properties, Lemma 01Q1. Hence (5) holds.

Let us show that (4) implies (1). Assume (4). Denote \(\pi : \underline{\text{Proj}}_S(\mathcal{A}) \to S\) the structure morphism. Choose \(V \subset S\) affine open. By Constructions, Definition 01O0 we see that \(\pi^{-1}(V) \subset \underline{\text{Proj}}_S(\mathcal{A})\) is equal to \(\text{Proj}(A)\) where \(A = \mathcal{A}(V)\) as a graded ring. Hence \(r_{\mathcal{L}, \psi}\) maps \(f^{-1}(V)\) isomorphically onto a quasi-compact open of \(\text{Proj}(A)\). Moreover, \(\mathcal{L}^{\otimes d}\) is isomorphic to the pullback of \(\mathcal{O}_{\text{Proj}(A)}(d)\) for some \(d \geq 1\). (See part (3) of Constructions, Lemma 01O9 and the final statement of Constructions, Lemma 01NK.) This implies that \(\mathcal{L}|_{f^{-1}(V)}\) is ample by Properties, Lemmas 01Q2 and 01PT.

Assume (6). By the equivalence of (1) - (5) above we see that the property of being relatively ample on \(X/S\) is local on \(S\). Hence we may assume that \(S\) is affine, and we have to show that \(\mathcal{L}\) is ample on \(X\). In this case the morphism \(r_{\mathcal{L}, \psi}\) is identified with the morphism, also denoted \(r_{\mathcal{L}, \psi} : X \to \text{Proj}(A)\) associated to the map \(\psi : A = \mathcal{A}(V) \to \Gamma_*(X, \mathcal{L})\). (See references above.) As above we also see that \(\mathcal{L}^{\otimes d}\) is the pullback of the sheaf \(\mathcal{O}_{\text{Proj}(A)}(d)\) for some \(d \geq 1\). Moreover, since \(X\) is quasi-compact we see that \(X\) gets identified with a closed subscheme of a quasi-compact open subscheme \(Y \subset \text{Proj}(A)\). By Constructions, Lemma 01MW (see also Properties, Lemma 01Q2) we see that \(\mathcal{O}_Y(d')\) is an ample invertible sheaf on \(Y\) for some \(d' \geq 1\). Since the restriction of an ample sheaf to a closed subscheme is ample, see Properties, Lemma 01PU we conclude that the pullback of \(\mathcal{O}_Y(d')\) is ample. Combining these results with Properties, Lemma 01PT we conclude that \(\mathcal{L}\) is ample as desired.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. Assume \(S\) affine. Then \(\mathcal{L}\) is \(f\)-relatively ample if and only if \(\mathcal{L}\) is ample on \(X\).

Proof

Immediate from Lemma 01VJ and the definitions.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Then \(f\) is quasi-affine if and only if \(\mathcal{O}_X\) is \(f\)-relatively ample.

Proof

Follows from Properties, Lemma 01QE and the definitions.

Lemma

Let \(f : X \to Y\) be a morphism of schemes, \(\mathcal{M}\) an invertible \(\mathcal{O}_Y\)-module, and \(\mathcal{L}\) an invertible \(\mathcal{O}_X\)-module.

If \(\mathcal{L}\) is \(f\)-ample and \(\mathcal{M}\) is ample, then \(\mathcal{L} \otimes f^*\mathcal{M}^{\otimes a}\) is ample for \(a \gg 0\).

If \(\mathcal{M}\) is ample and \(f\) quasi-affine, then \(f^*\mathcal{M}\) is ample.

Proof

Assume \(\mathcal{L}\) is \(f\)-ample and \(\mathcal{M}\) ample. By assumption \(Y\) and \(f\) are quasi-compact (see Definition 01VH and Properties, Definition 01PS). Hence \(X\) is quasi-compact. By Properties, Lemma 09MP the scheme \(Y\) is separated and by Lemma 01VI the morphism \(f\) is separated. Hence \(X\) is separated by Schemes, Lemma 01KU. Pick \(x \in X\). We can choose \(m \geq 1\) and \(t \in \Gamma(Y, \mathcal{M}^{\otimes m})\) such that \(Y_t\) is affine and \(f(x) \in Y_t\). Since \(\mathcal{L}\) restricts to an ample invertible sheaf on \(f^{-1}(Y_t) = X_{f^*t}\) we can choose \(n \geq 1\) and \(s \in \Gamma(X_{f^*t}, \mathcal{L}^{\otimes n})\) with \(x \in (X_{f^*t})_s\) with \((X_{f^*t})_s\) affine. By Properties, Lemma 01PW part (2) whose assumptions are satisfied by the above, there exists an integer \(e \geq 1\) and a section \(s' \in \Gamma(X, \mathcal{L}^{\otimes n} \otimes f^*\mathcal{M}^{\otimes em})\) which restricts to \(s(f^*t)^e\) on \(X_{f^*t}\). For any \(b > 0\) consider the section \(s'' = s'(f^*t)^b\) of \(\mathcal{L}^{\otimes n} \otimes f^*\mathcal{M}^{\otimes (e + b)m}\). Then \(X_{s''} = (X_{f^*t})_s\) is an affine open of \(X\) containing \(x\). Picking \(b\) such that \(n\) divides \(e + b\) we see \(\mathcal{L}^{\otimes n} \otimes f^*\mathcal{M}^{\otimes (e + b)m}\) is the \(n\)th power of \(\mathcal{L} \otimes f^*\mathcal{M}^{\otimes a}\) for some \(a\) and we can get any \(a\) divisible by \(m\) and big enough. Since \(X\) is quasi-compact a finite number of these affine opens cover \(X\). We conclude that for some \(a\) sufficiently divisible and large enough the invertible sheaf \(\mathcal{L} \otimes f^*\mathcal{M}^{\otimes a}\) is ample on \(X\). On the other hand, we know that \(\mathcal{M}^{\otimes c}\) (and hence its pullback to \(X\)) is globally generated for all \(c \gg 0\) by Properties, Proposition 01Q3. Thus \(\mathcal{L} \otimes f^*\mathcal{M}^{\otimes a + c}\) is ample (Properties, Lemma 0890) for \(c \gg 0\) and (1) is proved.

Part (2) follows from Lemma 0891, Properties, Lemma 01PT, and part (1).

Lemma

Let \(g : Y \to S\) and \(f : X \to Y\) be morphisms of schemes. Let \(\mathcal{M}\) be an invertible \(\mathcal{O}_Y\)-module. Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. If \(S\) is quasi-compact, \(\mathcal{M}\) is \(g\)-ample, and \(\mathcal{L}\) is \(f\)-ample, then \(\mathcal{L} \otimes f^*\mathcal{M}^{\otimes a}\) is \(g \circ f\)-ample for \(a \gg 0\).

Proof

Let \(S = \bigcup_{i = 1, \ldots, n} V_i\) be a finite affine open covering. By Lemma 01VJ it suffices to prove that \(\mathcal{L} \otimes f^*\mathcal{M}^{\otimes a}\) is ample on \((g \circ f)^{-1}(V_i)\) for \(i = 1, \ldots, n\). Thus the lemma follows from Lemma 0892.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. Let \(S' \to S\) be a morphism of schemes. Let \(f' : X' \to S'\) be the base change of \(f\) and denote \(\mathcal{L}'\) the pullback of \(\mathcal{L}\) to \(X'\). If \(\mathcal{L}\) is \(f\)-ample, then \(\mathcal{L}'\) is \(f'\)-ample.

Proof

By Lemma 01VJ it suffices to find an affine open covering \(S' = \bigcup U'_i\) such that \(\mathcal{L}'\) restricts to an ample invertible sheaf on \((f')^{-1}(U_i')\) for all \(i\). We may choose \(U'_i\) mapping into an affine open \(U_i \subset S\). In this case the morphism \((f')^{-1}(U'_i) \to f^{-1}(U_i)\) is affine as a base change of the affine morphism \(U'_i \to U_i\) (Lemma 01SD). Thus \(\mathcal{L}'|_{(f')^{-1}(U'_i)}\) is ample by Lemma 0892.

Lemma

Let \(g : Y \to S\) and \(f : X \to Y\) be morphisms of schemes. Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. If \(\mathcal{L}\) is \(g \circ f\)-ample and \(f\) is quasi-compact¹² then \(\mathcal{L}\) is \(f\)-ample.

Proof

Assume \(f\) is quasi-compact and \(\mathcal{L}\) is \(g \circ f\)-ample. Let \(U \subset S\) be an affine open and let \(V \subset Y\) be an affine open with \(g(V) \subset U\). Then \(\mathcal{L}|_{(g \circ f)^{-1}(U)}\) is ample on \((g \circ f)^{-1}(U)\) by assumption. Since \(f^{-1}(V) \subset (g \circ f)^{-1}(U)\) we see that \(\mathcal{L}|_{f^{-1}(V)}\) is ample on \(f^{-1}(V)\) by Properties, Lemma 0B3E. Namely, \(f^{-1}(V) \to (g \circ f)^{-1}(U)\) is a quasi-compact open immersion by Schemes, Lemma 03GI as \((g \circ f)^{-1}(U)\) is separated (Properties, Lemma 09MP) and \(f^{-1}(V)\) is quasi-compact (as \(f\) is quasi-compact). Thus we conclude that \(\mathcal{L}\) is \(f\)-ample by Lemma 01VJ.

Very ample sheaves

Recall that given a quasi-coherent sheaf \(\mathcal{E}\) on a scheme \(S\) the projective bundle associated to \(\mathcal{E}\) is the morphism \(\mathbf{P}(\mathcal{E}) \to S\), where \(\mathbf{P}(\mathcal{E}) = \underline{\text{Proj}}_S(\text{Sym}(\mathcal{E}))\), see Constructions, Definition 01OB.

Definition

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. We say \(\mathcal{L}\) is relatively very ample or more precisely \(f\)-relatively very ample, or very ample on \(X/S\), or \(f\)-very ample if there exist a quasi-coherent \(\mathcal{O}_S\)-module \(\mathcal{E}\) and an immersion \(i : X \to \mathbf{P}(\mathcal{E})\) over \(S\) such that \(\mathcal{L} \cong i^*\mathcal{O}_{\mathbf{P}(\mathcal{E})}(1)\).

Since there is no assumption of quasi-compactness in this definition it is not true in general that a relatively very ample invertible sheaf is a relatively ample invertible sheaf.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. If \(f\) is quasi-compact and \(\mathcal{L}\) is a relatively very ample invertible sheaf, then \(\mathcal{L}\) is a relatively ample invertible sheaf.

Proof

By definition there exists quasi-coherent \(\mathcal{O}_S\)-module \(\mathcal{E}\) and an immersion \(i : X \to \mathbf{P}(\mathcal{E})\) over \(S\) such that \(\mathcal{L} \cong i^*\mathcal{O}_{\mathbf{P}(\mathcal{E})}(1)\). Set \(\mathcal{A} = \text{Sym}(\mathcal{E})\), so \(\mathbf{P}(\mathcal{E}) = \underline{\text{Proj}}_S(\mathcal{A})\) by definition. The graded \(\mathcal{O}_S\)-algebra \(\mathcal{A}\) comes equipped with a map \[\psi : \mathcal{A} \to \bigoplus\nolimits_{n \geq 0} \pi_*\mathcal{O}_{\mathbf{P}(\mathcal{E})}(n) \to \bigoplus\nolimits_{n \geq 0} f_*\mathcal{L}^{\otimes n}\] where the second arrow uses the identification \(\mathcal{L} \cong i^*\mathcal{O}_{\mathbf{P}(\mathcal{E})}(1)\). By adjointness of \(f_*\) and \(f^*\) we get a morphism \(\psi : f^*\mathcal{A} \to \bigoplus_{n \geq 0}\mathcal{L}^{\otimes n}\). We omit the verification that the morphism \(r_{\mathcal{L}, \psi}\) associated to this map is exactly the immersion \(i\). Hence the result follows from part (6) of Lemma 01VJ.

To arrive at the correct converse of this lemma we ask whether given a relatively ample invertible sheaf \(\mathcal{L}\) there exists an integer \(n \geq 1\) such that \(\mathcal{L}^{\otimes n}\) is relatively very ample? In general this is false. There are several things that prevent this from being true:

Even if \(S\) is affine, it can happen that no finite integer \(n\) works because \(X \to S\) is not of finite type, see Example 01VO.
The base not being quasi-compact means the result can be prevented from being true even with \(f\) finite type. Namely, given a field \(k\) there exists a scheme \(X_d\) of finite type over \(k\) with an ample invertible sheaf \(\mathcal{O}_{X_d}(1)\) so that the smallest tensor power of \(\mathcal{O}_{X_d}(1)\) which is very ample is the \(d\)th power. See Example 01VP. Taking \(f\) to be the disjoint union of the schemes \(X_d\) mapping to the disjoint union of copies of \(\Spec(k)\) gives an example.

To see our version of the converse take a look at Lemma 01VU below. We will do some preliminary work before proving it.

Example

Let \(S\) be a scheme. Let \(\mathcal{A}\) be a quasi-coherent graded \(\mathcal{O}_S\)-algebra generated by \(\mathcal{A}_1\) over \(\mathcal{A}_0\). Set \(X = \underline{\text{Proj}}_S(\mathcal{A})\). In this case \(\mathcal{O}_X(1)\) is invertible and very ample on \(X/S\). Namely, the morphism associated to the graded \(\mathcal{O}_S\)-algebra map \[\text{Sym}_{\mathcal{O}_X}^*(\mathcal{A}_1) \longrightarrow \mathcal{A}\] is a closed immersion \(X \to \mathbf{P}(\mathcal{A}_1)\) which pulls back \(\mathcal{O}_{\mathbf{P}(\mathcal{A}_1)}(1)\) to \(\mathcal{O}_X(1)\), see Constructions, Lemma 07ZK.

Example

Let \(k\) be a field. Consider the graded \(k\)-algebra \[A = k[U, V, Z_1, Z_2, Z_3, \ldots]/I \quad \text{with} \quad I = (U^2 - Z_1^2, U^4 - Z_2^2, U^6 - Z_3^2, \ldots)\] with grading given by \(\deg(U) = \deg(V) = \deg(Z_1) = 1\) and \(\deg(Z_d) = d\). Note that \(X = \text{Proj}(A)\) is covered by \(D_{+}(U)\) and \(D_{+}(V)\). Hence the sheaves \(\mathcal{O}_X(n)\) are all invertible and isomorphic to \(\mathcal{O}_X(1)^{\otimes n}\). In particular \(\mathcal{O}_X(1)\) is ample and \(f\)-ample for the morphism \(f : X \to \Spec(k)\). We claim that no power of \(\mathcal{O}_X(1)\) is \(f\)-relatively very ample. Namely, it is easy to see that \(\Gamma(X, \mathcal{O}_X(n))\) is the degree \(n\) summand of the algebra \(A\). Hence if \(\mathcal{O}_X(n)\) were very ample, then \(X\) would be a closed subscheme of a projective space over \(k\) and hence of finite type over \(k\). On the other hand \(D_{+}(V)\) is the spectrum of \(k[t, t_1, t_2, \ldots]/(t^2 - t_1^2, t^4 - t_2^2, t^6 - t_3^2, \ldots)\) which is not of finite type over \(k\).

Example

Let \(k\) be an infinite field. Let \(\lambda_1, \lambda_2, \lambda_3, \ldots\) be pairwise distinct elements of \(k^*\). (This is not strictly necessary, and in fact the example works perfectly well even if all \(\lambda_i\) are equal to \(1\).) Consider the graded \(k\)-algebra \[A_d = k[U, V, Z]/I_d \quad \text{with} \quad I_d = (Z^2 - \prod\nolimits_{i = 1}^{2d} (U - \lambda_i V)).\] with grading given by \(\deg(U) = \deg(V) = 1\) and \(\deg(Z) = d\). Then \(X_d = \text{Proj}(A_d)\) has ample invertible sheaf \(\mathcal{O}_{X_d}(1)\). We claim that if \(\mathcal{O}_{X_d}(n)\) is very ample, then \(n \geq d\). The reason for this is that \(Z\) has degree \(d\), and hence \(\Gamma(X_d, \mathcal{O}_{X_d}(n)) = k[U, V]_n\) for \(n < d\). Details omitted.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{L}\) be an invertible sheaf on \(X\). If \(\mathcal{L}\) is relatively very ample on \(X/S\) then \(f\) is separated.

Proof

Being separated is local on the base (see Schemes, Section 01KH). An immersion is separated (see Schemes, Lemma 01L7). Hence the lemma follows since locally \(X\) has an immersion into the homogeneous spectrum of a graded ring which is separated, see Constructions, Lemma 01MC.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{L}\) be an invertible sheaf on \(X\). Assume \(f\) is quasi-compact. The following are equivalent

\(\mathcal{L}\) is relatively very ample on \(X/S\),

there exists an open covering \(S = \bigcup V_j\) such that \(\mathcal{L}|_{f^{-1}(V_j)}\) is relatively very ample on \(f^{-1}(V_j)/V_j\) for all \(j\),

there exists a quasi-coherent sheaf of graded \(\mathcal{O}_S\)-algebras \(\mathcal{A}\) generated in degree \(1\) over \(\mathcal{O}_S\) and a map of graded \(\mathcal{O}_X\)-algebras \(\psi : f^*\mathcal{A} \to \bigoplus_{n \geq 0} \mathcal{L}^{\otimes n}\) such that \(f^*\mathcal{A}_1 \to \mathcal{L}\) is surjective and the associated morphism \(r_{\mathcal{L}, \psi} : X \to \underline{\text{Proj}}_S(\mathcal{A})\) is an immersion, and

\(f\) is quasi-separated, the canonical map \(\psi : f^*f_*\mathcal{L} \to \mathcal{L}\) is surjective, and the associated map \(r_{\mathcal{L}, \psi} : X \to \mathbf{P}(f_*\mathcal{L})\) is an immersion.

Proof

It is clear that (1) implies (2). It is also clear that (4) implies (1); the hypothesis of quasi-separation in (4) is used to guarantee that \(f_*\mathcal{L}\) is quasi-coherent via Schemes, Lemma 01LC.

Assume (2). We will prove (4). Let \(S = \bigcup V_j\) be an open covering as in (2). Set \(X_j = f^{-1}(V_j)\) and \(f_j : X_j \to V_j\) the restriction of \(f\). We see that \(f\) is separated by Lemma 01VQ (as being separated is local on the base). By assumption there exists a quasi-coherent \(\mathcal{O}_{V_j}\)-module \(\mathcal{E}_j\) and an immersion \(i_j : X_j \to \mathbf{P}(\mathcal{E}_j)\) with \(\mathcal{L}|_{X_j} \cong i_j^*\mathcal{O}_{\mathbf{P}(\mathcal{E}_j)}(1)\). The morphism \(i_j\) corresponds to a surjection \(f_j^*\mathcal{E}_j \to \mathcal{L}|_{X_j}\), see Constructions, Section 01OA. This map is adjoint to a map \(\mathcal{E}_j \to f_*\mathcal{L}|_{V_j}\) such that the composition \[f_j^*\mathcal{E}_j \to (f^*f_*\mathcal{L})|_{X_j} \to \mathcal{L}|_{X_j}\] is surjective. We conclude that \(\psi : f^*f_*\mathcal{L} \to \mathcal{L}\) is surjective. Let \(r_{\mathcal{L}, \psi} : X \to \mathbf{P}(f_*\mathcal{L})\) be the associated morphism. We still have to show that \(r_{\mathcal{L}, \psi}\) is an immersion; we urge the reader to prove this for themselves. The \(\mathcal{O}_{V_j}\)-module map \(\mathcal{E}_j \to f_*\mathcal{L}|_{V_j}\) determines a homomorphism on symmetric algebras, which in turn defines a morphism \[\mathbf{P}(f_*\mathcal{L}|_{V_j}) \supset U_j \longrightarrow \mathbf{P}(\mathcal{E}_j)\] where \(U_j\) is the open subscheme of Constructions, Lemma 07ZG. The compatibility of \(\psi\) with \(\mathcal{E}_j \to f_*\mathcal{L}|_{V_j}\) shows that \(r_{\mathcal{L}, \psi}(X_j) \subset U_j\) and that there is a factorization \[\xymatrix{ X_j \ar[r]^-{r_{\mathcal{L}, \psi}} & U_j \ar[r] & \mathbf{P}(\mathcal{E}_j) }\] We omit the verification. This shows that \(r_{\mathcal{L}, \psi}\) is an immersion.

At this point we see that (1), (2) and (4) are equivalent. Clearly (4) implies (3). Assume (3). We will prove (1). Let \(\mathcal{A}\) be a quasi-coherent sheaf of graded \(\mathcal{O}_S\)-algebras generated in degree \(1\) over \(\mathcal{O}_S\). Consider the map of graded \(\mathcal{O}_S\)-algebras \(\text{Sym}(\mathcal{A}_1) \to \mathcal{A}\). This is surjective by hypothesis and hence induces a closed immersion \[\underline{\text{Proj}}_S(\mathcal{A}) \longrightarrow \mathbf{P}(\mathcal{A}_1)\] which pulls back \(\mathcal{O}(1)\) to \(\mathcal{O}(1)\), see Constructions, Lemma 07ZK. Hence it is clear that (3) implies (1).

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. Let \(S' \to S\) be a morphism of schemes. Let \(f' : X' \to S'\) be the base change of \(f\) and denote \(\mathcal{L}'\) the pullback of \(\mathcal{L}\) to \(X'\). If \(\mathcal{L}\) is \(f\)-very ample, then \(\mathcal{L}'\) is \(f'\)-very ample.

Proof

By Definition 01VM there exists there exist a quasi-coherent \(\mathcal{O}_S\)-module \(\mathcal{E}\) and an immersion \(i : X \to \mathbf{P}(\mathcal{E})\) over \(S\) such that \(\mathcal{L} \cong i^*\mathcal{O}_{\mathbf{P}(\mathcal{E})}(1)\). The base change of \(\mathbf{P}(\mathcal{E})\) to \(S'\) is the projective bundle associated to the pullback \(\mathcal{E}'\) of \(\mathcal{E}\) and the pullback of \(\mathcal{O}_{\mathbf{P}(\mathcal{E})}(1)\) is \(\mathcal{O}_{\mathbf{P}(\mathcal{E}')}(1)\), see Constructions, Lemma 01O3. Finally, the base change of an immersion is an immersion (Schemes, Lemma 01JY).

Ample and very ample sheaves relative to finite type morphisms

In fact most of the material in this section is about the notion of a (quasi-)projective morphism which we have not defined yet.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{L}\) be an invertible sheaf on \(X\). Assume that

the invertible sheaf \(\mathcal{L}\) is very ample on \(X/S\),

the morphism \(X \to S\) is of finite type, and

\(S\) is affine.

Then there exist an \(n \geq 0\) and an immersion \(i : X \to \mathbf{P}^n_S\) over \(S\) such that \(\mathcal{L} \cong i^*\mathcal{O}_{\mathbf{P}^n_S}(1)\).

Proof

Assume (1), (2) and (3). Condition (3) means \(S = \Spec(R)\) for some ring \(R\). Condition (1) means by definition there exists a quasi-coherent \(\mathcal{O}_S\)-module \(\mathcal{E}\) and an immersion \(\alpha : X \to \mathbf{P}(\mathcal{E})\) such that \(\mathcal{L} = \alpha^*\mathcal{O}_{\mathbf{P}(\mathcal{E})}(1)\). Write \(\mathcal{E} = \widetilde{M}\) for some \(R\)-module \(M\). Thus we have \[\mathbf{P}(\mathcal{E}) = \text{Proj}(\text{Sym}_R(M)).\] Since \(\alpha\) is an immersion, and since the topology of \(\text{Proj}(\text{Sym}_R(M))\) is generated by the standard opens \(D_{+}(f)\), \(f \in \text{Sym}_R^d(M)\), \(d \geq 1\), we can find for each \(x \in X\) an \(f \in \text{Sym}_R^d(M)\), \(d \geq 1\), with \(\alpha(x) \in D_{+}(f)\) such that \[\alpha|_{\alpha^{-1}(D_{+}(f))} : \alpha^{-1}(D_{+}(f)) \to D_{+}(f)\] is a closed immersion. Condition (2) implies \(X\) is quasi-compact. Hence we can find a finite collection of elements \(f_j \in \text{Sym}_R^{d_j}(M)\), \(d_j \geq 1\) such that for each \(f = f_j\) the displayed map above is a closed immersion and such that \(\alpha(X) \subset \bigcup D_{+}(f_j)\). Write \(U_j = \alpha^{-1}(D_{+}(f_j))\). Note that \(U_j\) is affine as a closed subscheme of the affine scheme \(D_{+}(f_j)\). Write \(U_j = \Spec(A_j)\). Condition (2) also implies that \(A_j\) is of finite type over \(R\), see Lemma 01T2. Choose finitely many \(x_{j, k} \in A_j\) which generate \(A_j\) as a \(R\)-algebra. Since \(\alpha|_{U_j}\) is a closed immersion we see that \(x_{j, k}\) is the image of an element \[f_{j, k}/f_j^{e_{j, k}} \in \text{Sym}_R(M)_{(f_j)} = \Gamma(D_{+}(f_j), \mathcal{O}_{\text{Proj}(\text{Sym}_R(M))}).\] Finally, choose \(n \geq 1\) and elements \(y_0, \ldots, y_n \in M\) such that each of the polynomials \(f_j, f_{j, k} \in \text{Sym}_R(M)\) is a polynomial in the elements \(y_t\) with coefficients in \(R\). Consider the graded ring map \[\psi : R[Y_0, \ldots, Y_n] \longrightarrow \text{Sym}_R(M), \quad Y_i \longmapsto y_i.\] Denote \(F_j\), \(F_{j, k}\) the elements of \(R[Y_0, \ldots, Y_n]\) such that \(\psi(F_j) = f_j\) and \(\psi(F_{j, k}) = f_{j, k}\). By Constructions, Lemma 01MY we obtain an open subscheme \[U(\psi) \subset \text{Proj}(\text{Sym}_R(M))\] and a morphism \(r_\psi : U(\psi) \to \mathbf{P}^n_R\). This morphism satisfies \(r_\psi^{-1}(D_{+}(F_j)) = D_{+}(f_j)\), and hence we see that \(\alpha(X) \subset U(\psi)\). Moreover, it is clear that \[i = r_\psi \circ \alpha : X \longrightarrow \mathbf{P}^n_R\] is still an immersion since \(i^\sharp(F_{j, k}/F_j^{e_{j, k}}) = x_{j, k} \in A_j = \Gamma(U_j, \mathcal{O}_X)\) by construction. Moreover, the morphism \(r_\psi\) comes equipped with a map \(\theta : r_\psi^*\mathcal{O}_{\mathbf{P}^n_R}(1) \to \mathcal{O}_{\text{Proj}(\text{Sym}_R(M))}(1)|_{U(\psi)}\) which is an isomorphism in this case (for construction \(\theta\) see lemma cited above; some details omitted). Since the original map \(\alpha\) was assumed to have the property that \(\mathcal{L} = \alpha^*\mathcal{O}_{\text{Proj}(\text{Sym}_R(M))}(1)\) we win.

Lemma

Let \(\pi : X \to S\) be a morphism of schemes. Assume that \(X\) is quasi-affine and that \(\pi\) is locally of finite type. Then there exist \(n \geq 0\) and an immersion \(i : X \to \mathbf{A}^n_S\) over \(S\).

Proof

Let \(A = \Gamma(X, \mathcal{O}_X)\). By assumption \(X\) is quasi-compact and is identified with an open subscheme of \(\Spec(A)\), see Properties, Lemma 01P9. Moreover, the set of opens \(X_f\), for those \(f \in A\) such that \(X_f\) is affine, forms a basis for the topology of \(X\), see the proof of Properties, Lemma 01P9. Hence we can find a finite number of \(f_j \in A\), \(j = 1, \ldots, m\) such that \(X = \bigcup X_{f_j}\), and such that \(\pi(X_{f_j}) \subset V_j\) for some affine open \(V_j \subset S\). By Lemma 01T2 the ring maps \(\mathcal{O}(V_j) \to \mathcal{O}(X_{f_j}) = A_{f_j}\) are of finite type. Thus we may choose \(a_1, \ldots, a_N \in A\) such that the elements \(a_1, \ldots, a_N, 1/f_j\) generate \(A_{f_j}\) over \(\mathcal{O}(V_j)\) for each \(j\). Take \(n = m + N\) and let \[i : X \longrightarrow \mathbf{A}^n_S\] be the morphism given by the global sections \(f_1, \ldots, f_m, a_1, \ldots, a_N\) of the structure sheaf of \(X\). Let \(D(x_j) \subset \mathbf{A}^n_S\) be the open subscheme where the \(j\)th coordinate function is nonzero. Then for \(1 \leq j \leq m\) we have \(i^{-1}(D(x_j)) = X_{f_j}\) and the induced morphism \(X_{f_j} \to D(x_j)\) factors through the affine open \(\Spec(\mathcal{O}(V_j)[x_1, \ldots, x_n, 1/x_j])\) of \(D(x_j)\). Since the ring map \(\mathcal{O}(V_j)[x_1, \ldots, x_n, 1/x_j] \to A_{f_j}\) is surjective by construction we conclude that \(i^{-1}(D(x_j)) \to D(x_j)\) is an immersion as desired.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{L}\) be an invertible sheaf on \(X\). Assume that

the invertible sheaf \(\mathcal{L}\) is ample on \(X\), and

the morphism \(X \to S\) is locally of finite type.

Then there exists a \(d_0 \geq 1\) such that for every \(d \geq d_0\) there exist an \(n \geq 0\) and an immersion \(i : X \to \mathbf{P}^n_S\) over \(S\) such that \(\mathcal{L}^{\otimes d} \cong i^*\mathcal{O}_{\mathbf{P}^n_S}(1)\).

Proof

Let \(A = \Gamma_*(X, \mathcal{L}) = \bigoplus_{d \geq 0} \Gamma(X, \mathcal{L}^{\otimes d})\). By Properties, Proposition 01Q3 the set of affine opens \(X_a\) with \(a \in A_{+}\) homogeneous forms a basis for the topology of \(X\). Hence we can find finitely many such elements \(a_0, \ldots, a_n \in A_{+}\) such that

we have \(X = \bigcup_{i = 0, \ldots, n} X_{a_i}\),

each \(X_{a_i}\) is affine, and

each \(X_{a_i}\) maps into an affine open \(V_i \subset S\).

By Lemma 01T2 we see that the ring maps \(\mathcal{O}_S(V_i) \to \mathcal{O}_X(X_{a_i})\) are of finite type. Hence we can find finitely many elements \(f_{ij} \in \mathcal{O}_X(X_{a_i})\), \(j = 1, \ldots, n_i\) which generate \(\mathcal{O}_X(X_{a_i})\) as an \(\mathcal{O}_S(V_i)\)-algebra. By Properties, Lemma 01PW we may write each \(f_{ij}\) as \(a_{ij}/a_i^{e_{ij}}\) for some \(a_{ij} \in A_{+}\) homogeneous. Let \(N\) be a positive integer which is a common multiple of all the degrees of the elements \(a_i\), \(a_{ij}\). Consider the elements \[a_i^{N/\deg(a_i)}, \ a_{ij}a_i^{(N/\deg(a_i)) - e_{ij}} \in A_N.\] By construction these generate the invertible sheaf \(\mathcal{L}^{\otimes N}\) over \(X\). Hence they give rise to a morphism \[j : X \longrightarrow \mathbf{P}_S^{m} \quad \text{with } m = n + \sum n_i\] over \(S\), see Constructions, Lemma 01NE and Definition 01NF. Moreover, \(j^*\mathcal{O}_{\mathbf{P}_S}(1) = \mathcal{L}^{\otimes N}\). We name the homogeneous coordinates \(T_0, \ldots, T_n, T_{ij}\) instead of \(T_0, \ldots, T_m\). For \(i = 0, \ldots, n\) we have \(j^{-1}(D_{+}(T_i)) = X_{a_i}\). Moreover, pulling back the element \(T_{ij}/T_i\) via \(j^\sharp\) we get the element \(f_{ij} \in \mathcal{O}_X(X_{a_i})\). Hence the morphism \(j\) restricted to \(X_{a_i}\) gives a closed immersion of \(X_{a_i}\) into the affine open \(D_{+}(T_i) \cap \mathbf{P}^m_{V_i}\) of \(\mathbf{P}^N_S\). Hence we conclude that the morphism \(j\) is an immersion. This implies the lemma holds for some \(d\) and \(n\) which is enough in virtually all applications.

This proves that for one \(d_2 \geq 1\) (namely \(d_2 = N\) above), some \(m \geq 0\) there exists some immersion \(j : X \to \mathbf{P}^m_S\) given by global sections \(s'_0, \ldots, s'_m \in \Gamma(X, \mathcal{L}^{\otimes d_2})\). By Properties, Proposition 01Q3 we know there exists an integer \(d_1\) such that \(\mathcal{L}^{\otimes d}\) is globally generated for all \(d \geq d_1\). Set \(d_0 = d_1 + d_2\). We claim that the lemma holds with this value of \(d_0\). Namely, given an integer \(d \geq d_0\) we may choose \(s''_1, \ldots, s''_t \in \Gamma(X, \mathcal{L}^{\otimes d - d_2})\) which generate \(\mathcal{L}^{\otimes d - d_2}\) over \(X\). Set \(k = (m + 1)t\) and denote \(s_0, \ldots, s_k\) the collection of sections \(s'_\alpha s''_\beta\), \(\alpha = 0, \ldots, m\), \(\beta = 1, \ldots, t\). These generate \(\mathcal{L}^{\otimes d}\) over \(X\) and therefore define a morphism \[i : X \longrightarrow \mathbf{P}^{k - 1}_S\] such that \(i^*\mathcal{O}_{\mathbf{P}^n_S}(1) \cong \mathcal{L}^{\otimes d}\). To see that \(i\) is an immersion, observe that \(i\) is the composition \[X \longrightarrow \mathbf{P}^m_S \times_S \mathbf{P}^{t - 1}_S \longrightarrow \mathbf{P}^{k - 1}_S\] where the first morphism is \((j, j')\) with \(j'\) given by \(s''_1, \ldots, s''_t\) and the second morphism is the Segre embedding (Constructions, Lemma 01WD). Since \(j\) is an immersion, so is \((j, j')\) (apply Lemma 07RK to \(X \to \mathbf{P}^m_S \times_S \mathbf{P}^{t - 1}_S \to \mathbf{P}^m_S\)). Thus \(i\) is a composition of immersions and hence an immersion (Schemes, Lemma 02V0).

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. Assume \(S\) affine and \(f\) of finite type. The following are equivalent

\(\mathcal{L}\) is ample on \(X\),

\(\mathcal{L}\) is \(f\)-ample,

\(\mathcal{L}^{\otimes d}\) is \(f\)-very ample for some \(d \geq 1\),

\(\mathcal{L}^{\otimes d}\) is \(f\)-very ample for all \(d \gg 1\),

for some \(d \geq 1\) there exist \(n \geq 1\) and an immersion \(i : X \to \mathbf{P}^n_S\) such that \(\mathcal{L}^{\otimes d} \cong i^*\mathcal{O}_{\mathbf{P}^n_S}(1)\), and

for all \(d \gg 1\) there exist \(n \geq 1\) and an immersion \(i : X \to \mathbf{P}^n_S\) such that \(\mathcal{L}^{\otimes d} \cong i^*\mathcal{O}_{\mathbf{P}^n_S}(1)\).

Proof

The equivalence of (1) and (2) is Lemma 01VK. The implication (2) \(\Rightarrow\) (6) is Lemma 01VS. Trivially (6) implies (5). As \(\mathbf{P}^n_S\) is a projective bundle over \(S\) (see Constructions, Lemma 01OE) we see that (5) implies (3) and (6) implies (4) from the definition of a relatively very ample sheaf. Trivially (4) implies (3). To finish we have to show that (3) implies (2) which follows from Lemma 01VN and Lemma 02NN.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. Assume \(S\) quasi-compact and \(f\) of finite type. The following are equivalent

\(\mathcal{L}\) is \(f\)-ample,

\(\mathcal{L}^{\otimes d}\) is \(f\)-very ample for some \(d \geq 1\),

\(\mathcal{L}^{\otimes d}\) is \(f\)-very ample for all \(d \gg 1\).

Proof

Trivially (3) implies (2). Lemma 01VN guarantees that (2) implies (1) since a morphism of finite type is quasi-compact by definition. Assume that \(\mathcal{L}\) is \(f\)-ample. Choose a finite affine open covering \(S = V_1 \cup \ldots \cup V_m\). Write \(X_i = f^{-1}(V_i)\). By Lemma 01VT above we see there exists a \(d_0\) such that \(\mathcal{L}^{\otimes d}\) is relatively very ample on \(X_i/V_i\) for all \(d \geq d_0\). Hence we conclude (1) implies (3) by Lemma 01VR.

The following two lemmas provide the most used and most useful characterizations of relatively very ample and relatively ample invertible sheaves when the morphism is of finite type.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{L}\) be an invertible sheaf on \(X\). Assume \(f\) is of finite type. The following are equivalent:

\(\mathcal{L}\) is \(f\)-relatively very ample, and

there exist an open covering \(S = \bigcup V_j\), for each \(j\) an integer \(n_j\), and immersions \[i_j : X_j = f^{-1}(V_j) = V_j \times_S X \longrightarrow \mathbf{P}^{n_j}_{V_j}\] over \(V_j\) such that \(\mathcal{L}|_{X_j} \cong i_j^*\mathcal{O}_{\mathbf{P}^{n_j}_{V_j}}(1)\).

Proof

We see that (1) implies (2) by taking an affine open covering of \(S\) and applying Lemma 02NP to each of the restrictions of \(f\) and \(\mathcal{L}\). We see that (2) implies (1) by Lemma 01VR.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{L}\) be an invertible sheaf on \(X\). Assume \(f\) is of finite type. The following are equivalent:

\(\mathcal{L}\) is \(f\)-relatively ample, and

there exist an open covering \(S = \bigcup V_j\), for each \(j\) an integers \(d_j \geq 1\), \(n_j \geq 0\), and immersions \[i_j : X_j = f^{-1}(V_j) = V_j \times_S X \longrightarrow \mathbf{P}^{n_j}_{V_j}\] over \(V_j\) such that \(\mathcal{L}^{\otimes d_j}|_{X_j} \cong i_j^*\mathcal{O}_{\mathbf{P}^{n_j}_{V_j}}(1)\).

Proof

We see that (1) implies (2) by taking an affine open covering of \(S\) and applying Lemma 01VT to each of the restrictions of \(f\) and \(\mathcal{L}\). We see that (2) implies (1) by Lemma 01VJ.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(\mathcal{N}\), \(\mathcal{L}\) be invertible \(\mathcal{O}_X\)-modules. Assume \(S\) is quasi-compact, \(f\) is of finite type, and \(\mathcal{L}\) is \(f\)-ample. Then \(\mathcal{N} \otimes_{\mathcal{O}_X} \mathcal{L}^{\otimes d}\) is \(f\)-very ample for all \(d \gg 1\).

Proof

By Lemma 02NQ we reduce to the case \(S\) is affine. Combining Lemma 01VT and Properties, Proposition 01Q3 we can find an integer \(d_0\) such that \(\mathcal{N} \otimes \mathcal{L}^{\otimes d_0}\) is globally generated. Choose global sections \(s_0, \ldots, s_n\) of \(\mathcal{N} \otimes \mathcal{L}^{\otimes d_0}\) which generate it. This determines a morphism \(j : X \to \mathbf{P}^n_S\) over \(S\). By Lemma 01VT we can also pick an integer \(d_1\) such that for all \(d \geq d_1\) there exist sections \(t_{d, 0}, \ldots, t_{d, n(d)}\) of \(\mathcal{L}^{\otimes d}\) which generate it and define an immersion \[j_d = \varphi_{\mathcal{L}^{\otimes d}, t_{d, 0}, \ldots, t_{d, n(d)}} : X \longrightarrow \mathbf{P}^{n(d)}_S\] over \(S\). Then for \(d \geq d_0 + d_1\) we can consider the morphism \[\varphi_{\mathcal{N} \otimes \mathcal{L}^{\otimes d}, s_j \otimes t_{d - d_0, i}} : X \longrightarrow \mathbf{P}^{(n + 1)(n(d - d_0) + 1) - 1}_S\] This morphism is an immersion as it is the composition \[X \to \mathbf{P}^n_S \times_S \mathbf{P}^{n(d - d_0)}_S \to \mathbf{P}^{(n + 1)(n(d - d_0) + 1) - 1}_S\] where the first morphism is \((j, j_{d - d_0})\) and the second is the Segre embedding (Constructions, Lemma 01WD). Since \(j\) is an immersion, so is \((j, j_{d - d_0})\) (apply Lemma 07RK). We have a composition of immersions and hence an immersion (Schemes, Lemma 02V0).

Quasi-projective morphisms

The discussion in the previous section suggests the following definitions. We take our definition of quasi-projective from [EGA]. The version with the letter “H” is the definition in [H].

Definition

Let \(f : X \to S\) be a morphism of schemes.

We say \(f\) is quasi-projective if \(f\) is of finite type and there exists an \(f\)-relatively ample invertible \(\mathcal{O}_X\)-module.

We say \(f\) is H-quasi-projective if there exists a quasi-compact immersion \(X \to \mathbf{P}^n_S\) over \(S\) for some \(n\).¹³

We say \(f\) is locally quasi-projective if there exists an open covering \(S = \bigcup V_j\) such that each \(f^{-1}(V_j) \to V_j\) is quasi-projective.

As this definition suggests the property of being quasi-projective is not local on \(S\). At a later stage we will be able to say more about the category of quasi-projective schemes, see More on Morphisms, Section 0B41.

Lemma

A base change of a quasi-projective morphism is quasi-projective.

Proof

This follows from Lemmas 01T4 and 0893.

Lemma

Let \(f : X \to Y\) and \(g : Y \to S\) be morphisms of schemes. If \(S\) is quasi-compact and \(f\) and \(g\) are quasi-projective, then \(g \circ f\) is quasi-projective.

Proof

This follows from Lemmas 01T3 and 0C4K.

Lemma

Let \(f : X \to S\) be a morphism of schemes. If \(f\) is quasi-projective, or H-quasi-projective or locally quasi-projective, then \(f\) is separated of finite type.

Proof

Omitted.

Lemma

A H-quasi-projective morphism is quasi-projective.

Proof

Omitted.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent:

The morphism \(f\) is locally quasi-projective.

There exists an open covering \(S = \bigcup V_j\) such that each \(f^{-1}(V_j) \to V_j\) is H-quasi-projective.

Proof

By Lemma 01VY we see that (2) implies (1). Assume (1). The question is local on \(S\) and hence we may assume \(S\) is affine, \(X\) of finite type over \(S\) and \(\mathcal{L}\) is a relatively ample invertible sheaf on \(X/S\). By Lemma 01VT we may assume \(\mathcal{L}\) is ample on \(X\). By Lemma 01VS we see that there exists an immersion of \(X\) into a projective space over \(S\), i.e., \(X\) is H-quasi-projective over \(S\) as desired.

Lemma

A quasi-affine morphism of finite type is quasi-projective.

Proof

This follows from Lemma 0891.

Lemma

Let \(g : Y \to S\) and \(f : X \to Y\) be morphisms of schemes. If \(g \circ f\) is quasi-projective and \(f\) is quasi-compact¹⁴, then \(f\) is quasi-projective.

Proof

Observe that \(f\) is of finite type by Lemma 01T8. Thus the lemma follows from Lemma 0C4L and the definitions.

Proper morphisms

The notion of a proper morphism plays an important role in algebraic geometry. An important example of a proper morphism will be the structure morphism \(\mathbf{P}^n_S \to S\) of projective \(n\)-space, and this is in fact the motivating example leading to the definition.

Definition

Let \(f : X \to S\) be a morphism of schemes. We say \(f\) is proper if \(f\) is separated, finite type, and universally closed.

The morphism from the affine line with zero doubled to the affine line is of finite type and universally closed, so the separation condition is necessary in the definition above. In the rest of this section we prove some of the basic properties of proper morphisms and of universally closed morphisms.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent:

The morphism \(f\) is universally closed.

There exists an open covering \(S = \bigcup V_j\) such that \(f^{-1}(V_j) \to V_j\) is universally closed for all indices \(j\).

Proof

This is clear from the definition.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent:

The morphism \(f\) is proper.

There exists an open covering \(S = \bigcup V_j\) such that \(f^{-1}(V_j) \to V_j\) is proper for all indices \(j\).

Proof

Omitted.

Lemma

The composition of proper morphisms is proper. The same is true for universally closed morphisms.

Proof

A composition of closed morphisms is closed. If \(X \to Y \to Z\) are universally closed morphisms and \(Z' \to Z\) is any morphism, then we see that \(Z' \times_Z X = (Z' \times_Z Y) \times_Y X \to Z' \times_Z Y\) is closed and \(Z' \times_Z Y \to Z'\) is closed. Hence the result for universally closed morphisms. We have seen that “separated” and “finite type” are preserved under compositions (Schemes, Lemma 01KU and Lemma 01T3). Hence the result for proper morphisms.

Lemma

The base change of a proper morphism is proper. The same is true for universally closed morphisms.

Proof

This is true by definition for universally closed morphisms. It is true for separated morphisms (Schemes, Lemma 01KU). It is true for morphisms of finite type (Lemma 01T4). Hence it is true for proper morphisms.

Lemma

A closed immersion is proper, hence a fortiori universally closed.

Proof

The base change of a closed immersion is a closed immersion (Schemes, Lemma 01JY). Hence it is universally closed. A closed immersion is separated (Schemes, Lemma 01L7). A closed immersion is of finite type (Lemma 01T5). Hence a closed immersion is proper.

Lemma

Suppose given a commutative diagram of schemes \[\xymatrix{ X \ar[rr] \ar[rd] & & Y \ar[ld] \\ & S & }\] with \(Y\) separated over \(S\).

If \(X \to S\) is universally closed, then the morphism \(X \to Y\) is universally closed.

If \(X\) is proper over \(S\), then the morphism \(X \to Y\) is proper.

In particular, in both cases the image of \(X\) in \(Y\) is closed.

Proof

Assume that \(X \to S\) is universally closed (resp. proper). We factor the morphism as \(X \to X \times_S Y \to Y\). The first morphism is a closed immersion, see Schemes, Lemma 01KS. Hence the first morphism is proper (Lemma 01W5). The projection \(X \times_S Y \to Y\) is the base change of a universally closed (resp. proper) morphism and hence universally closed (resp. proper), see Lemma 01W4. Thus \(X \to Y\) is universally closed (resp. proper) as the composition of universally closed (resp. proper) morphisms (Lemma 01W3).

The proof of the following lemma is due to Bjorn Poonen, see this location.

Lemma

A universally closed morphism of schemes is quasi-compact.

Proof

Let \(f : X \to S\) be a morphism. Assume that \(f\) is not quasi-compact. Our goal is to show that \(f\) is not universally closed. By Schemes, Lemma 01K4 there exists an affine open \(V \subset S\) such that \(f^{-1}(V)\) is not quasi-compact. To achieve our goal it suffices to show that \(f^{-1}(V) \to V\) is not universally closed, hence we may assume that \(S = \Spec(A)\) for some ring \(A\).

Write \(X = \bigcup_{i \in I} X_i\) where the \(X_i\) are affine open subschemes of \(X\). Let \(T = \Spec(A[y_i ; i \in I])\). Let \(T_i = D(y_i) \subset T\). Let \(Z\) be the closed set \((X \times_S T) - \bigcup_{i \in I} (X_i \times_S T_i)\). It suffices to prove that the image \(f_T(Z)\) of \(Z\) under \(f_T : X \times_S T \to T\) is not closed.

There exists a point \(s \in S\) such that there is no neighborhood \(U\) of \(s\) in \(S\) such that \(X_U\) is quasi-compact. Otherwise we could cover \(S\) with finitely many such \(U\) and Schemes, Lemma 01K4 would imply \(f\) quasi-compact. Fix such an \(s \in S\).

First we check that \(f_T(Z_s) \ne T_s\). Let \(t \in T\) be the point lying over \(s\) with \(\kappa(t) = \kappa(s)\) such that \(y_i = 1\) in \(\kappa(t)\) for all \(i\). Then \(t \in T_i\) for all \(i\), and the fiber of \(Z_s \to T_s\) above \(t\) is isomorphic to \((X - \bigcup_{i \in I} X_i)_s\), which is empty. Thus \(t \in T_s - f_T(Z_s)\).

Assume \(f_T(Z)\) is closed in \(T\). Then there exists an element \(g \in A[y_i; i \in I]\) with \(f_T(Z) \subset V(g)\) but \(t \not \in V(g)\). Hence the image of \(g\) in \(\kappa(t)\) is nonzero. In particular some coefficient of \(g\) has nonzero image in \(\kappa(s)\). Hence this coefficient is invertible on some neighborhood \(U\) of \(s\). Let \(J\) be the finite set of \(j \in I\) such that \(y_j\) appears in \(g\). Since \(X_U\) is not quasi-compact, we may choose a point \(x \in X - \bigcup_{j \in J} X_j\) lying above some \(u \in U\). Since \(g\) has a coefficient that is invertible on \(U\), we can find a point \(t' \in T\) lying above \(u\) such that \(t' \not \in V(g)\) and \(t' \in V(y_i)\) for all \(i \notin J\). This is true because \(V(y_i; i \in I, i \not\in J) = \Spec(A[y_j; j\in J])\) and the set of points of this scheme lying over \(u\) is bijective with \(\Spec(\kappa(u)[y_j; j \in J])\). In other words \(t' \notin T_i\) for each \(i \notin J\). By Schemes, Lemma 01JT we can find a point \(z\) of \(X \times_S T\) mapping to \(x \in X\) and to \(t' \in T\). Since \(x \not \in X_j\) for \(j \in J\) and \(t' \not \in T_i\) for \(i \in I \setminus J\) we see that \(z \in Z\). On the other hand \(f_T(z) = t' \not \in V(g)\) which contradicts \(f_T(Z) \subset V(g)\). Thus the assumption “\(f_T(Z)\) closed” is wrong and we conclude indeed that \(f_T\) is not closed, as desired.

The following lemma says that the image of a proper scheme (in a separated scheme of finite type over the base) is proper.

Lemma

Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of schemes over \(S\). If \(X\) is universally closed over \(S\) and \(f\) is surjective then \(Y\) is universally closed over \(S\). In particular, if also \(Y\) is separated and locally of finite type over \(S\), then \(Y\) is proper over \(S\).

Proof

Assume \(X\) is universally closed and \(f\) surjective. Denote \(p : X \to S\), \(q : Y \to S\) the structure morphisms. Let \(S' \to S\) be a morphism of schemes. The base change \(f' : X_{S'} \to Y_{S'}\) is surjective (Lemma 01S1), and the base change \(p' : X_{S'} \to S'\) is closed. If \(T \subset Y_{S'}\) is closed, then \((f')^{-1}(T) \subset X_{S'}\) is closed, hence \(p'((f')^{-1}(T)) = q'(T)\) is closed. So \(q'\) is closed. This proves the first statement. Thus \(Y \to S\) is quasi-compact by Lemma 04XU and hence \(Y \to S\) is proper by definition if in addition \(Y \to S\) is locally of finite type and separated.

Lemma

Suppose given a commutative diagram of schemes \[\xymatrix{ X \ar[rr]_h \ar[rd]_f & & Y \ar[ld]^g \\ & S }\] Assume

\(X \to S\) is a universally closed (for example proper) morphism, and

\(Y \to S\) is separated and locally of finite type.

Then the scheme theoretic image \(Z \subset Y\) of \(h\) is proper over \(S\) and \(X \to Z\) is surjective.

Proof

The scheme theoretic image of \(h\) is constructed in Section 01R5. Since \(f\) is quasi-compact (Lemma 04XU) we find that \(h\) is quasi-compact (Schemes, Lemma 03GI). Hence \(h(X) \subset Z\) is dense (Lemma 01R8). On the other hand \(h(X)\) is closed in \(Y\) (Lemma 01W6) hence \(X \to Z\) is surjective. Thus \(Z \to S\) is a proper (Lemma 03GN).

The target of a separated scheme under a surjective universally closed morphism is separated.

Lemma

Let \(S\) be a scheme. Let \(f : X \to Y\) be a surjective universally closed morphism of schemes over \(S\).

If \(X\) is quasi-separated, then \(Y\) is quasi-separated.

If \(X\) is separated, then \(Y\) is separated.

If \(X\) is quasi-separated over \(S\), then \(Y\) is quasi-separated over \(S\).

If \(X\) is separated over \(S\), then \(Y\) is separated over \(S\).

Proof

Parts (1) and (2) are a consequence of (3) and (4) for \(S = \Spec(\mathbf{Z})\) (see Schemes, Definition 01KK). Consider the commutative diagram \[\xymatrix{ X \ar[d] \ar[rr]_{\Delta_{X/S}} & & X \times_S X \ar[d] \\ Y \ar[rr]^{\Delta_{Y/S}} & & Y \times_S Y }\] The left vertical arrow is surjective (i.e., universally surjective). The right vertical arrow is universally closed as a composition of the universally closed morphisms \(X \times_S X \to X \times_S Y \to Y \times_S Y\). Hence it is also quasi-compact, see Lemma 04XU.

Assume \(X\) is quasi-separated over \(S\), i.e., \(\Delta_{X/S}\) is quasi-compact. If \(V \subset Y \times_S Y\) is a quasi-compact open, then \(V \times_{Y \times_S Y} X \to \Delta_{Y/S}^{-1}(V)\) is surjective and \(V \times_{Y \times_S Y} X\) is quasi-compact by our remarks above. We conclude that \(\Delta_{Y/S}\) is quasi-compact, i.e., \(Y\) is quasi-separated over \(S\).

Assume \(X\) is separated over \(S\), i.e., \(\Delta_{X/S}\) is a closed immersion. Then \(X \to Y \times_S Y\) is closed as a composition of closed morphisms. Since \(X \to Y\) is surjective, it follows that \(\Delta_{Y/S}(Y)\) is closed in \(Y \times_S Y\). Hence \(Y\) is separated over \(S\) by the discussion following Schemes, Definition 01KK.

Valuative criteria

We have already discussed the valuative criterion for universal closedness and for separatedness in Schemes, Sections 01KA and 01KY. In this section we will discuss some consequences and variants. In Limits, Section 0CM1 we will show that it suffices to consider discrete valuation rings when working with locally Noetherian schemes and morphisms of finite type.

Lemma

Let \(S\) be a scheme. Let \(f : X \to Y\) be a morphism of schemes over \(S\). Assume \(f\) is of finite type and quasi-separated. Then the following are equivalent

\(f\) is proper,

\(f\) satisfies the valuative criterion (Schemes, Definition 01KD),

given any commutative solid diagram \[\xymatrix{ \Spec(K) \ar[r] \ar[d] & X \ar[d] \\ \Spec(A) \ar[r] \ar@{-->}[ru] & Y }\] where \(A\) is a valuation ring with field of fractions \(K\), there exists a unique dotted arrow making the diagram commute.

Proof

Part (3) is a reformulation of (2). Thus the lemma is a formal consequence of Schemes, Proposition 01KF and Lemma 01KZ, Lemma 01L0, and the definitions.

One usually does not have to consider all possible diagrams when testing the valuative criterion. We will call a valuative criterion as in the next lemma a “refined valuative criterion”.

Lemma

Let \(f : X \to S\) and \(h : U \to X\) be morphisms of schemes. Assume that \(f\) and \(h\) are quasi-compact and that \(h(U)\) is dense in \(X\). If given any commutative solid diagram \[\xymatrix{ \Spec(K) \ar[r] \ar[d] & U \ar[r]^h & X \ar[d]^f \\ \Spec(A) \ar[rr] \ar@{-->}[rru] & & S }\] where \(A\) is a valuation ring with field of fractions \(K\), there exists a unique dotted arrow making the diagram commute, then \(f\) is universally closed. If moreover \(f\) is quasi-separated, then \(f\) is separated.

Proof

To prove \(f\) is universally closed we will verify the existence part of the valuative criterion for \(f\) which suffices by Schemes, Proposition 01KF. To do this, consider a commutative diagram \[\xymatrix{ \Spec(K) \ar[r] \ar[d] & X \ar[d] \\ \Spec(A) \ar[r] & S }\] where \(A\) is a valuation ring and \(K\) is the fraction field of \(A\). Note that since valuation rings and fields are reduced, we may replace \(U\), \(X\), and \(S\) by their respective reductions by Schemes, Lemma 0356. In this case the assumption that \(h(U)\) is dense means that the scheme theoretic image of \(h : U \to X\) is \(X\), see Lemma 056B. We may also replace \(S\) by an affine open through which the morphism \(\Spec(A) \to S\) factors. Thus we may assume that \(S = \Spec(R)\).

Let \(\Spec(B) \subset X\) be an affine open through which the morphism \(\Spec(K) \to X\) factors. Choose a polynomial algebra \(P\) over \(B\) and a \(B\)-algebra surjection \(P \to K\). Then \(\Spec(P) \to X\) is flat. Hence the scheme theoretic image of the morphism \(U \times_X \Spec(P) \to \Spec(P)\) is \(\Spec(P)\) by Lemma 081I. By Lemma 02JQ we can find a commutative diagram \[\xymatrix{ \Spec(K') \ar[r] \ar[d] & U \times_X \Spec(P) \ar[d] \\ \Spec(A') \ar[r] & \Spec(P) }\] where \(A'\) is a valuation ring and \(K'\) is the fraction field of \(A'\) such that the closed point of \(\Spec(A')\) maps to \(\Spec(K) \subset \Spec(P)\). In other words, there is a \(B\)-algebra map \(\varphi : K \to A'/\mathfrak m_{A'}\). Choose a valuation ring \(A'' \subset A'/\mathfrak m_{A'}\) dominating \(\varphi(A)\) with field of fractions \(K'' = A'/\mathfrak m_{A'}\) (Algebra, Lemma 00IA). We set \[C = \{\lambda \in A' \mid \lambda \bmod \mathfrak m_{A'} \in A''\}.\] which is a valuation ring by Algebra, Lemma 088Z. As \(C\) is an \(R\)-algebra with fraction field \(K'\), we obtain a commutative diagram \[\xymatrix{ \Spec(K') \ar[r] \ar[d] & U \ar[r] & X \ar[d] \\ \Spec(C) \ar[rr] \ar@{-->}[rru] & & S }\] as in the statement of the lemma. Thus a dotted arrow fitting into the diagram as indicated. By the uniqueness assumption of the lemma the composition \(\Spec(A') \to \Spec(C) \to X\) agrees with the given morphism \(\Spec(A') \to \Spec(P) \to \Spec(B) \subset X\). Hence the restriction of the morphism to the spectrum of \(C/\mathfrak m_{A'} = A''\) induces the given morphism \(\Spec(K'') = \Spec(A'/\mathfrak m_{A'}) \to \Spec(K) \to X\). Let \(x \in X\) be the image of the closed point of \(\Spec(A'') \to X\). The image of the induced ring map \(\mathcal{O}_{X, x} \to A''\) is a local subring which is contained in \(K \subset K''\). Since \(A\) is maximal for the relation of domination in \(K\) and since \(A \subset A''\), we have \(A = K \cap A''\). We conclude that \(\mathcal{O}_{X, x} \to A''\) factors through \(A \subset A''\). In this way we obtain our desired arrow \(\Spec(A) \to X\).

Finally, assume \(f\) is quasi-separated. Then \(\Delta : X \to X \times_S X\) is quasi-compact. Given a solid diagram \[\xymatrix{ \Spec(K) \ar[r] \ar[d] & U \ar[r]^h & X \ar[d]^\Delta \\ \Spec(A) \ar[rr] \ar@{-->}[rru] & & X \times_S X }\] where \(A\) is a valuation ring with field of fractions \(K\), there exists a unique dotted arrow making the diagram commute. Namely, the lower horizontal arrow is the same thing as a pair of morphisms \(\Spec(A) \to X\) which can serve as the dotted arrow in the diagram of the lemma. Thus the required uniqueness shows that the lower horizontal arrow factors through \(\Delta\). Hence we can apply the result we just proved to \(\Delta : X \to X \times_S X\) and \(h : U \to X\) and conclude that \(\Delta\) is universally closed. Clearly this means that \(f\) is separated.

Remark

The assumption on uniqueness of the dotted arrows in Lemma 0894 is necessary (details omitted). Of course, uniqueness is guaranteed if \(f\) is separated (Schemes, Lemma 01KZ).

Lemma

Let \(S\) be a scheme. Let \(X\), \(Y\) be schemes over \(S\). Let \(s \in S\) and \(x \in X\), \(y \in Y\) points over \(s\).

Let \(f, g : X \to Y\) be morphisms over \(S\) such that \(f(x) = g(x) = y\) and \(f^\sharp_x = g^\sharp_x : \mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}\). Then there is an open neighbourhood \(U \subset X\) with \(f|_U = g|_U\) in the following cases

\(Y\) is locally of finite type over \(S\),

\(X\) is integral,

\(X\) is locally Noetherian, or

\(X\) is reduced with finitely many irreducible components.

Let \(\varphi : \mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}\) be a local \(\mathcal{O}_{S, s}\)-algebra map. Then there exists an open neighbourhood \(U \subset X\) of \(x\) and a morphism \(f : U \to Y\) mapping \(x\) to \(y\) with \(f^\sharp_x = \varphi\) in the following cases

\(Y\) is locally of finite presentation over \(S\),

\(Y\) is locally of finite type and \(X\) is integral,

\(Y\) is locally of finite type and \(X\) is locally Noetherian, or

\(Y\) is locally of finite type and \(X\) is reduced with finitely many irreducible components.

Proof

Proof of (1). We may replace \(X\), \(Y\), \(S\) by suitable affine open neighbourhoods of \(x\), \(y\), \(s\) and reduce to the following algebra problem: given a ring \(R\), two \(R\)-algebra maps \(\varphi, \psi : B \to A\) such that

\(R \to B\) is of finite type, or \(A\) is a domain, or \(A\) is Noetherian, or \(A\) is reduced and has finitely many minimal primes,

the two maps \(B \to A_\mathfrak p\) are the same for some prime \(\mathfrak p \subset A\),

show that \(\varphi, \psi\) define the same map \(B \to A_g\) for a suitable \(g \in A\), \(g \not \in \mathfrak p\). If \(R \to B\) is of finite type, let \(t_1, \ldots, t_m \in B\) be generators of \(B\) as an \(R\)-algebra. For each \(j\) we can find \(g_j \in A\), \(g_j \not \in \mathfrak p\) such that \(\varphi(t_j)\) and \(\psi(t_j)\) have the same image in \(A_{g_j}\). Then we set \(g = \prod g_j\). In the other cases (if \(A\) is a domain, Noetherian, or reduced with finitely many minimal primes), we can find a \(g \in A\), \(g \not \in \mathfrak p\) such that \(A_g \subset A_\mathfrak p\). See Algebra, Lemma 0BX1. Thus the maps \(B \to A_g\) are equal as desired.

Proof of (2). To do this we may replace \(X\), \(Y\), and \(S\) by suitable affine opens. Say \(X = \Spec(A)\), \(Y = \Spec(B)\), and \(S = \Spec(R)\). Let \(\mathfrak p \subset A\) be the prime ideal corresponding to \(x\). Let \(\mathfrak q \subset B\) be the prime corresponding to \(y\). Then \(\varphi\) is a local \(R\)-algebra map \(\varphi : B_\mathfrak q \to A_\mathfrak p\). If \(R \to B\) is a ring map of finite presentation, then there exists a \(g \in A \setminus \mathfrak p\) and an \(R\)-algebra map \(B \to A_g\) such that \[\xymatrix{ B_\mathfrak q \ar[r]_\varphi & A_\mathfrak p \\ B \ar[u] \ar[r] & A_g \ar[u] }\] commutes, see Algebra, Lemmas 00QO and 00CR. The induced morphism \(\Spec(A_g) \to \Spec(B)\) works. If \(B\) is of finite type over \(R\), let \(t_1, \ldots, t_m \in B\) be generators of \(B\) as an \(R\)-algebra. Then we can choose \(g_j \in A\), \(g_j \not \in \mathfrak p\) such that \(\varphi(t_j) \in \Im(A_{g_j} \to A_\mathfrak p)\). Thus after replacing \(A\) by \(A[1/\prod g_j]\) we may assume that \(B\) maps into the image of \(A \to A_\mathfrak p\). If we can find a \(g \in A\), \(g \not \in \mathfrak p\) such that \(A_g \to A_\mathfrak p\) is injective, then we’ll get the desired \(R\)-algebra map \(B \to A_g\). Thus the proof is finished by another application of See Algebra, Lemma 0BX1.

Lemma

Let \(S\) be a scheme. Let \(X\), \(Y\) be schemes over \(S\). Let \(x \in X\). Let \(U \subset X\) be an open and let \(f : U \to Y\) be a morphism over \(S\). Assume

\(x\) is in the closure of \(U\),

\(X\) is reduced with finitely many irreducible components or \(X\) is Noetherian,

\(\mathcal{O}_{X, x}\) is a valuation ring,

\(Y \to S\) is proper

Then there exists an open \(U \subset U' \subset X\) containing \(x\) and an \(S\)-morphism \(f' : U' \to Y\) extending \(f\).

Proof

It is harmless to replace \(X\) by an open neighbourhood of \(x\) in \(X\) (small detail omitted). By Properties, Lemma 0BX3 we may assume \(X\) is affine with \(\Gamma(X, \mathcal{O}_X) \subset \mathcal{O}_{X, x}\). In particular \(X\) is integral with a unique generic point \(\xi\) whose residue field is the fraction field \(K\) of the valuation ring \(\mathcal{O}_{X, x}\). Since \(x\) is in the closure of \(U\) we see that \(U\) is not empty, hence \(U\) contains \(\xi\). Thus by the valuative criterion of properness (Lemma 0BX5) there is a morphism \(t : \Spec(\mathcal{O}_{X, x}) \to Y\) fitting into a commutative diagram \[\xymatrix{ \Spec(K) \ar[d]_\xi \ar[r] & \Spec(\mathcal{O}_{X, x}) \ar[d]_t \\ U \ar[r]^f & Y }\] of morphisms of schemes over \(S\). Applying Lemma 0BX6 with \(y = t(x)\) and \(\varphi = t^\sharp_x\) we obtain an open neighbourhood \(V \subset X\) of \(x\) and a morphism \(g : V \to Y\) over \(S\) which sends \(x\) to \(y\) and such that \(g^\sharp_x = t^\sharp_x\). As \(Y \to S\) is separated, the equalizer \(E\) of \(f|_{U \cap V}\) and \(g|_{U \cap V}\) is a closed subscheme of \(U \cap V\), see Schemes, Lemma 01KM. Since \(f\) and \(g\) determine the same morphism \(\Spec(K) \to Y\) by construction we see that \(E\) contains the generic point of the integral scheme \(U \cap V\). Hence \(E = U \cap V\) and we conclude that \(f\) and \(g\) glue to a morphism \(U' = U \cup V \to Y\) as desired.

Projective morphisms

We will use the definition of a projective morphism from [EGA]. The version of the definition with the “H” is the one from [H]. The resulting definitions are different. Both are useful.

Definition

Let \(f : X \to S\) be a morphism of schemes.

We say \(f\) is projective if \(X\) is isomorphic as an \(S\)-scheme to a closed subscheme of a projective bundle \(\mathbf{P}(\mathcal{E})\) for some quasi-coherent, finite type \(\mathcal{O}_S\)-module \(\mathcal{E}\).

We say \(f\) is H-projective if there exists an integer \(n\) and a closed immersion \(X \to \mathbf{P}^n_S\) over \(S\).

We say \(f\) is locally projective if there exists an open covering \(S = \bigcup U_i\) such that each \(f^{-1}(U_i) \to U_i\) is projective.

As expected, a projective morphism is quasi-projective, see Lemma 07RL. Conversely, quasi-projective morphisms are often compositions of open immersions and projective morphisms, see Lemma 07RM. For an overview of properties of projective morphisms over a quasi-projective base, see More on Morphisms, Section 0B44.

Example

Let \(S\) be a scheme. Let \(\mathcal{A}\) be a quasi-coherent graded \(\mathcal{O}_S\)-algebra generated by \(\mathcal{A}_1\) over \(\mathcal{A}_0\). Assume furthermore that \(\mathcal{A}_1\) is of finite type over \(\mathcal{O}_S\). Set \(X = \underline{\text{Proj}}_S(\mathcal{A})\). In this case \(X \to S\) is projective. Namely, the morphism associated to the graded \(\mathcal{O}_S\)-algebra map \[\text{Sym}_{\mathcal{O}_X}^*(\mathcal{A}_1) \longrightarrow \mathcal{A}\] is a closed immersion, see Constructions, Lemma 07ZK.

Lemma

An H-projective morphism is H-quasi-projective. An H-projective morphism is projective.

Proof

The first statement is immediate from the definitions. The second holds as \(\mathbf{P}^n_S\) is a projective bundle over \(S\), see Constructions, Lemma 01OE.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent:

The morphism \(f\) is locally projective.

There exists an open covering \(S = \bigcup U_i\) such that each \(f^{-1}(U_i) \to U_i\) is H-projective.

Proof

By Lemma 01W9 we see that (2) implies (1). Assume (1). For every point \(s \in S\) we can find \(\Spec(R) = U \subset S\) an affine open neighbourhood of \(s\) such that \(X_U\) is isomorphic to a closed subscheme of \(\mathbf{P}(\mathcal{E})\) for some finite type, quasi-coherent sheaf of \(\mathcal{O}_U\)-modules \(\mathcal{E}\). Write \(\mathcal{E} = \widetilde{M}\) for some finite type \(R\)-module \(M\) (see Properties, Lemma 01PB). Choose generators \(x_0, \ldots, x_n \in M\) of \(M\) as an \(R\)-module. Consider the surjective graded \(R\)-algebra map \[R[X_0, \ldots, X_n] \longrightarrow \text{Sym}_R(M).\] According to Constructions, Lemma 01N0 the corresponding morphism \[\mathbf{P}(\mathcal{E}) \to \mathbf{P}^n_R\] is a closed immersion. Hence we conclude that \(f^{-1}(U)\) is isomorphic to a closed subscheme of \(\mathbf{P}^n_U\) (as a scheme over \(U\)). In other words: (2) holds.

Lemma

A locally projective morphism is proper.

Proof

Let \(f : X \to S\) be locally projective. In order to show that \(f\) is proper we may work locally on the base, see Lemma 01W2. Hence, by Lemma 01WB above we may assume there exists a closed immersion \(X \to \mathbf{P}^n_S\). By Lemmas 01W3 and 01W5 it suffices to prove that \(\mathbf{P}^n_S \to S\) is proper. Since \(\mathbf{P}^n_S \to S\) is the base change of \(\mathbf{P}^n_{\mathbf{Z}} \to \Spec(\mathbf{Z})\) it suffices to show that \(\mathbf{P}^n_{\mathbf{Z}} \to \Spec(\mathbf{Z})\) is proper, see Lemma 01W4. By Constructions, Lemma 01MC the scheme \(\mathbf{P}^n_{\mathbf{Z}}\) is separated. By Constructions, Lemma 01MD the scheme \(\mathbf{P}^n_{\mathbf{Z}}\) is quasi-compact. It is clear that \(\mathbf{P}^n_{\mathbf{Z}} \to \Spec(\mathbf{Z})\) is locally of finite type since \(\mathbf{P}^n_{\mathbf{Z}}\) is covered by the affine opens \(D_{+}(X_i)\) each of which is the spectrum of the finite type \(\mathbf{Z}\)-algebra \[\mathbf{Z}[X_0/X_i, \ldots, X_n/X_i].\] Finally, we have to show that \(\mathbf{P}^n_{\mathbf{Z}} \to \Spec(\mathbf{Z})\) is universally closed. This follows from Constructions, Lemma 01MF and the valuative criterion (see Schemes, Proposition 01KF).

Lemma

Let \(f : X \to S\) be a proper morphism of schemes. If there exists an \(f\)-ample invertible sheaf on \(X\), then \(f\) is locally projective.

Proof

If there exists an \(f\)-ample invertible sheaf, then we can locally on \(S\) find an immersion \(i : X \to \mathbf{P}^n_S\), see Lemma 01VT. Since \(X \to S\) is proper the morphism \(i\) is a closed immersion, see Lemma 01W6.

Lemma

A composition of H-projective morphisms is H-projective.

Proof

Suppose \(X \to Y\) and \(Y \to Z\) are H-projective. Then there exist closed immersions \(X \to \mathbf{P}^n_Y\) over \(Y\), and \(Y \to \mathbf{P}^m_Z\) over \(Z\). Consider the following diagram \[\xymatrix{ X \ar[r] \ar[d] & \mathbf{P}^n_Y \ar[r] \ar[dl] & \mathbf{P}^n_{\mathbf{P}^m_Z} \ar[dl] \ar@{=}[r] & \mathbf{P}^n_Z \times_Z \mathbf{P}^m_Z \ar[r] & \mathbf{P}^{nm + n + m}_Z \ar[ddllll] \\ Y \ar[r] \ar[d] & \mathbf{P}^m_Z \ar[dl] & \\ Z & & }\] Here the rightmost top horizontal arrow is the Segre embedding, see Constructions, Lemma 01WD. The diagram identifies \(X\) as a closed subscheme of \(\mathbf{P}^{nm + n + m}_Z\) as desired.

Lemma

A base change of a H-projective morphism is H-projective.

Proof

This is true because the base change of projective space over a scheme is projective space, and the fact that the base change of a closed immersion is a closed immersion, see Schemes, Lemma 01JY.

Lemma

A base change of a (locally) projective morphism is (locally) projective.

Proof

This is true because the base change of a projective bundle over a scheme is a projective bundle, the pullback of a finite type \(\mathcal{O}\)-module is of finite type (Modules, Lemma 01B6) and the fact that the base change of a closed immersion is a closed immersion, see Schemes, Lemma 01JY. Some details omitted.

Lemma

A projective morphism is quasi-projective.

Proof

Let \(f : X \to S\) be a projective morphism. Choose a closed immersion \(i : X \to \mathbf{P}(\mathcal{E})\) where \(\mathcal{E}\) is a quasi-coherent, finite type \(\mathcal{O}_S\)-module. Then \(\mathcal{L} = i^*\mathcal{O}_{\mathbf{P}(\mathcal{E})}(1)\) is \(f\)-very ample. Since \(f\) is proper (Lemma 01WC) it is quasi-compact. Hence Lemma 01VN implies that \(\mathcal{L}\) is \(f\)-ample. Since \(f\) is proper it is of finite type. Thus we’ve checked all the defining properties of quasi-projective holds and we win.

Lemma

Let \(f : X \to S\) be a H-quasi-projective morphism. Then \(f\) factors as \(X \to X' \to S\) where \(X \to X'\) is an open immersion and \(X' \to S\) is H-projective.

Proof

By definition we can factor \(f\) as a quasi-compact immersion \(i : X \to \mathbf{P}^n_S\) followed by the projection \(\mathbf{P}^n_S \to S\). By Lemma 01RG there exists a closed subscheme \(X' \subset \mathbf{P}^n_S\) such that \(i\) factors through an open immersion \(X \to X'\). The lemma follows.

Lemma

Let \(f : X \to S\) be a quasi-projective morphism with \(S\) quasi-compact and quasi-separated. Then \(f\) factors as \(X \to X' \to S\) where \(X \to X'\) is an open immersion and \(X' \to S\) is projective.

Proof

Let \(\mathcal{L}\) be \(f\)-ample. Since \(f\) is of finite type and \(S\) is quasi-compact \(\mathcal{L}^{\otimes n}\) is \(f\)-very ample for some \(n > 0\), see Lemma 01VU. Replace \(\mathcal{L}\) by \(\mathcal{L}^{\otimes n}\). Write \(\mathcal{F} = f_*\mathcal{L}\). This is a quasi-coherent \(\mathcal{O}_S\)-module by Schemes, Lemma 01LC (quasi-projective morphisms are quasi-compact and separated, see Lemma 01VX). By Properties, Lemma 01PK we can find a directed set \(I\) and a system of finite type quasi-coherent \(\mathcal{O}_S\)-modules \(\mathcal{E}_i\) over \(I\) such that \(\mathcal{F} = \colim \mathcal{E}_i\). Consider the compositions \(\psi_i : f^*\mathcal{E}_i \to f^*\mathcal{F} \to \mathcal{L}\). Choose a finite affine open covering \(S = \bigcup_{j = 1, \ldots, m} V_j\). For each \(j\) we can choose sections \[s_{j, 0}, \ldots, s_{j, n_j} \in \Gamma(f^{-1}(V_j), \mathcal{L}) = f_*\mathcal{L}(V_j) = \mathcal{F}(V_j)\] which generate \(\mathcal{L}\) over \(f^{-1}V_j\) and define an immersion \[f^{-1}V_j \longrightarrow \mathbf{P}^{n_j}_{V_j},\] see Lemma 02NP. Choose \(i\) such that there exist sections \(e_{j, t} \in \mathcal{E}_i(V_j)\) mapping to \(s_{j, t}\) in \(\mathcal{F}\) for all \(j = 1, \ldots, m\) and \(t = 1, \ldots, n_j\). Then the map \(\psi_i\) is surjective as the sections \(f^*e_{j, t}\) have the same image as the sections \(s_{j, t}\) which generate \(\mathcal{L}|_{f^{-1}V_j}\). Whence we obtain a morphism \[r_{\mathcal{L}, \psi_i} : X \longrightarrow \mathbf{P}(\mathcal{E}_i)\] over \(S\) such that over \(V_j\) we have a factorization \[f^{-1}V_j \to \mathbf{P}(\mathcal{E}_i)|_{V_j} \to \mathbf{P}^{n_j}_{V_j}\] of the immersion given above. It follows that \(r_{\mathcal{L}, \psi_i}|_{V_j}\) is an immersion, see Lemma 07RK. Since \(S = \bigcup V_j\) we conclude that \(r_{\mathcal{L}, \psi_i}\) is an immersion. Note that \(r_{\mathcal{L}, \psi_i}\) is quasi-compact as \(X \to S\) is quasi-compact and \(\mathbf{P}(\mathcal{E}_i) \to S\) is separated (see Schemes, Lemma 03GI). By Lemma 01RG there exists a closed subscheme \(X' \subset \mathbf{P}(\mathcal{E}_i)\) such that \(i\) factors through an open immersion \(X \to X'\). Then \(X' \to S\) is projective by definition and we win.

Lemma

Let \(S\) be a quasi-compact and quasi-separated scheme. Let \(f : X \to S\) be a morphism of schemes. Then

\(f\) is projective if and only if \(f\) is quasi-projective and proper, and

\(f\) is H-projective if and only if \(f\) is H-quasi-projective and proper.

Proof

If \(f\) is projective, then \(f\) is quasi-projective by Lemma 07RL and proper by Lemma 01WC. Conversely, if \(X \to S\) is quasi-projective and proper, then we can choose an open immersion \(X \to X'\) with \(X' \to S\) projective by Lemma 07RM. Since \(X \to S\) is proper, we see that \(X\) is closed in \(X'\) (Lemma 01W6), i.e., \(X \to X'\) is a (open and) closed immersion. Since \(X'\) is isomorphic to a closed subscheme of a projective bundle over \(S\) (Definition 01W8) we see that the same thing is true for \(X\), i.e., \(X \to S\) is a projective morphism. This proves (1). The proof of (2) is the same, except it uses Lemmas 01W9 and 01WA.

Lemma

Let \(f : X \to Y\) and \(g : Y \to S\) be morphisms of schemes. If \(S\) is quasi-compact and quasi-separated and \(f\) and \(g\) are projective, then \(g \circ f\) is projective.

Proof

By Lemmas 07RL and 01WC we see that \(f\) and \(g\) are quasi-projective and proper. By Lemmas 01W3 and 0C4M we see that \(g \circ f\) is proper and quasi-projective. Thus \(g \circ f\) is projective by Lemma 0BCL.

Lemma

Let \(g : Y \to S\) and \(f : X \to Y\) be morphisms of schemes. If \(g \circ f\) is projective and \(g\) is separated, then \(f\) is projective.

Proof

Choose a closed immersion \(X \to \mathbf{P}(\mathcal{E})\) where \(\mathcal{E}\) is a quasi-coherent, finite type \(\mathcal{O}_S\)-module. Then we get a morphism \(X \to \mathbf{P}(\mathcal{E}) \times_S Y\). This morphism is a closed immersion because it is the composition \[X \to X \times_S Y \to \mathbf{P}(\mathcal{E}) \times_S Y\] where the first morphism is a closed immersion by Schemes, Lemma 01KS (and the fact that \(g\) is separated) and the second as the base change of a closed immersion. Finally, the fibre product \(\mathbf{P}(\mathcal{E}) \times_S Y\) is isomorphic to \(\mathbf{P}(g^*\mathcal{E})\) and pullback preserves quasi-coherent, finite type modules.

Lemma

Let \(S\) be a scheme which admits an ample invertible sheaf. Then

any projective morphism \(X \to S\) is H-projective, and

any quasi-projective morphism \(X \to S\) is H-quasi-projective.

Proof

The assumptions on \(S\) imply that \(S\) is quasi-compact and separated, see Properties, Definition 01PS and Lemma 01Q1 and Constructions, Lemma 01MC. Hence Lemma 07RM applies and we see that (1) implies (2). Let \(\mathcal{E}\) be a finite type quasi-coherent \(\mathcal{O}_S\)-module. By our definition of projective morphisms it suffices to show that \(\mathbf{P}(\mathcal{E}) \to S\) is H-projective. If \(\mathcal{E}\) is generated by finitely many global sections, then the corresponding surjection \(\mathcal{O}_S^{\oplus n} \to \mathcal{E}\) induces a closed immersion \[\mathbf{P}(\mathcal{E}) \longrightarrow \mathbf{P}(\mathcal{O}_S^{\oplus n}) = \mathbf{P}^{n - 1}_S\] as desired. In general, let \(\mathcal{L}\) be an ample invertible sheaf on \(S\). By Properties, Proposition 01Q3 there exists an integer \(n\) such that \(\mathcal{E} \otimes_{\mathcal{O}_S} \mathcal{L}^{\otimes n}\) is globally generated by finitely many sections. Since \(\mathbf{P}(\mathcal{E}) = \mathbf{P}(\mathcal{E} \otimes_{\mathcal{O}_S} \mathcal{L}^{\otimes n})\) by Constructions, Lemma 02NC this finishes the proof.

Lemma

Let \(f : X \to S\) be a universally closed morphism. Let \(\mathcal{L}\) be an \(f\)-ample invertible \(\mathcal{O}_X\)-module. Then the canonical morphism \[r : X \longrightarrow \underline{\text{Proj}}_S \left( \bigoplus\nolimits_{d \geq 0} f_*\mathcal{L}^{\otimes d} \right)\] of Lemma 01VJ is an isomorphism.

Proof

Observe that \(f\) is quasi-compact because the existence of an \(f\)-ample invertible module forces \(f\) to be quasi-compact. By the lemma cited the morphism \(r\) is an open immersion. On the other hand, the image of \(r\) is closed by Lemma 01W6 (the target of \(r\) is separated over \(S\) by Constructions, Lemma 01O2). Finally, the image of \(r\) is dense by Properties, Lemma 01Q1 (here we also use that it was shown in the proof of Lemma 01VJ that the morphism \(r\) over affine opens of \(S\) is given by the canonical morphism of Properties, Lemma 01PZ). Thus we conclude that \(r\) is a surjective open immersion, i.e., an isomorphism.

Lemma

Let \(f : X \to S\) be a universally closed morphism. Let \(\mathcal{L}\) be an \(f\)-ample invertible \(\mathcal{O}_X\)-module. Let \(s \in \Gamma(X, \mathcal{L})\). Then \(X_s \to S\) is an affine morphism.

Proof

The question is local on \(S\) (Lemma 01S8) hence we may assume \(S\) is affine. By Lemma 0C6J we can write \(X = \text{Proj}(A)\) where \(A\) is a graded ring and \(s\) corresponds to \(f \in A_1\) and \(X_s = D_+(f)\) (Properties, Lemma 01PZ) which proves the lemma by construction of \(\text{Proj}(A)\), see Constructions, Section 01M3.

Integral and finite morphisms

Recall that a ring map \(R \to A\) is said to be integral if every element of \(A\) satisfies a monic equation with coefficients in \(R\). Recall that a ring map \(R \to A\) is said to be finite if \(A\) is finite as an \(R\)-module. See Algebra, Definition 00GI.

Definition

Let \(f : X \to S\) be a morphism of schemes.

We say that \(f\) is integral if \(f\) is affine and if for every affine open \(\Spec(R) = V \subset S\) with inverse image \(\Spec(A) = f^{-1}(V) \subset X\) the associated ring map \(R \to A\) is integral.

We say that \(f\) is finite if \(f\) is affine and if for every affine open \(\Spec(R) = V \subset S\) with inverse image \(\Spec(A) = f^{-1}(V) \subset X\) the associated ring map \(R \to A\) is finite.

It is clear that integral/finite morphisms are separated and quasi-compact. It is also clear that a finite morphism is a morphism of finite type. Most of the lemmas in this section are completely standard. But note the fun Lemma 01WM at the end of the section.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent:

The morphism \(f\) is integral.

There exists an affine open covering \(S = \bigcup U_i\) such that each \(f^{-1}(U_i)\) is affine and \(\mathcal{O}_S(U_i) \to \mathcal{O}_X(f^{-1}(U_i))\) is integral.

There exists an open covering \(S = \bigcup U_i\) such that each \(f^{-1}(U_i) \to U_i\) is integral.

Moreover, if \(f\) is integral then for every open subscheme \(U \subset S\) the morphism \(f : f^{-1}(U) \to U\) is integral.

Proof

See Algebra, Lemma 02JL. Some details omitted.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent:

The morphism \(f\) is finite.

There exists an affine open covering \(S = \bigcup U_i\) such that each \(f^{-1}(U_i)\) is affine and \(\mathcal{O}_S(U_i) \to \mathcal{O}_X(f^{-1}(U_i))\) is finite.

There exists an open covering \(S = \bigcup U_i\) such that each \(f^{-1}(U_i) \to U_i\) is finite.

Moreover, if \(f\) is finite then for every open subscheme \(U \subset S\) the morphism \(f : f^{-1}(U) \to U\) is finite.

Proof

See Algebra, Lemma 02JL. Some details omitted.

Lemma

A finite morphism is integral. An integral morphism which is locally of finite type is finite.

Proof

See Algebra, Lemma 00GK and Lemma 02JJ.

Lemma

A composition of finite morphisms is finite. Same is true for integral morphisms.

Proof

See Algebra, Lemmas 00GL and 00GN.

Lemma

A base change of a finite morphism is finite. Same is true for integral morphisms.

Proof

See Algebra, Lemma 02JK.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent

\(f\) is integral, and

\(f\) is affine and universally closed.

Proof

Assume (1). An integral morphism is affine by definition. A base change of an integral morphism is integral so in order to prove (2) it suffices to show that an integral morphism is closed. This follows from Algebra, Lemmas 00GU and 00HZ.

Assume (2). We may assume \(f\) is the morphism \(f : \Spec(A) \to \Spec(R)\) coming from a ring map \(R \to A\). Let \(a\) be an element of \(A\). We have to show that \(a\) is integral over \(R\), i.e. that in the kernel \(I\) of the map \(R[x] \to A\) sending \(x\) to \(a\) there is a monic polynomial. Consider the ring \(B = A[x]/(ax -1)\) and let \(J\) be the kernel of the composition \(R[x]\to A[x] \to B\). If \(f\in J\) there exists \(q\in A[x]\) such that \(f = (ax-1)q\) in \(A[x]\) so if \(f = \sum_i f_ix^i\) and \(q = \sum_iq_ix^i\), for all \(i \geq 0\) we have \(f_i = aq_{i-1} - q_i\). For \(n \geq \deg q + 1\) the polynomial \[\sum\nolimits_{i \geq 0} f_i x^{n - i} = \sum\nolimits_{i \geq 0} (a q_{i - 1} - q_i) x^{n - i} = (a - x) \sum\nolimits_{i \geq 0} q_i x^{n - i - 1}\] is clearly in \(I\); if \(f_0 = 1\) this polynomial is also monic, so we are reduced to prove that \(J\) contains a polynomial with constant term \(1\). We do it by proving \(\Spec(R[x]/(J + (x))\) is empty.

Since \(f\) is universally closed the base change \(\Spec(A[x]) \to \Spec(R[x])\) is closed. Hence the image of the closed subset \(\Spec(B) \subset \Spec(A[x])\) is the closed subset \(\Spec(R[x]/J) \subset \Spec(R[x])\), see Example 056A and Lemma 01R8. In particular \(\Spec(B) \to \Spec(R[x]/J)\) is surjective. Consider the following diagram where every square is a pullback: \[\xymatrix{ \Spec(B) \ar@{->>}[r]^g & \Spec(R[x]/J) \ar[r] & \Spec(R[x])\\ \emptyset \ar[u] \ar[r] & \Spec(R[x]/(J + (x)))\ar[u] \ar[r] & \Spec(R) \ar[u]^0 }\] The bottom left corner is empty because it is the spectrum of \(R\otimes_{R[x]} B\) where the map \(R[x]\to B\) sends \(x\) to an invertible element and \(R[x]\to R\) sends \(x\) to \(0\). Since \(g\) is surjective this implies \(\Spec(R[x]/(J + (x)))\) is empty, as we wanted to show.

Lemma

Let \(f : X \to S\) be an integral morphism. Then every point of \(X\) is closed in its fibre.

Proof

See Algebra, Lemma 00GT.

Lemma

Let \(f : X \to Y\) be an integral morphism. Then \(\dim(X) \leq \dim(Y)\). If \(f\) is surjective then \(\dim(X) = \dim(Y)\).

Proof

Since the dimension of \(X\) and \(Y\) is the supremum of the dimensions of the members of an affine open covering, we may assume \(Y\) and \(X\) are affine. The inequality follows from Algebra, Lemma 00OJ. The equality then follows from Algebra, Lemmas 00OH and 00GU.

Lemma

A finite morphism is quasi-finite.

Proof

This is implied by Algebra, Lemma 00PM and Lemma 01TJ. Alternatively, all points in fibres are closed points by Lemma 02NT (and the fact that a finite morphism is integral) and use Lemma 01TH (3) to see that \(f\) is quasi-finite at \(x\) for all \(x \in X\).

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent

\(f\) is finite, and

\(f\) is affine and proper.

Proof

This follows formally from Lemma 01WM, the fact that a finite morphism is integral and separated, the fact that a proper morphism is the same thing as a finite type, separated, universally closed morphism, and the fact that an integral morphism of finite type is finite (Lemma 01WJ).

Lemma

A closed immersion is finite (and a fortiori integral).

Proof

True because a closed immersion is affine (Lemma 01SE) and a surjective ring map is finite and integral.

Lemma

Let \(X_i \to Y\), \(i = 1, \ldots, n\) be finite morphisms of schemes. Then \(X_1 \amalg \ldots \amalg X_n \to Y\) is finite too.

Proof

Follows from the algebra fact that if \(R \to A_i\), \(i = 1, \ldots, n\) are finite ring maps, then \(R \to A_1 \times \ldots \times A_n\) is finite too.

Lemma

Let \(f : X \to Y\) and \(g : Y \to Z\) be morphisms.

If \(g \circ f\) is finite and \(g\) separated then \(f\) is finite.

If \(g \circ f\) is integral and \(g\) separated then \(f\) is integral.

Proof

Assume \(g \circ f\) is finite (resp. integral) and \(g\) separated. The base change \(X \times_Z Y \to Y\) is finite (resp. integral) by Lemma 01WL. The morphism \(X \to X \times_Z Y\) is a closed immersion as \(Y \to Z\) is separated, see Schemes, Lemma 01KT. A closed immersion is finite (resp. integral), see Lemma 035C. The composition of finite (resp. integral) morphisms is finite (resp. integral), see Lemma 01WK. Thus we win.

Lemma

Let \(f : X \to Y\) be a morphism of schemes. If \(f\) is finite and a monomorphism, then \(f\) is a closed immersion.

Proof

This reduces to Algebra, Lemma 04VT.

Lemma

A finite morphism is projective.

Proof

Let \(f : X \to S\) be a finite morphism. Then \(f_*\mathcal{O}_X\) is a quasi-coherent \(\mathcal{O}_S\)-module (Lemma 01SA) of finite type (by our definition of finite morphisms and Properties, Lemma 01PB). We claim there is a closed immersion \[\sigma : X \longrightarrow \mathbf{P}(f_*\mathcal{O}_X) = \underline{\text{Proj}}_S(\text{Sym}^*_{\mathcal{O}_S}(f_*\mathcal{O}_X))\] over \(S\), which finishes the proof. Namely, we let \(\sigma\) be the morphism which corresponds (via Constructions, Lemma 01O4) to the surjection \[f^*f_*\mathcal{O}_X \longrightarrow \mathcal{O}_X\] coming from the adjunction map \(f^*f_* \to \text{id}\). Then \(\sigma\) is a closed immersion. Namely, affine locally on \(S\) we can write \(X = \Spec(A)\) and \(S = \Spec(R)\). Since \(X\) is finite over \(S\) we may choose \(a_1, \ldots, a_n \in A\) generating \(A\) as an \(R\)-module. Then the \(R\)-algebra map \(R[T_0, T_1, \ldots, T_n] \to \text{Sym}_R^*(A)\) sending \(T_0\) to \(1\) and \(T_i\) to \(a_i\) for \(1 \leq i \leq n\) is surjective. Whence \(\mathbf{P}(f_*\mathcal{O}_X)\) is a closed subscheme of \(\mathbf{P}^n_S\) by Constructions, Lemma 01N0. Thus it suffices to prove that the induced morphism \(X \to \mathbf{P}^n_S\) is a closed immersion (see for example, Lemma 01QP). The reader checks that \(X \to \mathbf{P}^n_S\) has image contained in the open \(D_+(T_0) \cong \mathbf{A}^n_S\) and that \(X \to \mathbf{A}^n_S\) corresponds to the surjective \(R\)-algebra map \(R[x_1, \ldots, x_n] \to \text{Sym}_R^*(A)\) sending \(x_i\) to \(a_i\). Whence \(X \to \mathbf{P}^n_S\) is an immersion (as a composition of a closed immersion and an open immersion). Since \(X\) is finite over \(S\) the image of this immersion is closed (this uses Lemma 01WN, Lemma 01W6, and Constructions, Lemma 01NH). We conclude by Schemes, Lemma 01IQ.

Universal homeomorphisms

The following definition is really superfluous since a universal homeomorphism is really just an integral, universally injective and surjective morphism, see Lemma 04DF.

Definition

A morphism \(f : X \to Y\) of schemes is called a universal homeomorphism if the base change \(f' : Y' \times_Y X \to Y'\) is a homeomorphism for every morphism \(Y' \to Y\).

First we state the obligatory lemmas.

Lemma

The base change of a universal homeomorphism of schemes by any morphism of schemes is a universal homeomorphism.

Proof

This is immediate from the definition.

Lemma

The composition of a pair of universal homeomorphisms of schemes is a universal homeomorphism.

Proof

Omitted.

The following simple lemma is the key to characterizing universal homeomorphisms.

Lemma

Let \(f : X \to Y\) be a morphism of schemes. If \(f\) is a homeomorphism onto a closed subset of \(Y\) then \(f\) is affine.

Proof

Let \(y \in Y\) be a point. If \(y \not \in f(X)\), then there exists an affine neighbourhood of \(y\) which is disjoint from \(f(X)\). If \(y \in f(X)\), let \(x \in X\) be the unique point of \(X\) mapping to \(y\). Let \(y \in V\) be an affine open neighbourhood. Let \(U \subset X\) be an affine open neighbourhood of \(x\) which maps into \(V\). Since \(f(U) \subset V \cap f(X)\) is open in the induced topology by our assumption on \(f\) we may choose a \(h \in \Gamma(V, \mathcal{O}_Y)\) such that \(y \in D(h)\) and \(D(h) \cap f(X) \subset f(U)\). Denote \(h' \in \Gamma(U, \mathcal{O}_X)\) the restriction of \(f^\sharp(h)\) to \(U\). Then we see that \(D(h') \subset U\) is equal to \(f^{-1}(D(h))\). In other words, every point of \(Y\) has an open neighbourhood whose inverse image is affine. Thus \(f\) is affine, see Lemma 01S8.

Lemma

Let \(f : X \to Y\) be a morphism of schemes. The following are equivalent:

\(f\) is a universal homeomorphism, and

\(f\) is integral, universally injective and surjective.

Proof

Assume \(f\) is a universal homeomorphism. By Lemma 04DE we see that \(f\) is affine. Since \(f\) is clearly universally closed we see that \(f\) is integral by Lemma 01WM. It is also clear that \(f\) is universally injective and surjective.

Assume \(f\) is integral, universally injective and surjective. By Lemma 01WM \(f\) is universally closed. Since it is also universally bijective (see Lemma 01S1) we see that it is a universal homeomorphism.

Lemma

Let \(X\) be a scheme. The canonical closed immersion \(X_{red} \to X\) (see Schemes, Definition 01J4) is a universal homeomorphism.

Proof

Omitted.

Lemma

Let \(f : X \to Y\) be a morphism of schemes and let \(f_{red} : X_{red} \to Y_{red}\) be the morphism of Schemes, Lemma schemes-lemma-reduction-functorial. Then

\(f\) is surjective if and only if \(f_{red}\) is surjective,

\(f\) is universally injective if and only if \(f_{red}\) is universally injective, and

if \(f\) is an immersion, a closed immersion, or an open immersion, then the same is true of \(f_{red}\).

Proof

The reductions have the same underlying spaces as the original schemes, and the vertical maps in the defining square induce isomorphisms on residue fields. The first assertion follows immediately. The second follows from Lemma 01S4, which characterizes universal injectivity by the map on points and the residue field extensions.

Suppose that \(f\) is a closed immersion. Reduction does not change the underlying topological map. The induced map on stalks for \(f_{red}\) is obtained from the surjective stalk map for \(f\) by dividing source and target by their nilradicals, and hence is surjective. The map on structure sheaves is therefore surjective by Sheaves, Lemma 0H7H. Thus \(f_{red}\) is a closed immersion by Schemes, Lemma 01LD. If \(f\) is an open immersion, then its stalk maps are isomorphisms and its image is open; both properties remain true after reduction, so \(f_{red}\) is an open immersion. Finally, factor an immersion as a closed immersion followed by an open immersion. Functoriality of reduction and Schemes, Lemma 02V0 prove the general immersion case.

Lemma

Let \(X \to S\) and \(Y \to S\) be morphisms of schemes. There is a canonical isomorphism \[X_{red} \times_{S_{red}} Y_{red} \longrightarrow X_{red} \times_S Y_{red}.\] Via this isomorphism, the canonical morphism \[X_{red} \times_{S_{red}} Y_{red} \longrightarrow X \times_S Y\] is a closed immersion and a universal homeomorphism.

Proof

The maps from \(X_{red}\) and \(Y_{red}\) to \(S\) factor through \(S_{red}\) by Schemes, Lemma schemes-lemma-reduction-functorial. Since \(S_{red} \to S\) is a monomorphism, the two fibre products in the statement have the same universal property. This gives the canonical isomorphism.

After replacing the source by \(X_{red} \times_S Y_{red}\), the morphism to \(X \times_S Y\) factors as \[X_{red} \times_S Y_{red} \longrightarrow X \times_S Y_{red} \longrightarrow X \times_S Y.\] Both arrows are base changes of reduction morphisms, and hence are closed immersions by Schemes, Lemma 01JU. Their composition is a closed immersion by Schemes, Lemma 02V0. Both arrows are universal homeomorphisms by Lemmas 054M and 0CEU; their composition is a universal homeomorphism by Lemma 0CEV.

Lemma

Let \(f : X \to S\) and \(S' \to S\) be morphisms of schemes. Assume

\(S' \to S\) is a closed immersion,

\(S' \to S\) is bijective on points,

\(X \times_S S' \to S'\) is a closed immersion, and

\(X \to S\) is of finite type or \(S' \to S\) is of finite presentation.

Then \(f : X \to S\) is a closed immersion.

Proof

Assumptions (1) and (2) imply that \(S' \to S\) is a universal homeomorphism (for example because \(S_{red} = S'_{red}\) and using Lemma 054M). Hence (3) implies that \(X \to S\) is homeomorphism onto a closed subset of \(S\). Then \(X \to S\) is affine by Lemma 04DE. Let \(U \subset S\) be an affine open, say \(U = \Spec(A)\). Then \(S' = \Spec(A/I)\) by (1) for a locally nilpotent ideal \(I\) by (2). As \(f\) is affine we see that \(f^{-1}(U) = \Spec(B)\). Assumption (4) tells us \(B\) is a finite type \(A\)-algebra (Lemma 01T2) or that \(I\) is finitely generated (Lemma 01TV). Assumption (3) is that \(A/I \to B/IB\) is surjective. From Algebra, Lemma 07RD if \(A \to B\) is of finite type or Algebra, Lemma 00DV if \(I\) is finitely generated and hence nilpotent we deduce that \(A \to B\) is surjective. This means that \(f\) is a closed immersion, see Lemma 01QO.

Lemma

Let \(f : X \to Z\) be the composition of two morphisms \(g : X \to Y\) and \(h : Y \to Z\). If two of the morphisms \(\{f, g, h\}\) are universal homeomorphisms, so is the third morphism.

Proof

If both of \(g\) and \(h\) are universal homeomorphisms, so is \(f\) by Lemma 0CEV.

Suppose both of \(f\) and \(g\) are universal homeomorphisms. We want to show that \(h\) is also. Now base change the diagram along an arbitrary morphism \(\alpha : Z' \to Z\) of schemes, we get the following diagram with all squares Cartesian: \[\xymatrix{ X' \ar[r]^{g'} \ar[d] & Y' \ar[r]^{h'} \ar[d] & Z' \ar[d] \\ X \ar[r]^{g} & Y \ar[r]^{h} & Z. }\] Our assumption implies that the composition \(f'= h' \circ g' : X' \to Z'\) and \(g' : X' \to Y'\) are homeomorphisms, therefore so is \(h'\). This finishes the proof of \(h\) being a universal homeomorphism.

Finally, assume \(f\) and \(h\) are universal homeomorphisms. We want to show that \(g\) is a universal homeomorphism. Let \(\beta : Y' \to Y\) be an arbitrary morphism of schemes. We get the following diagram with all squares Cartesian: \[\xymatrix{ X' \ar[r]^{g'} \ar[d] & Y' \ar[d]^{\gamma} & \\ X'' \ar[r]^{g''} \ar[d] & Y'' \ar[r]^{h''} \ar[d] & Y' \ar[d]^{h \circ \beta} \\ X \ar[r]^{g} & Y \ar[r]^{h} & Z. }\] Here the morphism \(\gamma : Y' \to Y''\) is defined by the universal property of fiber products and the two morphisms \(id_{Y'} : Y' \to Y'\) and \(\beta : Y' \to Y\). We shall prove that \(g'\) is a homeomorphism. Since the property of being a homeomorphism has 2-out-of-3 property, we see that \(g''\) is a homeomorphism. Staring at the top square, it suffices to prove that \(\gamma\) is a universal homeomorphism. Since \(h''\) is a homeomorphism, we see that it is an affine morphism by Lemma 04DE and a fortiori separated (Lemma 01S7). Since \(h'' \circ \gamma\) is the identity, we see that \(\gamma\) is a closed immersion by Schemes, Lemma 01KT. Since \(h''\) is bijective, it follows that \(\gamma\) is a bijective closed immersion and hence a universal homeomorphism (for example by the characterization in Lemma 04DF) as desired.

Universal homeomorphisms of affine schemes

In this section we characterize universal homeomorphisms of affine schemes.

Lemma

Let \(A \to B\) be a ring map such that the induced morphism of schemes \(f : \Spec(B) \to \Spec(A)\) is a universal homeomorphism, resp. a universal homeomorphism inducing isomorphisms on residue fields, resp. universally closed, resp. universally closed and universally injective. Then for any \(A\)-subalgebra \(B' \subset B\) the same thing is true for \(f' : \Spec(B') \to \Spec(A)\).

Proof

If \(f\) is universally closed, then \(B\) is integral over \(A\) by Lemma 01WM. Hence \(B'\) is integral over \(A\) and \(f'\) is universally closed (by the same lemma). This proves the case where \(f\) is universally closed.

Continuing, we see that \(B\) is integral over \(B'\) (Algebra, Lemma 02JM) which implies \(\Spec(B) \to \Spec(B')\) is surjective (Algebra, Lemma 00GQ). Thus if \(A \to B\) induces purely inseparable extensions of residue fields, then the same is true for \(A \to B'\). This proves the case where \(f\) is universally closed and universally injective, see Lemma 01S4.

The case where \(f\) is a universal homeomorphism follows from the remarks above, Lemma 04DF, and the obvious observation that if \(f\) is surjective, then so is \(f'\).

If \(A \to B\) induces isomorphisms on residue fields, then so does \(A \to B'\) (see argument in second paragraph). In this way we see that the lemma holds in the remaining case.

Lemma

Let \(A\) be a ring. Let \(B = \colim B_\lambda\) be a filtered colimit of \(A\)-algebras. If each \(f_\lambda : \Spec(B_\lambda) \to \Spec(A)\) is a universal homeomorphism, resp. a universal homeomorphism inducing isomorphisms on residue fields, resp. universally closed, resp. universally closed and universally injective, then the same thing is true for \(f : \Spec(B) \to \Spec(A)\).

Proof

If \(f_\lambda\) is universally closed, then \(B_\lambda\) is integral over \(A\) by Lemma 01WM. Hence \(B\) is integral over \(A\) and \(f\) is universally closed (by the same lemma). This proves the case where each \(f_\lambda\) is universally closed.

For a prime \(\mathfrak q \subset B\) lying over \(\mathfrak p \subset A\) denote \(\mathfrak q_\lambda \subset B_\lambda\) the inverse image. Then \(\kappa(\mathfrak q) = \colim \kappa(\mathfrak q_\lambda)\). Thus if \(A \to B_\lambda\) induces purely inseparable extensions of residue fields, then the same is true for \(A \to B\). This proves the case where \(f_\lambda\) is universally closed and universally injective, see Lemma 01S4.

The case where \(f\) is a universal homeomorphism follows from the remarks above and Lemma 04DF combined with the fact that prime ideals in \(B\) are the same thing as compatible sequences of prime ideals in all of the \(B_\lambda\).

If \(A \to B_\lambda\) induces isomorphisms on residue fields, then so does \(A \to B\) (see argument in second paragraph). In this way we see that the lemma holds in the remaining case.

Lemma

Let \(A \subset B\) be a ring extension. Let \(S \subset A\) be a multiplicative subset. Let \(n \geq 1\) and \(b_i \in B\) for \(1 \leq i \leq n\). Any \(x \in S^{-1}B\) such that \[x \not \in S^{-1}A\text{ and } b_i x^i \in S^{-1}A\text{ for }i = 1, \ldots, n\] is equal to \(s^{-1}y\) with \(s \in S\) and \(y \in B\) such that \[y \not \in A\text{ and } b_i y^i \in A\text{ for }i = 1, \ldots, n\]

Proof

Omitted. Hint: clear denominators.

Lemma

Let \(A \subset B\) be a ring extension. If there exists \(b \in B\), \(b \not \in A\) and an integer \(n \geq 2\) with \(b^n \in A\) and \(b^{n + 1} \in A\), then there exists a \(b' \in B\), \(b' \not \in A\) with \((b')^2 \in A\) and \((b')^3 \in A\).

Proof

Let \(b\) and \(n\) be as in the lemma. Then all sufficiently large powers of \(b\) are in \(A\). Namely, \((b^n)^k(b^{n + 1})^i = b^{(k + i)n + i}\) which implies any power \(b^m\) with \(m \geq n^2\) is in \(A\). Hence if \(i \geq 1\) is the largest integer such that \(b^i \not \in A\), then \((b^i)^2 \in A\) and \((b^i)^3 \in A\).

Lemma

Let \(A \subset B\) be a ring extension such that \(\Spec(B) \to \Spec(A)\) is a universal homeomorphism inducing isomorphisms on residue fields. If \(A \not = B\), then there exists a \(b \in B\), \(b \not \in A\) with \(b^2 \in A\) and \(b^3 \in A\).

Proof

Recall that \(A \subset B\) is integral (Lemma 01WM). By Lemma 0CN7 we may assume that \(B\) is generated by a single element over \(A\). Hence \(B\) is finite over \(A\) (Algebra, Lemma 02JJ). Hence the support of \(B/A\) as an \(A\)-module is closed and not empty (Algebra, Lemmas 00L2 and 0585). Let \(\mathfrak p \subset A\) be a minimal prime of the support. After replacing \(A \subset B\) by \(A_\mathfrak p \subset B_\mathfrak p\) (permissible by Lemma 0CN9) we may assume that \((A, \mathfrak m)\) is a local ring, that \(B\) is finite over \(A\), and that \(B/A\) has support \(\{\mathfrak m\}\) as an \(A\)-module. Since \(B/A\) is a finite module, we see that \(I = \text{Ann}_A(B/A)\) satisfies \(\mathfrak m = \sqrt{I}\) (Algebra, Lemma 00L2). Let \(\mathfrak m' \subset B\) be the unique prime ideal lying over \(\mathfrak m\). Because \(\Spec(B) \to \Spec(A)\) is a homeomorphism, we find that \(\mathfrak m' = \sqrt{IB}\). For \(f \in \mathfrak m'\) pick \(n \geq 1\) such that \(f^n \in IB\). Then also \(f^{n + 1} \in IB\). Since \(IB \subset A\) by our choice of \(I\) we conclude that \(f^n, f^{n + 1} \in A\). Using Lemma 0CNA we conclude our lemma is true if \(\mathfrak m' \not \subset A\). However, if \(\mathfrak m' \subset A\), then \(\mathfrak m' = \mathfrak m\) and we conclude that \(A = B\) as the residue fields are isomorphic as well by assumption. This contradiction finishes the proof.

Lemma

Let \(A \subset B\) be a ring extension such that \(\Spec(B) \to \Spec(A)\) is a universal homeomorphism. If \(A \not = B\), then either there exists a \(b \in B\), \(b \not \in A\) with \(b^2 \in A\) and \(b^3 \in A\) or there exists a prime number \(p\) and a \(b \in B\), \(b \not \in A\) with \(pb \in A\) and \(b^p \in A\).

Proof

The argument is almost exactly the same as in the proof of Lemma 0CNB but we write everything out to make sure it works.

Recall that \(A \subset B\) is integral (Lemma 01WM). By Lemma 0CN7 we may assume that \(B\) is generated by a single element over \(A\). Hence \(B\) is finite over \(A\) (Algebra, Lemma 02JJ). Hence the support of \(B/A\) as an \(A\)-module is closed and not empty (Algebra, Lemmas 00L2 and 0585). Let \(\mathfrak p \subset A\) be a minimal prime of the support. After replacing \(A \subset B\) by \(A_\mathfrak p \subset B_\mathfrak p\) (permissible by Lemma 0CN9) we may assume that \((A, \mathfrak m)\) is a local ring, that \(B\) is finite over \(A\), and that \(B/A\) has support \(\{\mathfrak m\}\) as an \(A\)-module. Since \(B/A\) is a finite module, we see that \(I = \text{Ann}_A(B/A)\) satisfies \(\mathfrak m = \sqrt{I}\) (Algebra, Lemma 00L2). Let \(\mathfrak m' \subset B\) be the unique prime ideal lying over \(\mathfrak m\). Because \(\Spec(B) \to \Spec(A)\) is a homeomorphism, we find that \(\mathfrak m' = \sqrt{IB}\). For \(f \in \mathfrak m'\) pick \(n \geq 1\) such that \(f^n \in IB\). Then also \(f^{n + 1} \in IB\). Since \(IB \subset A\) by our choice of \(I\) we conclude that \(f^n, f^{n + 1} \in A\). Using Lemma 0CNA we conclude our lemma is true if \(\mathfrak m' \not \subset A\). If \(\mathfrak m' \subset A\), then \(\mathfrak m' = \mathfrak m\). Since \(A \not = B\) we conclude the map \(\kappa = A/\mathfrak m \to B/\mathfrak m' = \kappa'\) of residue fields cannot be an isomorphism. By Lemma 01S4 we conclude that the characteristic of \(\kappa\) is a prime number \(p\) and that the extension \(\kappa'/\kappa\) is purely inseparable. Pick \(b \in B\) whose image in \(\kappa'\) is an element not contained in \(\kappa\) but whose \(p\)th power is in \(\kappa\). Then \(b \not \in A\), \(b^p \in A\), and \(pb \in A\) (because \(pb \in \mathfrak m' = \mathfrak m \subset A\)) as desired.

Proposition

Let \(A \subset B\) be a ring extension. The following are equivalent

\(\Spec(B) \to \Spec(A)\) is a universal homeomorphism inducing isomorphisms on residue fields, and

every finite subset \(E \subset B\) is contained in an extension \[A[b_1, \ldots, b_n] \subset B\] such that \(b_i^2, b_i^3 \in A[b_1, \ldots, b_{i - 1}]\) for \(i = 1, \ldots, n\).

Proof

Assume (1). Using transfinite recursion we construct for each ordinal \(\alpha\) an \(A\)-subalgebra \(B_\alpha \subset B\) as follows. Set \(B_0 = A\). If \(\alpha\) is a limit ordinal, then we set \(B_\alpha = \colim_{\beta < \alpha} B_\beta\). If \(\alpha = \beta + 1\), then either \(B_\beta = B\) in which case we set \(B_\alpha = B_\beta\) or \(B_\beta \not = B\), in which case we apply Lemma 0CNB to choose a \(b_\alpha \in B\), \(b_\alpha \not \in B_\beta\) with \(b_\alpha^2, b_\alpha^3 \in B_\beta\) and we set \(B_\alpha = B_\beta[b_\alpha] \subset B\). Clearly, \(B = \colim B_\alpha\) (in fact \(B = B_\alpha\) for some ordinal \(\alpha\) as one sees by looking at cardinalities). We will prove, by transfinite induction, that (2) holds for \(A \to B_\alpha\) for every ordinal \(\alpha\). It is clear for \(\alpha = 0\). Assume the statement holds for every \(\beta < \alpha\) and let \(E \subset B_\alpha\) be a finite subset. If \(\alpha\) is a limit ordinal, then \(B_\alpha = \bigcup_{\beta < \alpha} B_\beta\) and we see that \(E \subset B_\beta\) for some \(\beta < \alpha\) which proves the result in this case. If \(\alpha = \beta + 1\), then \(B_\alpha = B_\beta[b_\alpha]\). Thus any \(e \in E\) can be written as a polynomial \(e = \sum d_{e, i}b_\alpha^i\) with \(d_{e, i} \in B_\beta\). Let \(D \subset B_\beta\) be the set \(D = \{d_{e, i}\} \cup \{b_\alpha^2, b_\alpha^3\}\). By induction assumption there exists an \(A\)-subalgebra \(A[b_1, \ldots, b_n] \subset B_\beta\) as in the statement of the lemma containing \(D\). Then \(A[b_1, \ldots, b_n, b_\alpha] \subset B_\alpha\) is an \(A\)-subalgebra of \(B_\alpha\) as in the statement of the lemma containing \(E\).

Assume (2). Write \(B = \colim B_\lambda\) as the colimit of its finite \(A\)-subalgebras. By Lemma 0CN8 it suffices to show that \(\Spec(B_\lambda) \to \Spec(A)\) is a universal homeomorphism inducing isomorphisms on residue fields. Compositions of universally closed morphisms are universally closed and the same thing for morphisms which induce isomorphisms on residue fields. Thus it suffices to show that if \(A \subset B\) and \(B\) is generated by a single element \(b\) with \(b^2, b^3 \in A\), then (1) holds. Such an extension is integral and hence \(\Spec(B) \to \Spec(A)\) is universally closed and surjective (Lemma 01WM and Algebra, Lemma 00GQ). Note that \((b^2)^3 = (b^3)^2\) in \(A\). For any ring map \(\varphi : A \to K\) to a field \(K\) we see that there exists a \(\lambda \in K\) with \(\varphi(b^2) = \lambda^2\) and \(\varphi(b^3) = \lambda^3\). Namely, \(\lambda = 0\) if \(\varphi(b^2) = 0\) and \(\lambda = \varphi(b^3)/\varphi(b^2)\) if not. Thus \(B \otimes_A K\) is a quotient of \(K[x]/(x^2 - \lambda^2, x^3 - \lambda^3)\). This ring has exactly one prime with residue field \(K\). This implies that \(\Spec(B) \to \Spec(A)\) is bijective and induces isomorphisms on residue fields. Combined with universal closedness this shows (1) is true, see Lemmas 04DF and 01S4.

Proposition

Let \(A \subset B\) be a ring extension. The following are equivalent

\(\Spec(B) \to \Spec(A)\) is a universal homeomorphism, and

every finite subset \(E \subset B\) is contained in an extension \[A[b_1, \ldots, b_n] \subset B\] such that for \(i = 1, \ldots, n\) we have

\(b_i^2, b_i^3 \in A[b_1, \ldots, b_{i - 1}]\), or

there exists a prime number \(p\) with \(pb_i, b_i^p \in A[b_1, \ldots, b_{i - 1}]\).

Proof

The proof is exactly the same as the proof of Proposition 0CND except for the following changes:

Use Lemma 0CNC instead of Lemma 0CNB which means that for each successor ordinal \(\alpha = \beta + 1\) we either have \(b_\alpha^2, b_\alpha^3 \in B_\beta\) or we have a prime \(p\) and \(pb_\alpha, b_\alpha^p \in B_\beta\).

If \(\alpha\) is a successor ordinal, then take \(D = \{d_{e, i}\} \cup \{b_\alpha^2, b_\alpha^3\}\) or take \(D = \{d_{e, i}\} \cup \{pb_\alpha, b_\alpha^p\}\) depending on which case \(\alpha\) falls into.

In the proof of (2) \(\Rightarrow\) (1) we also need to consider the case where \(B\) is generated over \(A\) by a single element \(b\) with \(pb, b^p \in B\) for some prime number \(p\). Here \(A \subset B\) induces a universal homeomorphism on spectra for example by Algebra, Lemma 0BRA.

This finishes the proof.

Lemma

Let \(p\) be a prime number. Let \(A \to B\) be a ring map which induces an isomorphism \(A[1/p] \to B[1/p]\) (for example if \(p\) is nilpotent in \(A\)). The following are equivalent

\(\Spec(B) \to \Spec(A)\) is a universal homeomorphism, and

the kernel of \(A \to B\) is a locally nilpotent ideal and for every \(b \in B\) there exists a \(p\)-power \(q\) with \(qb\) and \(b^q\) in the image of \(A \to B\).

Proof

If (2) holds, then (1) holds by Algebra, Lemma 0BRA. Assume (1). Then the kernel of \(A \to B\) consists of nilpotent elements by Algebra, Lemma 00FL. Thus we may replace \(A\) by the image of \(A \to B\) and assume that \(A \subset B\). By Algebra, Lemma 0545 the set \[B' = \{b \in B \mid p^nb, b^{p^n} \in A\text{ for some }n \geq 0\}\] is an \(A\)-subalgebra of \(B\) (being closed under products is trivial). We have to show \(B' = B\). If not, then according to Lemma 0CNC there exists a \(b \in B\), \(b \not \in B'\) with either \(b^2, b^3 \in B'\) or there exists a prime number \(\ell\) with \(\ell b, b^\ell \in B'\). We will show both cases lead to a contradiction, thereby proving the lemma.

Since \(A[1/p] = B[1/p]\) we can choose a \(p\)-power \(q\) such that \(qb \in A\).

If \(b^2, b^3 \in B'\) then also \(b^q \in B'\). By definition of \(B'\) we find that \((b^q)^{q'} \in A\) for some \(p\)-power \(q'\). Then \(qq'b, b^{qq'} \in A\) whence \(b \in B'\) which is a contradiction.

Assume now there exists a prime number \(\ell\) with \(\ell b, b^\ell \in B'\). If \(\ell \not = p\) then \(\ell b \in B'\) and \(qb \in A \subset B'\) imply \(b \in B'\) a contradiction. Thus \(\ell = p\) and \(b^p \in B'\) and we get a contradiction exactly as before.

Lemma

Let \(A\) be a ring. Let \(x, y \in A\).

If \(x^3 = y^2\) in \(A\), then \(A \to B = A[t]/(t^2 - x, t^3 - y)\) induces bijections on residue fields and a universal homeomorphism on spectra.

If there is a prime number \(p\) such that \(p^px = y^p\) in \(A\), then \(A \to B = A[t]/(t^p - x, pt - y)\) induces a universal homeomorphism on spectra.

Proof

We will use the criterion of Lemma 04DF to check this. In both cases the ring map is integral. Thus it suffices to show that given a field \(k\) and a ring map \(\varphi : A \to k\) the \(k\)-algebra \(B \otimes_A k\) has a unique prime ideal whose residue field is equal to \(k\) in case (1) and purely inseparable over \(k\) in case (2). See Lemma 01S4.

In case (1) set \(\lambda = 0\) if \(\varphi(x) = 0\) and set \(\lambda = \varphi(y)/\varphi(x)\) if not. Then \(B = k[t]/(t^2 - \lambda^2, t^3 - \lambda^2)\). Thus the result is clear.

In case (2) if the characteristic of \(k\) is \(p\), then we obtain \(\varphi(y) = 0\) and \(B = k[t]/(t^p - \varphi(x))\) which is a local Artinian \(k\)-algebra whose residue field is either \(k\) or a degree \(p\) purely inseparable extension of \(k\). If the characteristic of \(k\) is not \(p\), then setting \(\lambda = \varphi(y)/p\) we see \(B = k[t]/(t - \lambda) = k\) and we conclude as well.

Lemma

Let \(A \to B\) be a ring map.

If \(A \to B\) induces a universal homeomorphism on spectra, then \(B = \colim B_i\) is a filtered colimit of finitely presented \(A\)-algebras \(B_i\) such that \(A \to B_i\) induces a universal homeomorphism on spectra.

If \(A \to B\) induces isomorphisms on residue fields and a universal homeomorphism on spectra, then \(B = \colim B_i\) is a filtered colimit of finitely presented \(A\)-algebras \(B_i\) such that \(A \to B_i\) induces isomorphisms on residue fields and a universal homeomorphism on spectra.

Proof

Proof of (1). We will use the criterion of Algebra, Lemma 07C3. Let \(A \to C\) be of finite presentation and let \(\varphi : C \to B\) be an \(A\)-algebra map. Let \(B' = \varphi(C) \subset B\) be the image. Then \(A \to B'\) induces a universal homeomorphism on spectra by Lemma 0CN7. By Algebra, Lemma 00QN we can write \(B' = \colim_{i \in I} B_i\) with \(A \to B_i\) of finite presentation and surjective transition maps. By Algebra, Lemma 00QO we can choose an index \(0 \in I\) and a factorization \(C \to B_0 \to B'\) of the map \(C \to B'\). We claim that \(\Spec(B_i) \to \Spec(A)\) is a universal homeomorphism for \(i\) sufficiently large. The claim finishes the proof of (1).

Proof of the claim. By Lemma 054M the ring map \(A_{red} \to B'_{red}\) induces a universal homeomorphism on spectra. Thus \(A_{red} \subset B'_{red}\) by Algebra, Lemma 00FL. Setting \(A' = \Im(A \to B')\) we have surjections \(A \to A' \to A_{red}\) inducing bijections \(\Spec(A_{red}) = \Spec(A') = \Spec(A)\). Thus \(A' \subset B'\) induces a universal homeomorphism on spectra. By Proposition 0CNE and the fact that \(B'\) is finite type over \(A'\) we can find \(n\) and \(b'_1, \ldots, b'_n \in B'\) such that \(B' = A'[b'_1, \ldots, b'_n]\) and such that for \(j = 1, \ldots, n\) we have

\((b'_j)^2, (b'_j)^3 \in A'[b'_1, \ldots, b'_{j - 1}]\), or

there exists a prime number \(p\) with \(pb'_j, (b'_j)^p \in A'[b'_1, \ldots, b'_{j - 1}]\).

Choose \(b_1, \ldots, b_n \in B_0\) lifting \(b'_1, \ldots, b'_n\). For \(i \geq 0\) denote \(b_{j, i}\) the image of \(b_j\) in \(B_i\). For large enough \(i\) we will have for \(j = 1, \ldots, n\)

\(b_{j, i}^2, b_{j, i}^3 \in A_i[b_{1, i}, \ldots, b_{j - 1, i}]\), or

there exists a prime number \(p\) with \(pb_{j, i}, b_{j, i}^p \in A_i[b_{1, i}, \ldots, b_{j - 1, i}]\).

Here \(A_i \subset B_i\) is the image of \(A \to B_i\). Observe that \(A \to A_i\) is a surjective ring map whose kernel is a locally nilpotent ideal. After increasing \(i\) more if necessary, we may assume \(B_i\) is generated by \(b_1, \ldots, b_n\) over \(A_i\), in other words \(B_i = A_i[b_1, \ldots, b_n]\). By Algebra, Lemmas 0BRA and 0EUH we conclude that \(A \to A_i \to A_i[b_1] \to \ldots \to A_i[b_1, \ldots, b_n] = B_i\) induce universal homeomorphisms on spectra. This finishes the proof of the claim.

The proof of (2) is exactly the same.

Absolute weak normalization and seminormalization

Motivated by the results proved in the previous section we give the following definition.

Definition

Let \(A\) be a ring.

We say \(A\) is seminormal if for all \(x, y \in A\) with \(x^3 = y^2\) there is a unique \(a \in A\) with \(x = a^2\) and \(y = a^3\).

We say \(A\) is absolutely weakly normal if (a) \(A\) is seminormal and (b) for any prime number \(p\) and \(x, y \in A\) with \(p^px = y^p\) there is a unique \(a \in A\) with \(x = a^p\) and \(y = pa\).

An amusing observation, see [Costa], is that in the definition of seminormal rings it suffices¹⁵ to assume the existence of \(a\). Absolutely weakly normal schemes were defined in [rydh_descent, Appendix B].

Lemma

Being seminormal or being absolutely weakly normal is a local property of rings, see Properties, Definition 01OP.

Proof

Suppose that \(A\) is seminormal and \(f \in A\). Let \(x', y' \in A_f\) with \((x')^3 = (y')^2\). Write \(x' = x/f^{2n}\) and \(y' = y/f^{3n}\) for some \(n \geq 0\) and \(x, y \in A\). After replacing \(x, y\) by \(f^{2m}x, f^{3m}y\) and \(n\) by \(n + m\), we see that \(x^3 = y^2\) in \(A\). Then we find a unique \(a \in A\) with \(x = a^2\) and \(y = a^3\). Setting \(a' = a/f^n\) we get \(x' = (a')^2\) and \(y' = (a')^3\) as desired. Uniqueness of \(a'\) follows from uniqueness of \(a\). In exactly the same manner the reader shows that if \(A\) is absolutely weakly normal, then \(A_f\) is absolutely weakly normal.

Assume \(A\) is a ring and \(f_1, \ldots, f_n \in A\) generate the unit ideal. Assume \(A_{f_i}\) is seminormal for each \(i\). Let \(x, y \in A\) with \(x^3 = y^2\). For each \(i\) we find a unique \(a_i \in A_{f_i}\) with \(x = a_i^2\) and \(y = a_i^3\) in \(A_{f_i}\). By the uniqueness and the result of the first paragraph (which tells us that \(A_{f_if_j}\) is seminormal) we see that \(a_i\) and \(a_j\) map to the same element of \(A_{f_if_j}\). By Algebra, Lemma 00EJ we find a unique \(a \in A\) mapping to \(a_i\) in \(A_{f_i}\) for all \(i\). Then \(x = a^2\) and \(y = a^3\) by the same token. Clearly this \(a\) is unique. Thus \(A\) is seminormal. If we assume \(A_{f_i}\) is absolutely weakly normal, then the exact same argument shows that \(A\) is absolutely weakly normal.

Next we define seminormal schemes and absolutely weakly normal schemes.

Definition

Let \(X\) be a scheme.

We say \(X\) is seminormal if every \(x \in X\) has an affine open neighbourhood \(\Spec(R) = U \subset X\) such that the ring \(R\) is seminormal.

We say \(X\) is absolutely weakly normal if every \(x \in X\) has an affine open neighbourhood \(\Spec(R) = U \subset X\) such that the ring \(R\) is absolutely weakly normal.

Here is the obligatory lemma.

Lemma

Let \(X\) be a scheme. The following are equivalent:

The scheme \(X\) is seminormal.

For every affine open \(U \subset X\) the ring \(\mathcal{O}_X(U)\) is seminormal.

There exists an affine open covering \(X = \bigcup U_i\) such that each \(\mathcal{O}_X(U_i)\) is seminormal.

There exists an open covering \(X = \bigcup X_j\) such that each open subscheme \(X_j\) is seminormal.

Moreover, if \(X\) is seminormal then every open subscheme is seminormal. The same statements are true with “seminormal” replaced by “absolutely weakly normal”.

Proof

Combine Properties, Lemma 01OR and Lemma 0EUM.

Lemma

A seminormal scheme or ring is reduced. A fortiori the same is true for absolutely weakly normal schemes or rings.

Proof

Let \(A\) be a ring. If \(a \in A\) is nonzero but \(a^2 = 0\), then \(a^2 = 0^2\) and \(a^3 = 0^3\) and hence \(A\) is not seminormal.

Lemma

Let \(A\) be a ring.

The category of ring maps \(A \to B\) inducing a universal homeomorphism on spectra has a final object \(A \to A^{awn}\).

Given \(A \to B\) in the category of (1) the resulting map \(B \to A^{awn}\) is an isomorphism if and only if \(B\) is absolutely weakly normal.

The category of ring maps \(A \to B\) inducing isomorphisms on residue fields and a universal homeomorphism on spectra has a final object \(A \to A^{sn}\).

Given \(A \to B\) in the category of (3) the resulting map \(B \to A^{sn}\) is an isomorphism if and only if \(B\) is seminormal.

For any ring map \(\varphi : A \to A'\) there are unique maps \(\varphi^{awn} : A^{awn} \to (A')^{awn}\) and \(\varphi^{sn} : A^{sn} \to (A')^{sn}\) compatible with \(\varphi\).

Proof

We prove (1) and (2) and we omit the proof of (3) and (4) and the final statement. Consider the category of \(A\)-algebras of the form \[B = A[x_1, \ldots, x_n]/J\] where \(J\) is a finitely generated ideal such that \(A \to B\) defines a universal homeomorphism on spectra. We claim this category is directed (Categories, Definition 002V). Namely, given \[B = A[x_1, \ldots, x_n]/J \quad\text{and}\quad B' = A[x_1, \ldots, x_{n'}]/J'\] then we can consider \[B'' = A[x_1, \ldots, x_{n + n'}]/J''\] where \(J''\) is generated by the elements of \(J\) and the elements \(f(x_{n + 1}, \ldots, x_{n + n'})\) where \(f \in J'\). Then we have \(A\)-algebra homomorphisms \(B \to B''\) and \(B' \to B''\) which induce an isomorphism \(B \otimes_A B' \to B''\). It follows from Lemmas 0CEU and 0CEV that \(\Spec(B'') \to \Spec(A)\) is a universal homeomorphism and hence \(A \to B''\) is in our category. Finally, given \(\varphi, \varphi' : B \to B'\) in our category with \(B\) as displayed above, then we consider the quotient \(B''\) of \(B'\) by the ideal generated by \(\varphi(x_i) - \varphi'(x_i)\), \(i = 1, \ldots, n\). Since \(\Spec(B') = \Spec(B)\) we see that \(\Spec(B'') \to \Spec(B')\) is a bijective closed immersion hence a universal homeomorphism. Thus \(B''\) is in our category and \(\varphi, \varphi'\) are equalized by \(B' \to B''\). This completes the proof of our claim. We set \[A^{awn} = \colim B\] where the colimit is over the category just described. Observe that \(A \to A^{awn}\) induces a universal homeomorphism on spectra by Lemma 0CN8 (this is where we use the category is directed).

Given a ring map \(A \to B\) of finite presentation inducing a universal homeomorphism on spectra, we get a canonical map \(B \to A^{awn}\) by the very construction of \(A^{awn}\). Since every \(A \to B\) as in (1) is a filtered colimit of \(A \to B\) as in (1) of finite presentation (Lemma 0EUJ), we see that \(A \to A^{awn}\) is final in the category (1).

Let \(x, y \in A^{awn}\) be elements such that \(x^3 = y^2\). Then \(A^{awn} \to A^{awn}[t]/(t^2 - x, t^3 - y)\) induces a universal homeomorphism on spectra by Lemma 0EUI. Thus \(A \to A^{awn}[t]/(t^2 - x, t^3 - y)\) is in the category (1) and we obtain a unique \(A\)-algebra map \(A^{awn}[t]/(t^2 - x, t^3 - y) \to A^{awn}\). The image \(a \in A^{awn}\) of \(t\) is therefore the unique element such that \(a^2 = x\) and \(a^3 = y\) in \(A^{awn}\). In exactly the same manner, given a prime \(p\) and \(x, y \in A^{awn}\) with \(p^px = y^p\) we find a unique \(a \in A^{awn}\) with \(a^p = x\) and \(pq = y\). Thus \(A^{awn}\) is absolutely weakly normal by definition.

Finally, let \(A \to B\) be in the category (1) with \(B\) absolutely weakly normal. Since \(A^{awn} \to B^{awn}\) induces a universal homeomorphism on spectra and since \(A^{awn}\) is reduced (Lemma 0EUQ) we find \(A^{awn} \subset B^{awn}\) (see Algebra, Lemma 00FL). If this inclusion is not an equality, then Lemma 0CNC implies there is an element \(b \in B^{awn}\), \(b \not \in A^{awn}\) such that either \(b^2, b^3 \in A^{awn}\) or \(pb, b^p \in A^{awn}\) for some prime number \(p\). However, by the existence and uniqueness in Definition 0EUL this forces \(b \in A^{awn}\) and hence we obtain the contradiction that finishes the proof.

Lemma

Let \(X\) be a scheme.

The category of universal homeomorphisms \(Y \to X\) has an initial object \(X^{awn} \to X\).

Given \(Y \to X\) in the category of (1) the resulting morphism \(X^{awn} \to Y\) is an isomorphism if and only if \(Y\) is absolutely weakly normal.

The category of universal homeomorphisms \(Y \to X\) which induce ismomorphisms on residue fields has an initial object \(X^{sn} \to X\).

Given \(Y \to X\) in the category of (3) the resulting morphism \(X^{sn} \to Y\) is an isomorphism if and only if \(Y\) is seminormal.

For any morphism \(h : X' \to X\) of schemes there are unique morphisms \(h^{awn} : (X')^{awn} \to X^{awn}\) and \(h^{sn} : (X')^{sn} \to X^{sn}\) compatible with \(h\).

Proof

We will prove (1) and (2) and omit the proof of (3) and (4). Let \(h : X' \to X\) be a morphism of schemes. If (1) holds for \(X\) and \(X'\), then \(X' \times_X X^{awn} \to X'\) is a universal homeomorphism and hence we get a unique morphism \((X')^{awn} \to X' \times_X X^{awn}\) over \(X'\) by the universal property of \((X')^{awn} \to X'\). Composed with the projection \(X' \times_X X^{awn} \to X^{awn}\) we obtain \(h^{awn}\). If in addition (2) holds for \(X\) and \(X'\) and \(h\) is an open immersion, then \(X' \times_X X^{awn}\) is absolutely weakly normal (Lemma 0EUP) and we deduce that \((X')^{awn} \to X' \times_X X^{awn}\) is an isomorphism.

Recall that any universal homeomorphism is affine, see Lemma 04DE. Thus if \(X\) is affine then (1) and (2) follow immediately from Lemma 0EUR. Let \(X\) be a scheme and let \(\mathcal{B}\) be the set of affine opens of \(X\). For each \(U \in \mathcal{B}\) we obtain \(U^{awn} \to U\) and for \(V \subset U\), \(V, U \in \mathcal{B}\) we obtain a canonical isomorphism \(\rho_{V, U} : V^{awn} \to V \times_U U^{awn}\) by the discussion in the previous paragraph. Thus by relative glueing (Constructions, Lemma 01LH) we obtain a morphism \(X^{awn} \to X\) which restricts to \(U^{awn}\) over \(U\) compatibly with the \(\rho_{V, U}\). Next, let \(Y \to X\) be a universal homeomorphism. Then \(U \times_X Y \to U\) is a universal homeomorphism for \(U \in \mathcal{B}\) and we obtain a unique morphism \(g_U : U^{awn} \to U \times_X Y\) over \(U\). These \(g_U\) are compatible with the morphisms \(\rho_{V, U}\); details omitted. Hence there is a unique morphism \(g : X^{awn} \to Y\) over \(X\) agreeing with \(g_U\) over \(U\), see Constructions, Remark 01LK. This proves (1) for \(X\). Part (2) follows because it holds affine locally.

Definition

Let \(X\) be a scheme.

The morphism \(X^{sn} \to X\) constructed in Lemma 0EUS is the seminormalization of \(X\).

The morphism \(X^{awn} \to X\) constructed in Lemma 0EUS is the absolute weak normalization of \(X\).

To be sure, the seminormalization \(X^{sn}\) of \(X\) is a seminormal scheme and the absolute weak normalization \(X^{awn}\) is an absolutely weakly normal scheme. Moreover, for any morphism \(h : Y \to X\) of schemes we obtain a canonical commutative diagram \[\xymatrix{ Y^{awn} \ar[d]^{h^{awn}} \ar[r] & Y^{sn} \ar[d]^{h^{sn}} \ar[r] & Y \ar[d]^h \\ X^{awn} \ar[r] & X^{sn} \ar[r] & X }\] of schemes; the arrows \(h^{sn}\) and \(h^{awn}\) are the unique ones compatible with \(h\).

Lemma

Let \(X\) be a scheme. The following are equivalent

\(X\) is seminormal,

\(X\) is equal to its own seminormalization, i.e., the morphism \(X^{sn} \to X\) is an isomorphism,

if \(\pi : Y \to X\) is a universal homeomorphism inducing isomorphisms on residue fields with \(Y\) reduced, then \(\pi\) is an isomorphism.

Proof

The equivalence of (1) and (2) is clear from Lemma 0EUS. If (3) holds, then \(X^{sn} \to X\) is an isomorphism and we see that (2) holds.

Assume (2) holds and let \(\pi : Y \to X\) be a universal homeomorphism inducing isomorphisms on residue fields with \(Y\) reduced. Then there exists a factorization \(X \to Y \to X\) of \(\text{id}_X\) by Lemma 0EUS. Then \(X \to Y\) is a closed immersion (by Schemes, Lemma 01KT and the fact that \(\pi\) is separated for example by Lemma 05VE). Since \(X \to Y\) is also a bijection on points, the reducedness of \(Y\) shows that it has to be an isomorphism. This finishes the proof.

Lemma

Let \(X\) be a scheme. The following are equivalent

\(X\) is absolutely weakly normal,

\(X\) is equal to its own absolute weak normalization, i.e., the morphism \(X^{awn} \to X\) is an isomorphism,

if \(\pi : Y \to X\) is a universal homeomorphism with \(Y\) reduced, then \(\pi\) is an isomorphism.

Proof

This is proved in exactly the same manner as Lemma 0H3G.

Finite locally free morphisms

In many papers the authors use finite flat morphisms when they really mean finite locally free morphisms. The reason is that if the base is locally Noetherian then this is the same thing. But in general it is not, see Exercises, Exercise 02CS.

Definition

Let \(f : X \to S\) be a morphism of schemes. We say \(f\) is finite locally free if \(f\) is affine and \(f_*\mathcal{O}_X\) is a finite locally free \(\mathcal{O}_S\)-module. In this case we say \(f\) is has rank or degree \(d\) if the sheaf \(f_*\mathcal{O}_X\) is finite locally free of degree \(d\).

Note that if \(f : X \to S\) is finite locally free then \(S\) is the disjoint union of open and closed subschemes \(S_d\) such that \(f^{-1}(S_d) \to S_d\) is finite locally free of degree \(d\).

Lemma

Let \(f : X \to S\) be a morphism of schemes. The following are equivalent:

\(f\) is finite locally free,

\(f\) is finite, flat, and locally of finite presentation.

If \(S\) is locally Noetherian these are also equivalent to

\(f\) is finite and flat.

Proof

Let \(V \subset S\) be affine open. In all three cases the morphism is affine hence \(f^{-1}(V)\) is affine. Thus we may write \(V = \Spec(R)\) and \(f^{-1}(V) = \Spec(A)\) for some \(R\)-algebra \(A\). Assume (1). This means we can cover \(S\) by affine opens \(V = \Spec(R)\) such that \(A\) is finite free as an \(R\)-module. Then \(R \to A\) is of finite presentation by Algebra, Lemma 0D46. Thus (2) holds. Conversely, assume (2). For every affine open \(V = \Spec(R)\) of \(S\) the ring map \(R \to A\) is finite and of finite presentation and \(A\) is flat as an \(R\)-module. By Algebra, Lemma 0564 we see that \(A\) is finitely presented as an \(R\)-module. Thus Algebra, Lemma 00NX implies \(A\) is finite locally free. Thus (1) holds. The Noetherian case follows as a finite module over a Noetherian ring is a finitely presented module, see Algebra, Lemma 00FP.

Lemma

A composition of finite locally free morphisms is finite locally free.

Proof

Omitted.

Lemma

A base change of a finite locally free morphism is finite locally free.

Proof

Omitted.

Lemma

Let \(f : X \to S\) be a finite locally free morphism of schemes. There exists a disjoint union decomposition \(S = \coprod_{d \geq 0} S_d\) by open and closed subschemes such that setting \(X_d = f^{-1}(S_d)\) the restrictions \(f|_{X_d}\) are finite locally free morphisms \(X_d \to S_d\) of degree \(d\).

Proof

This is true because a finite locally free sheaf locally has a well defined rank. Details omitted.

Lemma

Let \(f : Y \to X\) be a finite morphism with \(X\) affine. There exists a diagram \[\xymatrix{ Z' \ar[rd] & Y' \ar[l]^i \ar[d] \ar[r] & Y \ar[d] \\ & X' \ar[r] & X }\] where

\(Y' \to Y\) and \(X' \to X\) are surjective finite locally free,

\(Y' = X' \times_X Y\),

\(i : Y' \to Z'\) is a closed immersion,

\(Z' \to X'\) is finite locally free, and

\(Z' = \bigcup_{j = 1, \ldots, m} Z'_j\) is a (set theoretic) finite union of closed subschemes, each of which maps isomorphically to \(X'\).

Proof

Write \(X = \Spec(A)\) and \(Y = \Spec(B)\). See also More on Algebra, Section 052Y. Let \(x_1, \ldots, x_n \in B\) be generators of \(B\) over \(A\). For each \(i\) we can choose a monic polynomial \(P_i(T) \in A[T]\) such that \(P(x_i) = 0\) in \(B\). By Algebra, Lemma 03HS (applied \(n\) times) there exists a finite locally free ring extension \(A \subset A'\) such that each \(P_i\) splits completely: \[P_i(T) = \prod\nolimits_{k = 1, \ldots, d_i} (T - \alpha_{ik})\] for certain \(\alpha_{ik} \in A'\). Set \[C = A'[T_1, \ldots, T_n]/(P_1(T_1), \ldots, P_n(T_n))\] and \(B' = A' \otimes_A B\). The map \(C \to B'\), \(T_i \mapsto 1 \otimes x_i\) is an \(A'\)-algebra surjection. Setting \(X' = \Spec(A')\), \(Y' = \Spec(B')\) and \(Z' = \Spec(C)\) we see that (1) – (4) hold. Part (5) holds because set theoretically \(\Spec(C)\) is the union of the closed subschemes cut out by the ideals \[(T_1 - \alpha_{1k_1}, T_2 - \alpha_{2k_2}, \ldots, T_n - \alpha_{nk_n})\] for any \(1 \leq k_i \leq d_i\).

The following lemma is stated in the correct generality in Lemma 03J2 below.

Lemma

Let \(f : Y \to X\) be a finite morphism of schemes. Let \(T \subset Y\) be a closed nowhere dense subset of \(Y\). Then \(f(T) \subset X\) is a closed nowhere dense subset of \(X\).

Proof

By Lemma 01WN we know that \(f(T) \subset X\) is closed. Let \(X = \bigcup X_i\) be an affine covering. Since \(T\) is nowhere dense in \(Y\), we see that also \(T \cap f^{-1}(X_i)\) is nowhere dense in \(f^{-1}(X_i)\). Hence if we can prove the theorem in the affine case, then we see that \(f(T) \cap X_i\) is nowhere dense. This then implies that \(T\) is nowhere dense in \(X\) by Topology, Lemma 03HP.

Assume \(X\) is affine. Choose a diagram \[\xymatrix{ Z' \ar[rd] & Y' \ar[l]^i \ar[d]^{f'} \ar[r]_a & Y \ar[d]^f \\ & X' \ar[r]^b & X }\] as in Lemma 03HW. The morphisms \(a\), \(b\) are open since they are finite locally free (Lemmas 02KB and 01UA). Hence \(T' = a^{-1}(T)\) is nowhere dense, see Topology, Lemma 03HR. The morphism \(b\) is surjective and open. Hence, if we can prove \(f'(T') = b^{-1}(f(T))\) is nowhere dense, then \(f(T)\) is nowhere dense, see Topology, Lemma 03HR. As \(i\) is a closed immersion, by Topology, Lemma 03HQ we see that \(i(T') \subset Z'\) is closed and nowhere dense. Thus we have reduced the problem to the case discussed in the following paragraph.

Assume that \(Y = \bigcup_{i = 1, \ldots, n} Y_i\) is a finite union of closed subsets, each mapping isomorphically to \(X\). Consider \(T_i = Y_i \cap T\). If each of the \(T_i\) is nowhere dense in \(Y_i\), then each \(f(T_i)\) is nowhere dense in \(X\) as \(Y_i \to X\) is an isomorphism. Hence \(f(T) = f(T_i)\) is a finite union of nowhere dense closed subsets of \(X\) and we win, see Topology, Lemma 03HO. Suppose not, say \(T_1\) contains a nonempty open \(V \subset Y_1\). We are going to show this leads to a contradiction. Consider \(Y_2 \cap V \subset V\). This is either a proper closed subset, or equal to \(V\). In the first case we replace \(V\) by \(V \setminus V \cap Y_2\), so \(V \subset T_1\) is open in \(Y_1\) and does not meet \(Y_2\). In the second case we have \(V \subset Y_1 \cap Y_2\) is open in both \(Y_1\) and \(Y_2\). Repeat sequentially with \(i = 3, \ldots, n\). The result is a disjoint union decomposition \[\{1, \ldots, n\} = I_1 \amalg I_2, \quad 1 \in I_1\] and an open \(V\) of \(Y_1\) contained in \(T_1\) such that \(V \subset Y_i\) for \(i \in I_1\) and \(V \cap Y_i = \emptyset\) for \(i \in I_2\). Set \(U = f(V)\). This is an open of \(X\) since \(f|_{Y_1} : Y_1 \to X\) is an isomorphism. Then \[f^{-1}(U) = V\ \amalg\ \bigcup\nolimits_{i \in I_2} (Y_i \cap f^{-1}(U))\] As \(\bigcup_{i \in I_2} Y_i\) is closed, this implies that \(V \subset f^{-1}(U)\) is open, hence \(V \subset Y\) is open. This contradicts the assumption that \(T\) is nowhere dense in \(Y\), as desired.

Rational maps

Let \(X\) be a scheme. Note that if \(U\), \(V\) are dense open in \(X\), then so is \(U \cap V\).

Definition

Let \(X\), \(Y\) be schemes.

Let \(f : U \to Y\), \(g : V \to Y\) be morphisms of schemes defined on dense open subsets \(U\), \(V\) of \(X\). We say that \(f\) is equivalent to \(g\) if \(f|_W = g|_W\) for some \(W \subset U \cap V\) dense open in \(X\).

A rational map from \(X\) to \(Y\) is an equivalence class for the equivalence relation defined in (1).

If \(X\), \(Y\) are schemes over a base scheme \(S\) we say that a rational map from \(X\) to \(Y\) is an \(S\)-rational map from \(X\) to \(Y\) if there exists a representative \(f : U \to Y\) of the equivalence class which is an \(S\)-morphism.

We say that two morphisms \(f\), \(g\) as in (1) of the definition define the same rational map instead of saying that they are equivalent. In some cases rational maps are determined by maps on local rings at generic points.

Lemma

Let \(S\) be a scheme. Let \(X\) and \(Y\) be schemes over \(S\). Assume \(X\) has finitely many irreducible components with generic points \(x_1, \ldots, x_n\). Let \(s_i \in S\) be the image of \(x_i\). Consider the map \[\left\{ \begin{matrix} S\text{-rational maps} \\ \text{from }X\text{ to }Y \end{matrix} \right\} \longrightarrow \left\{ \begin{matrix} (y_1, \varphi_1, \ldots, y_n, \varphi_n)\text{ where } y_i \in Y\text{ lies over }s_i\text{ and}\\ \varphi_i : \mathcal{O}_{Y, y_i} \to \mathcal{O}_{X, x_i} \text{ is a local }\mathcal{O}_{S, s_i}\text{-algebra map} \end{matrix} \right\}\] which sends \(f : U \to Y\) to the \(2n\)-tuple with \(y_i = f(x_i)\) and \(\varphi_i = f^\sharp_{x_i}\). Then

If \(Y \to S\) is locally of finite type, then the map is injective.

If \(Y \to S\) is locally of finite presentation, then the map is bijective.

If \(Y \to S\) is locally of finite type and \(X\) reduced, then the map is bijective.

Proof

Observe that any dense open of \(X\) contains the points \(x_i\) so the construction makes sense. To prove (1) or (2) we may replace \(X\) by any dense open. Thus if \(Z_1, \ldots, Z_n\) are the irreducible components of \(X\), then we may replace \(X\) by \(X \setminus \bigcup_{i \not = j} Z_i \cap Z_j\). After doing this \(X\) is the disjoint union of its irreducible components (viewed as open and closed subschemes). Then both the right hand side and the left hand side of the arrow are products over the irreducible components and we reduce to the case where \(X\) is irreducible.

Assume \(X\) is irreducible with generic point \(x\) lying over \(s \in S\). Part (1) follows from part (1) of Lemma 0BX6. Parts (2) and (3) follow from part (2) of the same lemma.

Definition

Let \(X\) be a scheme. A rational function on \(X\) is a rational map from \(X\) to \(\mathbf{A}^1_{\mathbf{Z}}\).

See Constructions, Definition 01M0 for the definition of the affine line \(\mathbf{A}^1\). Let \(X\) be a scheme over \(S\). For any open \(U \subset X\) a morphism \(U \to \mathbf{A}^1_{\mathbf{Z}}\) is the same as a morphism \(U \to \mathbf{A}^1_S\) over \(S\). Hence a rational function is also the same as a \(S\)-rational map from \(X\) into \(\mathbf{A}^1_S\).

Recall that we have the canonical identification \(\Mor(T, \mathbf{A}^1_{\mathbf{Z}}) = \Gamma(T, \mathcal{O}_T)\) for any scheme \(T\), see Schemes, Example 01JH. Hence \(\mathbf{A}^1_{\mathbf{Z}}\) is a ring-object in the category of schemes. More precisely, the morphisms \[\begin{eqnarray*} + : \mathbf{A}^1_{\mathbf{Z}} \times \mathbf{A}^1_{\mathbf{Z}} & \longrightarrow & \mathbf{A}^1_{\mathbf{Z}} \\ (f, g) & \longmapsto & f + g \\ * : \mathbf{A}^1_{\mathbf{Z}} \times \mathbf{A}^1_{\mathbf{Z}} & \longrightarrow & \mathbf{A}^1_{\mathbf{Z}} \\ (f, g) & \longmapsto & fg \end{eqnarray*}\] satisfy all the axioms of the addition and multiplication in a ring (commutative with \(1\) as always). Hence also the set of rational maps into \(\mathbf{A}^1_{\mathbf{Z}}\) has a natural ring structure.

Definition

Let \(X\) be a scheme. The ring of rational functions on \(X\) is the ring \(R(X)\) whose elements are rational functions with addition and multiplication as just described.

For schemes with finitely many irreducible components we can compute this.

Lemma

Let \(X\) be a scheme with finitely many irreducible components \(X_1, \ldots, X_n\). If \(\eta_i \in X_i\) is the generic point, then \[R(X) = \mathcal{O}_{X, \eta_1} \times \ldots \times \mathcal{O}_{X, \eta_n}\] If \(X\) is reduced this is equal to \(\prod \kappa(\eta_i)\). If \(X\) is integral then \(R(X) = \mathcal{O}_{X, \eta} = \kappa(\eta)\) is a field.

Proof

Let \(U \subset X\) be an open dense subset. Then \(U_i = (U \cap X_i) \setminus (\bigcup_{j \not = i} X_j)\) is nonempty open as it contained \(\eta_i\), contained in \(X_i\), and \(\bigcup U_i \subset U \subset X\) is dense. Thus the identification in the lemma comes from the string of equalities \[\begin{align*} R(X) & = \colim_{U \subset X\text{ open dense}} \Mor(U, \mathbf{A}^1_\mathbf{Z}) \\ & = \colim_{U \subset X\text{ open dense}} \mathcal{O}_X(U) \\ & = \colim_{\eta_i \in U_i \subset X\text{ open}} \prod \mathcal{O}_X(U_i) \\ & = \prod \colim_{\eta_i \in U_i \subset X\text{ open}} \mathcal{O}_X(U_i) \\ & = \prod \mathcal{O}_{X, \eta_i} \end{align*}\] where the second equality is Schemes, Example 01JH. The final statement follows from Algebra, Lemma 00EU.

Definition

Let \(X\) be an integral scheme. The function field, or the field of rational functions of \(X\) is the field \(R(X)\).

We may occasionally indicate this field \(k(X)\) instead of \(R(X)\). We can use the notion of the function field to elucidate the separation condition on an integral scheme. Note that by Lemma 01RV on an integral scheme every local ring \(\mathcal{O}_{X, x}\) may be viewed as a local subring of \(R(X)\).

Lemma

Let \(X\) be an integral scheme and let \(U \subset X\) be a nonempty open. Using the embeddings above, we have \[\Gamma(U, \mathcal{O}_X) = \bigcap_{x \in U} \mathcal{O}_{X, x}\] as subrings of \(R(X)\).

Proof

Let \(\eta\) be the generic point of \(X\). The map \(\Gamma(U, \mathcal{O}_X) \to \mathcal{O}_{X, \eta} = R(X)\) is injective. Indeed, this can be checked on an affine open covering of \(U\), where it is the injection of a domain into its fraction field. Its image is contained in every \(\mathcal{O}_{X, x}\) for \(x \in U\).

Conversely, let \(f \in R(X)\) belong to \(\mathcal{O}_{X, x}\) for every \(x \in U\). For each \(x\), choose an open neighbourhood \(U_x \subset U\) and a section \(f_x \in \Gamma(U_x, \mathcal{O}_X)\) representing \(f\). On every nonempty overlap \(U_x \cap U_y\), the sections \(f_x\) and \(f_y\) have the same image in \(R(X)\) and hence are equal by the injectivity just proved. Thus the \(f_x\) glue to a section over \(U\) whose image is \(f\).

Lemma

Let \(X\) be an integral separated scheme. Let \(Z_1\), \(Z_2\) be distinct irreducible closed subsets of \(X\). Let \(\eta_i\) be the generic point of \(Z_i\). If \(Z_1 \not\subset Z_2\), then \(\mathcal{O}_{X, \eta_1} \not \subset \mathcal{O}_{X, \eta_2}\) as subrings of \(R(X)\). In particular, if \(Z_1 = \{x\}\) consists of one closed point \(x\), there exists a function regular in a neighborhood of \(x\) which is not in \(\mathcal{O}_{X, \eta_{2}}\).

Proof

First observe that under the assumption of \(X\) being separated, there is a unique map of schemes \(\Spec(\mathcal{O}_{X, \eta_2}) \to X\) over \(X\) such that the composition \[\Spec(R(X)) \longrightarrow \Spec(\mathcal{O}_{X, \eta_2}) \longrightarrow X\] is the canonical map \(\Spec(R(X)) \to X\). Namely, there is the canonical map \(can : \Spec(\mathcal{O}_{X, \eta_2}) \to X\), see Schemes, Equation (02NA). Given a second morphism \(a\) to \(X\), we have that \(a\) agrees with \(can\) on the generic point of \(\Spec(\mathcal{O}_{X, \eta_2})\) by assumption. Now \(X\) being separated guarantees that the subset in \(\Spec(\mathcal{O}_{X, \eta_2})\) where these two maps agree is closed, see Schemes, Lemma 01KM. Hence \(a = can\) on all of \(\Spec(\mathcal{O}_{X, \eta_2})\).

Assume \(Z_1 \not \subset Z_2\) and assume on the contrary that \(\mathcal{O}_{X, \eta_{1}} \subset \mathcal{O}_{X, \eta_{2}}\) as subrings of \(R(X)\). Then we would obtain a second morphism \[\Spec(\mathcal{O}_{X, \eta_{2}}) \longrightarrow \Spec(\mathcal{O}_{X, \eta_{1}}) \longrightarrow X.\] By the above this composition would have to be equal to \(can\). This implies that \(\eta_2\) specializes to \(\eta_1\) (see Schemes, Lemma 01J7). But this contradicts our assumption \(Z_1 \not \subset Z_2\).

Definition

Let \(\varphi\) be a rational map between two schemes \(X\) and \(Y\). We say \(\varphi\) is defined in a point \(x \in X\) if there exists a representative \((U, f)\) of \(\varphi\) with \(x \in U\). The domain of definition of \(\varphi\) is the set of all points where \(\varphi\) is defined.

With this definition it isn’t true in general that \(\varphi\) has a representative which is defined on all of the domain of definition.

Lemma

Let \(X\) and \(Y\) be schemes. Assume \(X\) reduced and \(Y\) separated. Let \(\varphi\) be a rational map from \(X\) to \(Y\) with domain of definition \(U \subset X\). Then there exists a unique morphism \(f : U \to Y\) representing \(\varphi\). If \(X\) and \(Y\) are schemes over a separated scheme \(S\) and if \(\varphi\) is an \(S\)-rational map, then \(f\) is a morphism over \(S\).

Proof

Let \((V, g)\) and \((V', g')\) be representatives of \(\varphi\). Then \(g, g'\) agree on a dense open subscheme \(W \subset V \cap V'\). On the other hand, the equalizer \(E\) of \(g|_{V \cap V'}\) and \(g'|_{V \cap V'}\) is a closed subscheme of \(V \cap V'\) (Schemes, Lemma 01KM). Now \(W \subset E\) implies that \(E = V \cap V'\) set theoretically. As \(V \cap V'\) is reduced we conclude \(E = V \cap V'\) scheme theoretically, i.e., \(g|_{V \cap V'} = g'|_{V \cap V'}\). It follows that we can glue the representatives \(g : V \to Y\) of \(\varphi\) to a morphism \(f : U \to Y\), see Schemes, Lemma 01JB. We omit the proof of the final statement.

In general it does not make sense to compose rational maps. The reason is that the image of a representative of the first rational map may have empty intersection with the domain of definition of the second. However, if we assume that our schemes are irreducible and we look at dominant rational maps, then we can compose rational maps.

Definition

Let \(X\) and \(Y\) be irreducible schemes. A rational map from \(X\) to \(Y\) is called dominant if any representative \(f : U \to Y\) is a dominant morphism of schemes.

By Lemma 01RM it is equivalent to require that the generic point \(\eta \in X\) maps to the generic point \(\xi\) of \(Y\), i.e., \(f(\eta) = \xi\) for any representative \(f : U \to Y\). We can compose a dominant rational map \(\varphi\) between irreducible schemes \(X\) and \(Y\) with an arbitrary rational map \(\psi\) from \(Y\) to \(Z\). Namely, choose representatives \(f : U \to Y\) with \(U \subset X\) open dense and \(g : V \to Z\) with \(V \subset Y\) open dense. Then \(W = f^{-1}(V) \subset X\) is open nonempty (because it contains the generic point of \(X\)) and we let \(\psi \circ \varphi\) be the equivalence class of \(g \circ f|_W : W \to Z\). We omit the verification that this is well defined.

In this way we obtain a category whose objects are irreducible schemes and whose morphisms are dominant rational maps. Given a base scheme \(S\) we can similarly define a category whose objects are irreducible schemes over \(S\) and whose morphisms are dominant \(S\)-rational maps.

Definition

Let \(X\) and \(Y\) be irreducible schemes.

We say \(X\) and \(Y\) are birational if \(X\) and \(Y\) are isomorphic in the category of irreducible schemes and dominant rational maps.

Assume \(X\) and \(Y\) are schemes over a base scheme \(S\). We say \(X\) and \(Y\) are \(S\)-birational if \(X\) and \(Y\) are isomorphic in the category of irreducible schemes over \(S\) and dominant \(S\)-rational maps.

If \(X\) and \(Y\) are birational irreducible schemes, then the set of rational maps from \(X\) to \(Z\) is bijective with the set of rational map from \(Y\) to \(Z\) for all schemes \(Z\) (functorially in \(Z\)). For “general” irreducible schemes this is just one possible definition. Another would be to require \(X\) and \(Y\) have isomorphic rings of rational functions. For varieties these conditions are equivalent, see Lemma 0BAD.

Lemma

Let \(X\) and \(Y\) be irreducible schemes.

The schemes \(X\) and \(Y\) are birational if and only if they have isomorphic nonempty opens.

Assume \(X\) and \(Y\) are schemes over a base scheme \(S\). Then \(X\) and \(Y\) are \(S\)-birational if and only if there are nonempty opens \(U \subset X\) and \(V \subset Y\) which are \(S\)-isomorphic.

Proof

Assume \(X\) and \(Y\) are birational. Let \(f : U \to Y\) and \(g : V \to X\) define inverse dominant rational maps from \(X\) to \(Y\) and from \(Y\) to \(X\). We may assume \(V\) affine. We may replace \(U\) by an affine open of \(f^{-1}(V)\). As \(g \circ f\) is the identity as a dominant rational map, we see that the composition \(U \to V \to X\) is the identity on a dense open of \(U\). Thus after replacing \(U\) by a smaller affine open we may assume that \(U \to V \to X\) is the inclusion of \(U\) into \(X\). It follows that \(U \to V\) is an immersion (apply Schemes, Lemma 01KT to \(U \to g^{-1}(U) \to U\)). However, switching the roles of \(U\) and \(V\) and redoing the argument above, we see that there exists a nonempty affine open \(V' \subset V\) such that the inclusion factors as \(V' \to U \to V\). Then \(V' \to U\) is necessarily an open immersion. Namely, \(V' \to f^{-1}(V') \to V'\) are monomorphisms (Schemes, Lemma 01L7) composing to the identity, hence isomorphisms. Thus \(V'\) is isomorphic to an open of both \(X\) and \(Y\). In the \(S\)-rational maps case, the exact same argument works.

Remark

Here is a generalization of the category of irreducible schemes and dominant rational maps. For a scheme \(X\) denote \(X^0\) the set of points \(x \in X\) with \(\dim(\mathcal{O}_{X, x}) = 0\), in other words, \(X^0\) is the set of generic points of irreducible components of \(X\). Then we can consider the category with

objects are schemes \(X\) such that every quasi-compact open has finitely many irreducible components, and

morphisms from \(X\) to \(Y\) are rational maps \(f : U \to Y\) from \(X\) to \(Y\) such that \(f(U^0) = Y^0\).

If \(U \subset X\) is a dense open of a scheme, then \(U^0 \subset X^0\) need not be an equality, but if \(X\) is an object of our category, then this is the case. Thus given two morphisms in our category, the composition is well defined and a morphism in our category.

Remark

There is a variant of Definition 01RS where we consider only those morphism \(U \to Y\) defined on scheme theoretically dense open subschemes \(U \subset X\). We use Lemma 01RF to see that we obtain an equivalence relation. An equivalence class of these is called a pseudo-morphism from \(X\) to \(Y\). If \(X\) is reduced the two notions coincide.

Birational morphisms

You may be used to the notion of a birational map of varieties having the property that it is an isomorphism over an open subset of the target. However, in general a birational morphism may not be an isomorphism over any nonempty open, see Example 01RQ. Here is the formal definition.

Definition

Let \(X\), \(Y\) be schemes. Assume \(X\) and \(Y\) have finitely many irreducible components. We say a morphism \(f : X \to Y\) is birational if

\(f\) induces a bijection between the set of generic points of irreducible components of \(X\) and the set of generic points of the irreducible components of \(Y\), and

for every generic point \(\eta \in X\) of an irreducible component of \(X\) the local ring map \(\mathcal{O}_{Y, f(\eta)} \to \mathcal{O}_{X, \eta}\) is an isomorphism.

We will see below that the fibres of a birational morphism over generic points are singletons. Moreover, we will see that in most cases one encounters in practice the existence of a birational morphism between irreducible schemes \(X\) and \(Y\) implies \(X\) and \(Y\) are birational schemes.

Lemma

Let \(f : X \to Y\) be a morphism of schemes having finitely many irreducible components. If \(f\) is birational then \(f\) is dominant.

Proof

Follows from Lemma 01RK and the definition.

Lemma

Let \(f : X \to Y\) be a birational morphism of schemes having finitely many irreducible components. If \(y \in Y\) is the generic point of an irreducible component, then the base change \(X \times_Y \Spec(\mathcal{O}_{Y, y}) \to \Spec(\mathcal{O}_{Y, y})\) is an isomorphism.

Proof

We may assume \(Y = \Spec(B)\) is affine and irreducible. Then \(X\) is irreducible too. If we prove the result for any nonempty affine open \(U \subset X\), then the result holds for \(X\) (small argument omitted). Hence we may assume \(X\) is affine too, say \(X = \Spec(A)\). Let \(y \in Y\) correspond to the minimal prime \(\mathfrak q \subset B\). By assumption \(A\) has a unique minimal prime \(\mathfrak p\) lying over \(\mathfrak q\) and \(B_\mathfrak q \to A_\mathfrak p\) is an isomorphism. It follows that \(A_\mathfrak q \to \kappa(\mathfrak p)\) is surjective, hence \(\mathfrak p A_\mathfrak q\) is a maximal ideal. On the other hand \(\mathfrak p A_\mathfrak q\) is the unique minimal prime of \(A_\mathfrak q\). We conclude that \(\mathfrak p A_\mathfrak q\) is the unique prime of \(A_\mathfrak q\) and that \(A_\mathfrak q = A_\mathfrak p\). Since \(A_\mathfrak q = A \otimes_B B_\mathfrak q\) the lemma follows.

Example

Here are two examples of birational morphisms which are not isomorphisms over any open of the target.

First example. Let \(k\) be an infinite field. Let \(A = k[x]\). Let \(B = k[x, \{y_{\alpha}\}_{\alpha \in k}]/ ((x-\alpha)y_\alpha, y_\alpha y_\beta)\). There is an inclusion \(A \subset B\) and a retraction \(B \to A\) setting all \(y_\alpha\) equal to zero. Both the morphism \(\Spec(A) \to \Spec(B)\) and the morphism \(\Spec(B) \to \Spec(A)\) are birational but not an isomorphism over any open.

Second example. Let \(A\) be a domain. Let \(S \subset A\) be a multiplicative subset not containing \(0\). With \(B = S^{-1}A\) the morphism \(f : \Spec(B) \to \Spec(A)\) is birational. If there exists an open \(U\) of \(\Spec(A)\) such that \(f^{-1}(U) \to U\) is an isomorphism, then there exists an \(a \in A\) such that each every element of \(S\) becomes invertible in the principal localization \(A_a\). Taking \(A = \mathbf{Z}\) and \(S\) the set of odd integers give a counter example.

Lemma

Let \(f : X \to Y\) be a birational morphism of schemes having finitely many irreducible components over a base scheme \(S\). Assume one of the following conditions is satisfied

\(f\) is locally of finite type and \(Y\) reduced,

\(f\) is locally of finite presentation.

Then there exist dense opens \(U \subset X\) and \(V \subset Y\) such that \(f(U) \subset V\) and \(f|_U : U \to V\) is an isomorphism. In particular if \(X\) and \(Y\) are irreducible, then \(X\) and \(Y\) are \(S\)-birational.

Proof

There is an immediate reduction to the case where \(X\) and \(Y\) are irreducible which we omit. Moreover, after shrinking further and we may assume \(X\) and \(Y\) are affine, say \(X = \Spec(A)\) and \(Y = \Spec(B)\). By assumption \(A\), resp. \(B\) has a unique minimal prime \(\mathfrak p\), resp. \(\mathfrak q\), the prime \(\mathfrak p\) lies over \(\mathfrak q\), and \(B_\mathfrak q = A_\mathfrak p\). By Lemma 0BAB we have \(B_\mathfrak q = A_\mathfrak q = A_\mathfrak p\).

Suppose \(B \to A\) is of finite type, say \(A = B[x_1, \ldots, x_n]\). There exist a \(b_i \in B\) and \(g_i \in B \setminus \mathfrak q\) such that \(b_i/g_i\) maps to the image of \(x_i\) in \(A_\mathfrak q\). Hence \(b_i - g_ix_i\) maps to zero in \(A_{g_i'}\) for some \(g_i' \in B \setminus \mathfrak q\). Setting \(g = \prod g_i g'_i\) we see that \(B_g \to A_g\) is surjective. If moreover \(Y\) is reduced, then the map \(B_g \to B_\mathfrak q\) is injective and hence \(B_g \to A_g\) is injective as well. This proves case (1).

Proof of (2). By the argument given in the previous paragraph we may assume that \(B \to A\) is surjective. As \(f\) is locally of finite presentation the kernel \(J \subset B\) is a finitely generated ideal. Say \(J = (b_1, \ldots, b_r)\). Since \(B_\mathfrak q = A_\mathfrak q\) there exist \(g_i \in B \setminus \mathfrak q\) such that \(g_i b_i = 0\). Setting \(g = \prod g_i\) we see that \(B_g \to A_g\) is an isomorphism.

Lemma

Let \(S\) be a scheme. Let \(X\) and \(Y\) be irreducible schemes locally of finite presentation over \(S\). Let \(x \in X\) and \(y \in Y\) be the generic points. The following are equivalent

\(X\) and \(Y\) are \(S\)-birational,

there exist nonempty opens of \(X\) and \(Y\) which are \(S\)-isomorphic, and

\(x\) and \(y\) map to the same point \(s\) of \(S\) and \(\mathcal{O}_{X, x}\) and \(\mathcal{O}_{Y, y}\) are isomorphic as \(\mathcal{O}_{S, s}\)-algebras.

Proof

We have seen the equivalence of (1) and (2) in Lemma 0BAA. It is immediate that (2) implies (3). To finish we assume (3) holds and we prove (1). By Lemma 0BX8 there is a rational map \(f : U \to Y\) which sends \(x \in U\) to \(y\) and induces the given isomorphism \(\mathcal{O}_{Y, y} \cong \mathcal{O}_{X, x}\). Thus \(f\) is a birational morphism and hence induces an isomorphism on nonempty opens by Lemma 0BAC. This finishes the proof.

Lemma

Let \(S\) be a scheme. Let \(X\) and \(Y\) be integral schemes locally of finite type over \(S\). Let \(x \in X\) and \(y \in Y\) be the generic points. The following are equivalent

\(X\) and \(Y\) are \(S\)-birational,

there exist nonempty opens of \(X\) and \(Y\) which are \(S\)-isomorphic, and

\(x\) and \(y\) map to the same point \(s \in S\) and \(\kappa(x) \cong \kappa(y)\) as \(\kappa(s)\)-extensions.

Proof

We have seen the equivalence of (1) and (2) in Lemma 0BAA. It is immediate that (2) implies (3). To finish we assume (3) holds and we prove (1). Observe that \(\mathcal{O}_{X, x} = \kappa(x)\) and \(\mathcal{O}_{Y, y} = \kappa(y)\) by Algebra, Lemma 00EU. By Lemma 0BX8 there is a rational map \(f : U \to Y\) which sends \(x \in U\) to \(y\) and induces the given isomorphism \(\mathcal{O}_{Y, y} \cong \mathcal{O}_{X, x}\). Thus \(f\) is a birational morphism and hence induces an isomorphism on nonempty opens by Lemma 0BAC. This finishes the proof.

Generically finite morphisms

In this section we characterize maps between schemes which are locally of finite type and which are “generically finite” in some sense.

Lemma

Let \(X\), \(Y\) be schemes. Let \(f : X \to Y\) be locally of finite type. Let \(\eta \in Y\) be a generic point of an irreducible component of \(Y\). The following are equivalent:

the set \(f^{-1}(\{\eta\})\) is finite,

there exist affine opens \(U_i \subset X\), \(i = 1, \ldots, n\) and \(V \subset Y\) with \(f(U_i) \subset V\), \(\eta \in V\) and \(f^{-1}(\{\eta\}) \subset \bigcup U_i\) such that each \(f|_{U_i} : U_i \to V\) is finite.

If \(f\) is quasi-separated, then these are also equivalent to

there exist affine opens \(V \subset Y\), and \(U \subset X\) with \(f(U) \subset V\), \(\eta \in V\) and \(f^{-1}(\{\eta\}) \subset U\) such that \(f|_U : U \to V\) is finite.

If \(f\) is quasi-compact and quasi-separated, then these are also equivalent to

there exists an affine open \(V \subset Y\), \(\eta \in V\) such that \(f^{-1}(V) \to V\) is finite.

Proof

The question is local on the base. Hence we may replace \(Y\) by an affine neighbourhood of \(\eta\), and we may and do assume throughout the proof below that \(Y\) is affine, say \(Y = \Spec(R)\).

It is clear that (2) implies (1). Assume that \(f^{-1}(\{\eta\}) = \{\xi_1, \ldots, \xi_n\}\) is finite. Choose affine opens \(U_i \subset X\) with \(\xi_i \in U_i\). By Algebra, Lemma 02ML we see that after replacing \(Y\) by a standard open in \(Y\) each of the morphisms \(U_i \to Y\) is finite. In other words (2) holds.

It is clear that (3) implies (1). Assume \(f\) is quasi-separated and (1). Write \(f^{-1}(\{\eta\}) = \{\xi_1, \ldots, \xi_n\}\). There are no specializations among the \(\xi_i\) by Lemma 02NG. Since each \(\xi_i\) maps to the generic point \(\eta\) of an irreducible component of \(Y\), there cannot be a nontrivial specialization \(\xi \leadsto \xi_i\) in \(X\) (since \(\xi\) would map to \(\eta\) as well). We conclude each \(\xi_i\) is a generic point of an irreducible component of \(X\). Since \(Y\) is affine and \(f\) quasi-separated we see \(X\) is quasi-separated. By Properties, Lemma 01ZV we can find an affine open \(U \subset X\) containing each \(\xi_i\). By Algebra, Lemma 02ML we see that after replacing \(Y\) by a standard open in \(Y\) the morphisms \(U \to Y\) is finite. In other words (3) holds.

It is clear that (4) implies all of (1) – (3) with no further assumptions on \(f\). Suppose that \(f\) is quasi-compact and quasi-separated. We have to show that the equivalent conditions (1) – (3) imply (4). Let \(U\), \(V\) be as in (3). Replace \(Y\) by \(V\). Since \(f\) is quasi-compact and \(Y\) is quasi-compact (being affine) we see that \(X\) is quasi-compact. Hence \(Z = X \setminus U\) is quasi-compact, hence the morphism \(f|_Z : Z \to Y\) is quasi-compact. By construction of \(Z\) we see that \(\eta \not \in f(Z)\). Hence by Lemma 02NE we see that there exists an affine open neighbourhood \(V'\) of \(\eta\) in \(Y\) such that \(f^{-1}(V') \cap Z = \emptyset\). Then we have \(f^{-1}(V') \subset U\) and this means that \(f^{-1}(V') \to V'\) is finite.

Example

Let \(A = \prod_{n \in \mathbf{N}} \mathbf{F}_2\). Every element of \(A\) is an idempotent. Hence every prime ideal is maximal with residue field \(\mathbf{F}_2\). Thus the topology on \(X = \Spec(A)\) is totally disconnected and quasi-compact. The projection maps \(A \to \mathbf{F}_2\) define open points of \(\Spec(A)\). It cannot be the case that all the points of \(X\) are open since \(X\) is quasi-compact. Let \(x \in X\) be a closed point which is not open. Then we can form a scheme \(Y\) which is two copies of \(X\) glued along \(X \setminus \{x\}\). In other words, this is \(X\) with \(x\) doubled, compare Schemes, Example 01JD. The morphism \(f : Y \to X\) is quasi-compact, finite type and has finite fibres but is not quasi-separated. The point \(x \in X\) is a generic point of an irreducible component of \(X\) (since \(X\) is totally disconnected). But properties (3) and (4) of Lemma 02NW do not hold. The reason is that for any open neighbourhood \(x \in U \subset X\) the inverse image \(f^{-1}(U)\) is not affine because functions on \(f^{-1}(U)\) cannot separate the two points lying over \(x\) (proof omitted; this is a nice exercise). Hence the condition that \(f\) is quasi-separated is necessary in parts (3) and (4) of the lemma.

Remark

An alternative to Lemma 02NW is the statement that a quasi-finite morphism is finite over a dense open of the target. This will be shown in More on Morphisms, Lemma 03I1.

Lemma

Let \(X\), \(Y\) be schemes. Let \(f : X \to Y\) be locally of finite type. Let \(X^0\), resp. \(Y^0\) denote the set of generic points of irreducible components of \(X\), resp. \(Y\). Let \(\eta \in Y^0\). The following are equivalent

\(f^{-1}(\{\eta\}) \subset X^0\),

\(f\) is quasi-finite at all points lying over \(\eta\),

\(f\) is quasi-finite at all \(\xi \in X^0\) lying over \(\eta\).

Proof

Condition (1) implies there are no specializations among the points of the fibre \(X_\eta\). Hence (2) holds by Lemma 01TH. The implication (2) \(\Rightarrow\) (3) is immediate. Since \(\eta\) is a generic point of \(Y\), the generic points of \(X_\eta\) are generic points of \(X\). Hence (3) and Lemma 01TH imply the generic points of \(X_\eta\) are also closed. Thus all points of \(X_\eta\) are generic and we see that (1) holds.

Lemma

Let \(X\), \(Y\) be schemes. Let \(f : X \to Y\) be locally of finite type. Let \(X^0\), resp. \(Y^0\) denote the set of generic points of irreducible components of \(X\), resp. \(Y\). Assume

\(X^0\) and \(Y^0\) are finite and \(f^{-1}(Y^0) = X^0\),

either \(f\) is quasi-compact or \(f\) is separated.

Then there exists a dense open \(V \subset Y\) such that \(f^{-1}(V) \to V\) is finite.

Proof

Since \(Y\) has finitely many irreducible components, we can find a dense open which is a disjoint union of its irreducible components. Thus we may assume \(Y\) is irreducible affine with generic point \(\eta\). Then the fibre over \(\eta\) is finite as \(X^0\) is finite.

Assume \(f\) is separated and \(Y\) irreducible affine. Choose \(V \subset Y\) and \(U \subset X\) as in Lemma 02NW part (3). Since \(f|_U : U \to V\) is finite, we see that \(U \subset f^{-1}(V)\) is closed as well as open (Lemmas 01W6 and 01WN). Thus \(f^{-1}(V) = U \amalg W\) for some open subscheme \(W\) of \(X\). However, since \(U\) contains all the generic points of \(X\) we conclude that \(W = \emptyset\) as desired.

Assume \(f\) is quasi-compact and \(Y\) irreducible affine. Then \(X\) is quasi-compact, hence there exists a dense open subscheme \(U \subset X\) which is separated (Properties, Lemma 03J1). Since the set of generic points \(X^0\) is finite, we see that \(X^0 \subset U\). Thus \(\eta \not \in f(X \setminus U)\). Since \(X \setminus U \to Y\) is quasi-compact, we conclude that there is a nonempty open \(V \subset Y\) such that \(f^{-1}(V) \subset U\), see Lemma 01RL. After replacing \(X\) by \(f^{-1}(V)\) and \(Y\) by \(V\) we reduce to the separated case which we dealt with in the preceding paragraph.

Lemma

Let \(X\), \(Y\) be schemes. Let \(f : X \to Y\) be a birational morphism between schemes which have finitely many irreducible components. Assume

either \(f\) is quasi-compact or \(f\) is separated, and

either \(f\) is locally of finite type and \(Y\) is reduced or \(f\) is locally of finite presentation.

Then there exists a dense open \(V \subset Y\) such that \(f^{-1}(V) \to V\) is an isomorphism.

Proof

By Lemma 0BAI we may assume that \(f\) is finite. Since \(Y\) has finitely many irreducible components, we can find a dense open which is a disjoint union of its irreducible components. Thus we may assume \(Y\) is irreducible. By Lemma 0BAC we find a nonempty open \(U \subset X\) such that \(f|_U : U \to Y\) is an open immersion. After removing the closed (as \(f\) finite) subset \(f(X \setminus U)\) from \(Y\) we see that \(f\) is an isomorphism.

Lemma

Let \(X\), \(Y\) be integral schemes. Let \(f : X \to Y\) be locally of finite type. Assume \(f\) is dominant. The following are equivalent:

the extension \(R(Y) \subset R(X)\) has transcendence degree \(0\),

the extension \(R(Y) \subset R(X)\) is finite,

there exist nonempty affine opens \(U \subset X\) and \(V \subset Y\) such that \(f(U) \subset V\) and \(f|_U : U \to V\) is finite, and

the generic point of \(X\) is the only point of \(X\) mapping to the generic point of \(Y\).

If \(f\) is separated or if \(f\) is quasi-compact, then these are also equivalent to

there exists a nonempty affine open \(V \subset Y\) such that \(f^{-1}(V) \to V\) is finite.

Proof

Choose any affine opens \(\Spec(A) = U \subset X\) and \(\Spec(R) = V \subset Y\) such that \(f(U) \subset V\). Then \(R\) and \(A\) are domains by definition. The ring map \(R \to A\) is of finite type (Lemma 01T2). By Lemma 01RM the generic point of \(X\) maps to the generic point of \(Y\) hence \(R \to A\) is injective. Let \(K = R(Y)\) be the fraction field of \(R\) and \(L = R(X)\) the fraction field of \(A\). Then \(L/K\) is a finitely generated field extension. Hence we see that (1) is equivalent to (2).

Suppose (2) holds. Let \(x_1, \ldots, x_n \in A\) be generators of \(A\) over \(R\). By assumption there exist nonzero polynomials \(P_i(X) \in R[X]\) such that \(P_i(x_i) = 0\). Let \(f_i \in R\) be the leading coefficient of \(P_i\). Then we conclude that \(R_{f_1 \ldots f_n} \to A_{f_1 \ldots f_n}\) is finite, i.e., (3) holds. Note that (3) implies (2). So now we see that (1), (2) and (3) are all equivalent.

Let \(\eta\) be the generic point of \(X\), and let \(\eta' \in Y\) be the generic point of \(Y\). Assume (4). Then \(\dim_\eta(X_{\eta'}) = 0\) and we see that \(R(X) = \kappa(\eta)\) has transcendence degree \(0\) over \(R(Y) = \kappa(\eta')\) by Lemma 02FX. In other words (1) holds. Assume the equivalent conditions (1), (2) and (3). Suppose that \(x \in X\) is a point mapping to \(\eta'\). As \(x\) is a specialization of \(\eta\), this gives inclusions \(R(Y) \subset \mathcal{O}_{X, x} \subset R(X)\), which implies \(\mathcal{O}_{X, x}\) is a field, see Algebra, Lemma 00GS. Hence \(x = \eta\). Thus we see that (1) – (4) are all equivalent.

It is clear that (5) implies (3) with no additional assumptions on \(f\). What remains is to prove that if \(f\) is either separated or quasi-compact, then the equivalent conditions (1) – (4) imply (5). This follows from Lemma 0BAI.

Definition

Let \(X\) and \(Y\) be integral schemes. Let \(f : X \to Y\) be locally of finite type and dominant. Assume \([R(X) : R(Y)] < \infty\), or any other of the equivalent conditions (1) – (4) of Lemma 02NX. Then the positive integer \[\deg(X/Y) = [R(X) : R(Y)]\] is called the degree of \(X\) over \(Y\).

It is possible to extend this notion to a morphism \(f : X \to Y\) if (a) \(Y\) is integral with generic point \(\eta\), (b) \(f\) is locally of finite type, and (c) \(f^{-1}(\{\eta\})\) is finite. In this case we can define \[\deg(X/Y) = \sum\nolimits_{\xi \in X, \ f(\xi) = \eta} \dim_{R(Y)} (\mathcal{O}_{X, \xi}).\] Namely, given that \(R(Y) = \kappa(\eta) = \mathcal{O}_{Y, \eta}\) (Lemma 01RV) the dimensions above are finite by Lemma 02NW above. However, for most applications the definition given above is the right one.

Lemma

Let \(X\), \(Y\), \(Z\) be integral schemes. Let \(f : X \to Y\) and \(g : Y \to Z\) be dominant morphisms locally of finite type. Assume that \([R(X) : R(Y)] < \infty\) and \([R(Y) : R(Z)] < \infty\). Then \[\deg(X/Z) = \deg(X/Y) \deg(Y/Z).\]

Proof

This comes from the multiplicativity of degrees in towers of finite extensions of fields, see Fields, Lemma 09G9.

Remark

Let \(f : X \to Y\) be a morphism of schemes which is locally of finite type. There are (at least) two properties that we could use to define generically finite morphisms. These correspond to whether you want the property to be local on the source or local on the target:

(Local on the target; suggested by Ravi Vakil.) Assume every quasi-compact open of \(Y\) has finitely many irreducible components (for example if \(Y\) is locally Noetherian). The requirement is that the inverse image of each generic point is finite, see Lemma 02NW.

(Local on the source.) The requirement is that there exists a dense open \(U \subset X\) such that \(U \to Y\) is locally quasi-finite.

In case (1) the requirement can be formulated without the auxiliary condition on \(Y\), but probably doesn’t give the right notion for general schemes. Property (2) as formulated doesn’t imply that the fibres over generic points are finite; however, if \(f\) is quasi-compact and \(Y\) is as in (1) then it does.

Definition

Let \(X\) be an integral scheme. A modification of \(X\) is a birational proper morphism \(f : X' \to X\) with \(X'\) integral.

Let \(f : X' \to X\) be a modification as in the definition. By Lemma 02NX there exists a nonempty \(U \subset X\) such that \(f^{-1}(U) \to U\) is finite. By generic flatness (Proposition 052A) we may assume \(f^{-1}(U) \to U\) is flat and of finite presentation. So \(f^{-1}(U) \to U\) is finite locally free (Lemma 02KB). Since \(f\) is birational, the degree of \(X'\) over \(X\) is \(1\). Hence \(f^{-1}(U) \to U\) is finite locally free of degree \(1\), in other words it is an isomorphism. Thus we can redefine a modification to be a proper morphism \(f : X' \to X\) of integral schemes such that \(f^{-1}(U) \to U\) is an isomorphism for some nonempty open \(U \subset X\).

Definition

Let \(X\) be an integral scheme. An alteration of \(X\) is a proper dominant morphism \(f : Y \to X\) with \(Y\) integral such that \(f^{-1}(U) \to U\) is finite for some nonempty open \(U \subset X\).

This is the definition as given in [alterations], except that here we do not require \(X\) and \(Y\) to be Noetherian. Arguing as above we see that an alteration is a proper dominant morphism \(f : Y \to X\) of integral schemes which induces a finite extension of function fields, i.e., such that the equivalent conditions of Lemma 02NX hold.

The dimension formula

For morphisms between Noetherian schemes we can say a little more about dimensions of local rings. Here is an important (and not so hard to prove) result. Recall that \(R(X)\) denotes the function field of an integral scheme \(X\).

Lemma

Let \(S\) be a scheme. Let \(f : X \to S\) be a morphism of schemes. Let \(x \in X\), and set \(s = f(x)\). Assume

\(S\) is locally Noetherian,

\(f\) is locally of finite type,

\(X\) and \(S\) integral, and

\(f\) dominant.

We have [02JV]\[\begin{equation} \dim(\mathcal{O}_{X, x}) \leq \dim(\mathcal{O}_{S, s}) + \text{trdeg}_{R(S)}R(X) - \text{trdeg}_{\kappa(s)} \kappa(x). \end{equation}\] Moreover, equality holds if \(S\) is universally catenary.

Proof

The corresponding algebra statement is Algebra, Lemma 02IJ.

Lemma

Let \(S\) be a scheme. Let \(f : X \to S\) be a morphism of schemes. Let \(x \in X\), and set \(s = f(x)\). Assume \(S\) is locally Noetherian and \(f\) is locally of finite type, We have [0BAF]\[\begin{equation} \dim(\mathcal{O}_{X, x}) \leq \dim(\mathcal{O}_{S, s}) + E - \text{trdeg}_{\kappa(s)} \kappa(x). \end{equation}\] where \(E\) is the maximum of \(\text{trdeg}_{\kappa(f(\xi))}(\kappa(\xi))\) where \(\xi\) runs over the generic points of irreducible components of \(X\) containing \(x\).

Proof

Let \(X_1, \ldots, X_n\) be the irreducible components of \(X\) containing \(x\) endowed with their reduced induced scheme structure. These correspond to the minimal primes \(\mathfrak q_i\) of \(\mathcal{O}_{X, x}\) and hence there are finitely many of them (Schemes, Lemma 01J7 and Algebra, Lemma 00FR). Then \(\dim(\mathcal{O}_{X, x}) = \max \dim(\mathcal{O}_{X, x}/\mathfrak q_i) = \max \dim(\mathcal{O}_{X_i, x})\). The \(\xi\)’s occurring in the definition of \(E\) are exactly the generic points \(\xi_i \in X_i\). Let \(Z_i = \overline{\{f(\xi_i)\}} \subset S\) endowed with the reduced induced scheme structure. The composition \(X_i \to X \to S\) factors through \(Z_i\) (Schemes, Lemma 0356). Thus we may apply the dimension formula (Lemma 02JU) to see that \(\dim(\mathcal{O}_{X_i, x}) \leq \dim(\mathcal{O}_{Z_i, x}) + \text{trdeg}_{\kappa(f(\xi))}(\kappa(\xi)) - \text{trdeg}_{\kappa(s)} \kappa(x)\). Putting everything together we obtain the lemma.

An application is the construction of a dimension function on any scheme of finite type over a universally catenary scheme endowed with a dimension function. For the definition of dimension functions, see Topology, Definition 02I9.

Lemma

Let \(S\) be a locally Noetherian and universally catenary scheme. Let \(\delta : S \to \mathbf{Z}\) be a dimension function. Let \(f : X \to S\) be a morphism of schemes. Assume \(f\) locally of finite type. Then the map \[\begin{align*} \delta = \delta_{X/S} : X & \longrightarrow \mathbf{Z} \\ x & \longmapsto \delta(f(x)) + \text{trdeg}_{\kappa(f(x))} \kappa(x) \end{align*}\] is a dimension function on \(X\).

Proof

Let \(f : X \to S\) be locally of finite type. Let \(x \leadsto y\), \(x \not = y\) be a specialization in \(X\). We have to show that \(\delta_{X/S}(x) > \delta_{X/S}(y)\) and that \(\delta_{X/S}(x) = \delta_{X/S}(y) + 1\) if \(y\) is an immediate specialization of \(x\).

Choose an affine open \(V \subset S\) containing the image of \(y\) and choose an affine open \(U \subset X\) mapping into \(V\) and containing \(y\). We may clearly replace \(X\) by \(U\) and \(S\) by \(V\). Thus we may assume that \(X = \Spec(A)\) and \(S = \Spec(R)\) and that \(f\) is given by a ring map \(R \to A\). The ring \(R\) is universally catenary (Lemma 02J9) and the map \(R \to A\) is of finite type (Lemma 01T2).

Let \(\mathfrak q \subset A\) be the prime ideal corresponding to the point \(x\) and let \(\mathfrak p \subset R\) be the prime ideal corresponding to \(f(x)\). The restriction \(\delta'\) of \(\delta\) to \(S' = \Spec(R/\mathfrak p) \subset S\) is a dimension function. The ring \(R/\mathfrak p\) is universally catenary. The restriction of \(\delta_{X/S}\) to \(X' = \Spec(A/\mathfrak q)\) is clearly equal to the function \(\delta_{X'/S'}\) constructed using the dimension function \(\delta'\). Hence we may assume in addition to the above that \(R \subset A\) are domains, in other words that \(X\) and \(S\) are integral schemes, and that \(x\) is the generic point of \(X\) and \(f(x)\) is the generic point of \(S\).

Note that \(\mathcal{O}_{X, x} = R(X)\) and that since \(x \leadsto y\), \(x \not = y\), the spectrum of \(\mathcal{O}_{X, y}\) has at least two points (Schemes, Lemma 01J7) hence \(\dim(\mathcal{O}_{X, y}) > 0\) . If \(y\) is an immediate specialization of \(x\), then \(\Spec(\mathcal{O}_{X, y}) = \{x, y\}\) and \(\dim(\mathcal{O}_{X, y}) = 1\).

Write \(s = f(x)\) and \(t = f(y)\). We compute \[\begin{align*} \delta_{X/S}(x) - \delta_{X/S}(y) & = \delta(s) + \text{trdeg}_{\kappa(s)} \kappa(x) - \delta(t) - \text{trdeg}_{\kappa(t)} \kappa(y) \\ & = \delta(s) - \delta(t) + \text{trdeg}_{R(S)} R(X) - \text{trdeg}_{\kappa(t)} \kappa(y) \\ & = \delta(s) - \delta(t) + \dim(\mathcal{O}_{X, y}) - \dim(\mathcal{O}_{S, t}) \end{align*}\] where we use equality in (02JV) in the last step. Since \(\delta\) is a dimension function on the scheme \(S\) and \(s \in S\) is the generic point, the difference \(\delta(s) - \delta(t)\) is equal to \(\text{codim}(\overline{\{t\}}, S)\) by Topology, Lemma 02IA. This is equal to \(\dim(\mathcal{O}_{S, t})\) by Properties, Lemma 02IZ. Hence we conclude that \[\delta_{X/S}(x) - \delta_{X/S}(y) = \dim(\mathcal{O}_{X, y})\] and the lemma follows from what we said above about \(\dim(\mathcal{O}_{X, y})\).

Another application of the dimension formula is that the dimension does not change under “alterations” (to be defined later).

Lemma

Let \(f : X \to Y\) be a morphism of schemes. Assume that

\(Y\) is locally Noetherian,

\(X\) and \(Y\) are integral schemes,

\(f\) is dominant, and

\(f\) is locally of finite type.

Then we have \[\dim(X) \leq \dim(Y) + \text{trdeg}_{R(Y)} R(X).\] If \(f\) is closed¹⁶ then equality holds.

Proof

Let \(f : X \to Y\) be as in the lemma. Let \(\xi_0 \leadsto \xi_1 \leadsto \ldots \leadsto \xi_e\) be a sequence of specializations in \(X\). Set \(x = \xi_e\) and \(y = f(x)\). Observe that \(e \leq \dim(\mathcal{O}_{X, x})\) as the given specializations occur in the spectrum of \(\mathcal{O}_{X, x}\), see Schemes, Lemma 01J7. By the dimension formula, Lemma 02JU, we see that \[\begin{align*} e & \leq \dim(\mathcal{O}_{X, x}) \\ & \leq \dim(\mathcal{O}_{Y, y}) + \text{trdeg}_{R(Y)} R(X) - \text{trdeg}_{\kappa(y)} \kappa(x) \\ & \leq \dim(\mathcal{O}_{Y, y}) + \text{trdeg}_{R(Y)} R(X) \end{align*}\] Hence we conclude that \(e \leq \dim(Y) + \text{trdeg}_{R(Y)} R(X)\) as desired.

Next, assume \(f\) is also closed. Say \(\overline{\xi}_0 \leadsto \overline{\xi}_1 \leadsto \ldots \leadsto \overline{\xi}_d\) is a sequence of specializations in \(Y\). We want to show that \(\dim(X) \geq d + r\). We may assume that \(\overline{\xi}_0 = \eta\) is the generic point of \(Y\). The generic fibre \(X_\eta\) is a scheme locally of finite type over \(\kappa(\eta) = R(Y)\). It is nonempty as \(f\) is dominant. Hence by Lemma 02J6 it is a Jacobson scheme. Thus by Lemma 01TB we can find a closed point \(\xi_0 \in X_\eta\) and the extension \(\kappa(\eta) \subset \kappa(\xi_0)\) is a finite extension. Note that \(\mathcal{O}_{X, \xi_0} = \mathcal{O}_{X_\eta, \xi_0}\) because \(\eta\) is the generic point of \(Y\). Hence we see that \(\dim(\mathcal{O}_{X, \xi_0}) = r\) by Lemma 02JU applied to the scheme \(X_\eta\) over the universally catenary scheme \(\Spec(\kappa(\eta))\) (see Lemma 02JB) and the point \(\xi_0\). This means that we can find \(\xi_{-r} \leadsto \ldots \leadsto \xi_{-1} \leadsto \xi_0\) in \(X\). On the other hand, as \(f\) is closed specializations lift along \(f\), see Topology, Lemma 0066. Thus, as \(\xi_0\) lies over \(\eta = \overline{\xi}_0\) we can find specializations \(\xi_0 \leadsto \xi_1 \leadsto \ldots \leadsto \xi_d\) lying over \(\overline{\xi}_0 \leadsto \overline{\xi}_1 \leadsto \ldots \leadsto \overline{\xi}_d\). In other words we have \[\xi_{-r} \leadsto \ldots \leadsto \xi_{-1} \leadsto \xi_0 \leadsto \xi_1 \leadsto \ldots \leadsto \xi_d\] which means that \(\dim(X) \geq d + r\) as desired.

Lemma

Let \(f : X \to Y\) be a morphism of schemes. Assume that \(Y\) is locally Noetherian and \(f\) is locally of finite type. Then \[\dim(X) \leq \dim(Y) + E\] where \(E\) is the supremum of \(\text{trdeg}_{\kappa(f(\xi))}(\kappa(\xi))\) where \(\xi\) runs through the generic points of the irreducible components of \(X\).

Proof

Immediate consequence of Lemma 0BAE and Properties, Lemma 04MU.

Relative normalization

In this section we construct the normalization of one scheme in another.

Lemma

Let \(X\) be a scheme. Let \(\mathcal{A}\) be a quasi-coherent sheaf of \(\mathcal{O}_X\)-algebras. The subsheaf \(\mathcal{A}' \subset \mathcal{A}\) defined by the rule \[U \longmapsto \{f \in \mathcal{A}(U) \mid f_x \in \mathcal{A}_x \text{ integral over } \mathcal{O}_{X, x} \text{ for all }x \in U\}\] is a quasi-coherent \(\mathcal{O}_X\)-algebra, the stalk \(\mathcal{A}'_x\) is the integral closure of \(\mathcal{O}_{X, x}\) in \(\mathcal{A}_x\), and for any affine open \(U \subset X\) the ring \(\mathcal{A}'(U) \subset \mathcal{A}(U)\) is the integral closure of \(\mathcal{O}_X(U)\) in \(\mathcal{A}(U)\).

Proof

This is a subsheaf by the local nature of the conditions. It is an \(\mathcal{O}_X\)-algebra by Algebra, Lemma 00GO. Let \(U \subset X\) be an affine open. Say \(U = \Spec(R)\) and say \(\mathcal{A}\) is the quasi-coherent sheaf associated to the \(R\)-algebra \(A\). Then according to Algebra, Lemma 034K the value of \(\mathcal{A}'\) over \(U\) is given by the integral closure \(A'\) of \(R\) in \(A\). This proves the last assertion of the lemma. To prove that \(\mathcal{A}'\) is quasi-coherent, it suffices to show that \(\mathcal{A}'(D(f)) = A'_f\). This follows from the fact that integral closure and localization commute, see Algebra, Lemma 0307. The same fact shows that the stalks are as advertised.

Definition

Let \(X\) be a scheme. Let \(\mathcal{A}\) be a quasi-coherent sheaf of \(\mathcal{O}_X\)-algebras. The integral closure of \(\mathcal{O}_X\) in \(\mathcal{A}\) is the quasi-coherent \(\mathcal{O}_X\)-subalgebra \(\mathcal{A}' \subset \mathcal{A}\) constructed in Lemma 035F above.

In the setting of the definition above we can consider the morphism of relative spectra \[\xymatrix{ Y = \underline{\Spec}_X(\mathcal{A}) \ar[rr] \ar[rd] & & X' = \underline{\Spec}_X(\mathcal{A}') \ar[ld] \\ & X & }\] see Lemma 01SA. The scheme \(X' \to X\) will be the normalization of \(X\) in the scheme \(Y\). Here is a slightly more general setting. Suppose we have a quasi-compact and quasi-separated morphism \(f : Y \to X\) of schemes. In this case the sheaf of \(\mathcal{O}_X\)-algebras \(f_*\mathcal{O}_Y\) is quasi-coherent, see Schemes, Lemma 01LC. Taking the integral closure \(\mathcal{O}' \subset f_*\mathcal{O}_Y\) we obtain a quasi-coherent sheaf of \(\mathcal{O}_X\)-algebras whose relative spectrum is the normalization of \(X\) in \(Y\). Here is the formal definition.

Definition

Let \(f : Y \to X\) be a quasi-compact and quasi-separated morphism of schemes. Let \(\mathcal{O}'\) be the integral closure of \(\mathcal{O}_X\) in \(f_*\mathcal{O}_Y\). The normalization of \(X\) in \(Y\) is the scheme¹⁷ \[\nu : X' = \underline{\Spec}_X(\mathcal{O}') \to X\] over \(X\). It comes equipped with a natural factorization \[Y \xrightarrow{f'} X' \xrightarrow{\nu} X\] of the initial morphism \(f\).

The factorization is the composition of the canonical morphism \(Y \to \underline{\Spec}_X(f_*\mathcal{O}_Y)\) (see Constructions, Lemma 01LY) and the morphism of relative spectra coming from the inclusion map \(\mathcal{O}' \to f_*\mathcal{O}_Y\). We can characterize the normalization as follows.

Lemma

Let \(f : Y \to X\) be a quasi-compact and quasi-separated morphism of schemes. The factorization \(f = \nu \circ f'\), where \(\nu : X' \to X\) is the normalization of \(X\) in \(Y\) is characterized by the following two properties:

the morphism \(\nu\) is integral, and

for any factorization \(f = \pi \circ g\), with \(\pi : Z \to X\) integral, there exists a commutative diagram \[\xymatrix{ Y \ar[d]_{f'} \ar[r]_g & Z \ar[d]^\pi \\ X' \ar[ru]^h \ar[r]^\nu & X }\] for some unique morphism \(h : X' \to Z\).

Moreover, the morphism \(f' : Y \to X'\) is dominant and in (2) the morphism \(h : X' \to Z\) is the normalization of \(Z\) in \(Y\).

Proof

Let \(\mathcal{O}' \subset f_*\mathcal{O}_Y\) be the integral closure of \(\mathcal{O}_X\) as in Definition 035H. The morphism \(\nu\) is integral by construction, which proves (1). Assume given a factorization \(f = \pi \circ g\) with \(\pi : Z \to X\) integral as in (2). By Definition 01WH \(\pi\) is affine, and hence \(Z\) is the relative spectrum of a quasi-coherent sheaf of \(\mathcal{O}_X\)-algebras \(\mathcal{B}\). The morphism \(g : Y \to Z\) corresponds to a map of \(\mathcal{O}_X\)-algebras \(\chi : \mathcal{B} \to f_*\mathcal{O}_Y\). Since \(\mathcal{B}(U)\) is integral over \(\mathcal{O}_X(U)\) for every affine open \(U \subset X\) (by Definition 01WH) we see from Lemma 035F that \(\chi(\mathcal{B}) \subset \mathcal{O}'\). By the functoriality of the relative spectrum Lemma 01SA this provides us with a unique morphism \(h : X' \to Z\). We omit the verification that the diagram commutes.

It is clear that (1) and (2) characterize the factorization \(f = \nu \circ f'\) since it characterizes it as an initial object in a category.

From the universal property in (2) we see that \(f'\) does not factor through a proper closed subscheme of \(X'\). Hence the scheme theoretic image of \(f'\) is \(X'\). Since \(f'\) is quasi-compact (by Schemes, Lemma 03GI and the fact that \(\nu\) is separated as an affine morphism) we see that \(f'(Y)\) is dense in \(X'\). Hence \(f'\) is dominant.

Observe that \(g\) is quasi-compact and quasi-separated by Schemes, Lemmas 01KV and 03GI. Thus the last statement of the lemma makes sense. The morphism \(h\) in (2) is integral by Lemma 035D. Given a factorization \(g = \pi' \circ g'\) with \(\pi' : Z' \to Z\) integral, we get a factorization \(f = (\pi \circ \pi') \circ g'\) and we get a morphism \(h' : X' \to Z'\). Uniqueness implies that \(\pi' \circ h' = h\). Hence the characterization (1), (2) applies to the morphism \(h : X' \to Z\) which gives the final assertion of the lemma.

Lemma

Let \[\xymatrix{ Y_2 \ar[d]_{f_2} \ar[r] & Y_1 \ar[d]^{f_1} \\ X_2 \ar[r] & X_1 }\] be a commutative diagram of morphisms of schemes. Assume \(f_1\), \(f_2\) quasi-compact and quasi-separated. Let \(f_i = \nu_i \circ f_i'\), \(i = 1, 2\) be the canonical factorizations, where \(\nu_i : X_i' \to X_i\) is the normalization of \(X_i\) in \(Y_i\). Then there exists a unique arrow \(X'_2 \to X'_1\) fitting into a commutative diagram \[\xymatrix{ Y_2 \ar[d]_{f_2'} \ar[r] & Y_1 \ar[d]^{f_1'} \\ X_2' \ar[d]_{\nu_2} \ar[r] & X_1' \ar[d]^{\nu_1} \\ X_2 \ar[r] & X_1 }\]

Proof

By Lemmas 035I (1) and 01WL the base change \(X_2 \times_{X_1} X'_1 \to X_2\) is integral. Note that \(f_2\) factors through this morphism. Hence we get a unique morphism \(X'_2 \to X_2 \times_{X_1} X'_1\) from Lemma 035I (2). This gives the arrow \(X'_2 \to X'_1\) fitting into the commutative diagram and uniqueness follows as well.

Lemma

Let \(f : Y \to X\) be a quasi-compact and quasi-separated morphism of schemes. Let \(U \subset X\) be an open subscheme and set \(V = f^{-1}(U)\). Then the normalization of \(U\) in \(V\) is the inverse image of \(U\) in the normalization of \(X\) in \(Y\).

Proof

Clear from the construction.

Lemma

Let \(f : Y \to X\) be a quasi-compact and quasi-separated morphism of schemes. Let \(X'\) be the normalization of \(X\) in \(Y\). Then the normalization of \(X'\) in \(Y\) is \(X'\).

Proof

If \(Y \to X'' \to X'\) is the normalization of \(X'\) in \(Y\), then we can apply Lemma 035I to the composition \(X'' \to X\) to get a canonical morphism \(h : X' \to X''\) over \(X\). We omit the verification that the morphisms \(h\) and \(X'' \to X'\) are mutually inverse (using uniqueness of the factorization in the lemma).

Lemma

Let \(f : Y \to X\) be a quasi-compact and quasi-separated morphism of schemes. Let \(X' \to X\) be the normalization of \(X\) in \(Y\). If \(Y\) is reduced, so is \(X'\).

Proof

This follows from the fact that a subring of a reduced ring is reduced. Some details omitted.

Lemma

Let \(f : Y \to X\) be a quasi-compact and quasi-separated morphism of schemes. Let \(X' \to X\) be the normalization of \(X\) in \(Y\). Every generic point of an irreducible component of \(X'\) is the image of a generic point of an irreducible component of \(Y\).

Proof

By Lemma 035K we may assume \(X = \Spec(A)\) is affine. Choose a finite affine open covering \(Y = \bigcup \Spec(B_i)\). Then \(X' = \Spec(A')\) and the morphisms \(\Spec(B_i) \to Y \to X'\) jointly define an injective \(A\)-algebra map \(A' \to \prod B_i\). Thus the lemma follows from Algebra, Lemma 00FK.

Lemma

Let \(f : Y \to X\) be a quasi-compact and quasi-separated morphism of schemes. Suppose that \(Y = Y_1 \amalg Y_2\) is a disjoint union of two schemes. Write \(f_i = f|_{Y_i}\). Let \(X_i'\) be the normalization of \(X\) in \(Y_i\). Then \(X_1' \amalg X_2'\) is the normalization of \(X\) in \(Y\).

Proof

In terms of integral closures this corresponds to the following fact: Let \(A \to B\) be a ring map. Suppose that \(B = B_1 \times B_2\). Let \(A_i'\) be the integral closure of \(A\) in \(B_i\). Then \(A_1' \times A_2'\) is the integral closure of \(A\) in \(B\). The reason this works is that the elements \((1, 0)\) and \((0, 1)\) of \(B\) are idempotents and hence integral over \(A\). Thus the integral closure \(A'\) of \(A\) in \(B\) is a product and it is not hard to see that the factors are the integral closures \(A'_i\) as described above (some details omitted).

Lemma

Let \(f : X \to S\) be a quasi-compact, quasi-separated and universally closed morphisms of schemes. Then \(f_*\mathcal{O}_X\) is integral over \(\mathcal{O}_S\). In other words, the normalization of \(S\) in \(X\) is equal to the factorization \[X \longrightarrow \underline{\Spec}_S(f_*\mathcal{O}_X) \longrightarrow S\] of Constructions, Lemma 01LY.

Proof

The question is local on \(S\), hence we may assume \(S = \Spec(R)\) is affine. Let \(h \in \Gamma(X, \mathcal{O}_X)\). We have to show that \(h\) satisfies a monic equation over \(R\). Think of \(h\) as a morphism as in the following commutative diagram \[\xymatrix{ X \ar[rr]_h \ar[rd]_f & & \mathbf{A}^1_S \ar[ld] \\ & S & }\] Let \(Z \subset \mathbf{A}^1_S\) be the scheme theoretic image of \(h\), see Definition 01R7. The morphism \(h\) is quasi-compact as \(f\) is quasi-compact and \(\mathbf{A}^1_S \to S\) is separated, see Schemes, Lemma 03GI. By Lemma 01R8 the morphism \(X \to Z\) is dominant. By Lemma 01W6 the morphism \(X \to Z\) is closed. Hence \(h(X) = Z\) (set theoretically). Thus we can use Lemma 03GN to conclude that \(Z \to S\) is universally closed (and even proper). Since \(Z \subset \mathbf{A}^1_S\), we see that \(Z \to S\) is affine and proper, hence integral by Lemma 01WM. Writing \(\mathbf{A}^1_S = \Spec(R[T])\) we conclude that the ideal \(I \subset R[T]\) of \(Z\) contains a monic polynomial \(P(T) \in R[T]\). Hence \(P(h) = 0\) and we win.

Lemma

Let \(f : Y \to X\) be an integral morphism. Then the normalization of \(X\) in \(Y\) is equal to \(Y\).

Proof

By Lemma 01WM this is a special case of Lemma 03GQ.

Lemma

Let \(f : Y \to X\) be a quasi-compact and quasi-separated morphism of schemes. Let \(X'\) be the normalization of \(X\) in \(Y\). Assume

\(Y\) is a normal scheme,

quasi-compact opens of \(Y\) have finitely many irreducible components.

Then \(X'\) is a disjoint union of integral normal schemes. Moreover, the morphism \(Y \to X'\) is dominant and induces a bijection of irreducible components.

Proof

Let \(U \subset X\) be an affine open. Consider the inverse image \(U'\) of \(U\) in \(X'\). Set \(V = f^{-1}(U)\). By Lemma 035K we \(V \to U' \to U\) is the normalization of \(U\) in \(V\). Say \(U = \Spec(A)\). Then \(V\) is quasi-compact, and hence has a finite number of irreducible components by assumption. Hence \(V = \coprod_{i = 1, \ldots n} V_i\) is a finite disjoint union of normal integral schemes by Properties, Lemma 0357. By Lemma 03GO we see that \(U' = \coprod_{i = 1, \ldots, n} U_i'\), where \(U'_i\) is the normalization of \(U\) in \(V_i\). By Properties, Lemma 0358 we see that \(B_i = \Gamma(V_i, \mathcal{O}_{V_i})\) is a normal domain. Note that \(U_i' = \Spec(A_i')\), where \(A_i' \subset B_i\) is the integral closure of \(A\) in \(B_i\), see Lemma 035F. By Algebra, Lemma 034L we see that \(A_i' \subset B_i\) is a normal domain. Hence \(U' = \coprod U_i'\) is a finite union of normal integral schemes and hence is normal.

As \(X'\) has an open covering by the schemes \(U'\) we conclude from Properties, Lemma 033J that \(X'\) is normal. On the other hand, each \(U'\) is a finite disjoint union of irreducible schemes, hence every quasi-compact open of \(X'\) has finitely many irreducible components (by a topological argument which we omit). Thus \(X'\) is a disjoint union of normal integral schemes by Properties, Lemma 0357. It is clear from the description of \(X'\) above that \(Y \to X'\) is dominant and induces a bijection on irreducible components \(V \to U'\) for every affine open \(U \subset X\). The bijection of irreducible components for the morphism \(Y \to X'\) follows from this by a topological argument (omitted).

Lemma

Let \(f : X \to S\) be a morphism. Assume that

\(S\) is a Nagata scheme,

\(f\) is quasi-compact and quasi-separated,

quasi-compact opens of \(X\) have finitely many irreducible components,

if \(x \in X\) is a generic point of an irreducible component, then the field extension \(\kappa(x)/\kappa(f(x))\) is finitely generated, and

\(X\) is reduced.

Then the normalization \(\nu : S' \to S\) of \(S\) in \(X\) is finite.

Proof

There is an immediate reduction to the case \(S = \Spec(R)\) where \(R\) is a Nagata ring by assumption (1). We have to show that the integral closure \(A\) of \(R\) in \(\Gamma(X, \mathcal{O}_X)\) is finite over \(R\). Since \(f\) is quasi-compact by assumption (2) we can write \(X = \bigcup_{i = 1, \ldots, n} U_i\) with each \(U_i\) affine. Say \(U_i = \Spec(B_i)\). Each \(B_i\) is reduced by assumption (5) and has finitely many minimal primes \(\mathfrak q_{i1}, \ldots, \mathfrak q_{im_i}\) by assumption (3) and Algebra, Lemma 00ES. We have \[\Gamma(X, \mathcal{O}_X) \subset B_1 \times \ldots \times B_n \subset \prod\nolimits_{i = 1, \ldots, n} \prod\nolimits_{j = 1, \ldots, m_i} (B_i)_{\mathfrak q_{ij}}\] the second inclusion by Algebra, Lemma 00EW. We have \(\kappa(\mathfrak q_{ij}) = (B_i)_{\mathfrak q_{ij}}\) by Algebra, Lemma 00EU. Hence the integral closure \(A\) of \(R\) in \(\Gamma(X, \mathcal{O}_X)\) is contained in the product of the integral closures \(A_{ij}\) of \(R\) in \(\kappa(\mathfrak q_{ij})\). Since \(R\) is Noetherian it suffices to show that \(A_{ij}\) is a finite \(R\)-module for each \(i, j\). Let \(\mathfrak p_{ij} \subset R\) be the image of \(\mathfrak q_{ij}\). As \(\kappa(\mathfrak q_{ij})/\kappa(\mathfrak p_{ij})\) is a finitely generated field extension by assumption (4), we see that \(R \to \kappa(\mathfrak q_{ij})\) is essentially of finite type. Thus \(R \to A_{ij}\) is finite by Algebra, Lemma 03GH.

Lemma

Let \(f : X \to S\) be a morphism. Assume that

\(S\) is a Nagata scheme,

\(f\) is of finite type,

\(X\) is reduced.

Then the normalization \(\nu : S' \to S\) of \(S\) in \(X\) is finite.

Proof

This is a special case of Lemma 0AVK. Namely, (2) holds as the finite type morphism \(f\) is quasi-compact by definition and quasi-separated by Lemma 01T7. Condition (3) holds because \(X\) is locally Noetherian by Lemma 01T6. Finally, condition (4) holds because a finite type morphism induces finitely generated residue field extensions.

Lemma

Let \(f : Y \to X\) be a finite type morphism of schemes with \(Y\) reduced and \(X\) Nagata. Let \(X'\) be the normalization of \(X\) in \(Y\). Let \(x' \in X'\) be a point such that

\(\dim(\mathcal{O}_{X', x'}) = 1\), and

the fibre of \(Y \to X'\) over \(x'\) is empty.

Then \(\mathcal{O}_{X', x'}\) is a discrete valuation ring.

Proof

We can replace \(X\) by an affine neighbourhood of the image of \(x'\). Hence we may assume \(X = \Spec(A)\) with \(A\) Nagata. By Lemma 03GR the morphism \(X' \to X\) is finite. Hence we can write \(X' = \Spec(A')\) for a finite \(A\)-algebra \(A'\). By Lemma 0BXA after replacing \(X\) by \(X'\) we reduce to the case described in the next paragraph.

The case \(X = X' = \Spec(A)\) with \(A\) Noetherian. Let \(\mathfrak p \subset A\) be the prime ideal corresponding to our point \(x'\). Choose \(g \in \mathfrak p\) not contained in any minimal prime of \(A\) (use prime avoidance and the fact that \(A\) has finitely many minimal primes, see Algebra, Lemmas 00DS and 00FR). Set \(Z = f^{-1}V(g) \subset Y\); it is a closed subscheme of \(Y\). Then \(f(Z)\) does not contain any generic point by choice of \(g\) and does not contain \(x'\) because \(x'\) is not in the image of \(f\). The closure of \(f(Z)\) is the set of specializations of points of \(f(Z)\) by Lemma 02JQ. Thus the closure of \(f(Z)\) does not contain \(x'\) because the condition \(\dim(\mathcal{O}_{X', x'}) = 1\) implies only the generic points of \(X = X'\) specialize to \(x'\). In other words, after replacing \(X\) by an affine open neighbourhood of \(x'\) we may assume that \(f^{-1}V(g) = \emptyset\). Thus \(g\) maps to an invertible global function on \(Y\) and we obtain a factorization \[A \to A_g \to \Gamma(Y, \mathcal{O}_Y)\] Since \(X = X'\) this implies that \(A\) is equal to the integral closure of \(A\) in \(A_g\). By Algebra, Lemma 0307 we conclude that \(A_\mathfrak p\) is the integral closure of \(A_\mathfrak p\) in \(A_\mathfrak p[1/g]\). By our choice of \(g\), since \(\dim(A_\mathfrak p) = 1\) and since \(A\) is reduced we see that \(A_\mathfrak p[1/g]\) is a finite product of fields (the product of the residue fields of the minimal primes contained in \(\mathfrak p\)). Hence \(A_\mathfrak p\) is normal (Algebra, Lemma 030C) and the proof is complete. Some details omitted.

Normalization

Next, we come to the normalization of a scheme \(X\). We only define/construct it when \(X\) has locally finitely many irreducible components. Let \(X\) be a scheme such that every quasi-compact open has finitely many irreducible components. Let \(X^{(0)} \subset X\) be the set of generic points of irreducible components of \(X\). Let [035M]\[\begin{equation} f : Y = \coprod\nolimits_{\eta \in X^{(0)}} \Spec(\kappa(\eta)) \longrightarrow X \end{equation}\] be the inclusion of the generic points into \(X\) using the canonical maps of Schemes, Section 01J5. Note that this morphism is quasi-compact by assumption and quasi-separated as \(Y\) is separated (see Schemes, Section 01KH).

Definition

Let \(X\) be a scheme such that every quasi-compact open has finitely many irreducible components. We define the normalization of \(X\) as the morphism \[\nu : X^\nu \longrightarrow X\] which is the normalization of \(X\) in the morphism \(f : Y \to X\) (035M) constructed above.

Any locally Noetherian scheme has a locally finite set of irreducible components and the definition applies to it. Usually the normalization is defined only for reduced schemes. With the definition above the normalization of \(X\) is the same as the normalization of the reduction \(X_{red}\) of \(X\).

Lemma

Let \(X\) be a scheme such that every quasi-compact open has finitely many irreducible components. The normalization morphism \(\nu\) factors through the reduction \(X_{red}\) and \(X^\nu \to X_{red}\) is the normalization of \(X_{red}\).

Proof

Let \(f : Y \to X\) be the morphism (035M). We get a factorization \(Y \to X_{red} \to X\) of \(f\) from Schemes, Lemma 0356. By Lemma 035I we obtain a canonical morphism \(X^\nu \to X_{red}\) and that \(X^\nu\) is the normalization of \(X_{red}\) in \(Y\). The lemma follows as \(Y \to X_{red}\) is identical to the morphism (035M) constructed for \(X_{red}\).

If \(X\) is reduced, then the normalization of \(X\) is the same as the relative spectrum of the integral closure of \(\mathcal{O}_X\) in the sheaf of meromorphic functions \(\mathcal{K}_X\) (see Divisors, Section 01X1). Namely, \(\mathcal{K}_X = f_*\mathcal{O}_Y\) in this case, see Divisors, Lemma 02OW and its proof. We describe this here explicitly.

Lemma

Let \(X\) be a reduced scheme such that every quasi-compact open has finitely many irreducible components. Let \(\Spec(A) = U \subset X\) be an affine open. Then

\(A\) has finitely many minimal primes \(\mathfrak q_1, \ldots, \mathfrak q_t\),

the total ring of fractions \(Q(A)\) of \(A\) is \(Q(A/\mathfrak q_1) \times \ldots \times Q(A/\mathfrak q_t)\),

the integral closure \(A'\) of \(A\) in \(Q(A)\) is the product of the integral closures of the domains \(A/\mathfrak q_i\) in the fields \(Q(A/\mathfrak q_i)\), and

\(\nu^{-1}(U)\) is identified with the spectrum of \(A'\) where \(\nu : X^\nu \to X\) is the normalization morphism.

Proof

Minimal primes correspond to irreducible components (Algebra, Lemma 00ES), hence we have (1) by assumption. Then \((0) = \mathfrak q_1 \cap \ldots \cap \mathfrak q_t\) because \(A\) is reduced (Algebra, Lemma 00E0). Then we have \(Q(A) = \prod A_{\mathfrak q_i} = \prod \kappa(\mathfrak q_i)\) by Algebra, Lemmas 02LX and 00EU. This proves (2). Part (3) follows from Algebra, Lemma 030C, or Lemma 03GO. Part (4) holds because it is clear that \(f^{-1}(U) \to U\) is the morphism \[\Spec\left(\prod \kappa(\mathfrak q_i)\right) \longrightarrow \Spec(A)\] where \(f : Y \to X\) is the morphism (035M).

Lemma

Let \(X\) be a scheme such that every quasi-compact open has a finite number of irreducible components. Let \(\nu : X^\nu \to X\) be the normalization of \(X\). Let \(x \in X\). Then the following are canonically isomorphic as \(\mathcal{O}_{X, x}\)-algebras

the stalk \((\nu_*\mathcal{O}_{X^\nu})_x\),

the integral closure of \(\mathcal{O}_{X, x}\) in the total ring of fractions of \((\mathcal{O}_{X, x})_{red}\),

the integral closure of \(\mathcal{O}_{X, x}\) in the product of the residue fields of the minimal primes of \(\mathcal{O}_{X, x}\) (and there are finitely many of these).

Proof

After replacing \(X\) by an affine open neighbourhood of \(x\) we may assume that \(X\) has finitely many irreducible components and that \(x\) is contained in each of them. Then the stalk \((\nu_*\mathcal{O}_{X^\nu})_x\) is the integral closure of \(A = \mathcal{O}_{X, x}\) in the product \(L\) of the residue fields of the minimal primes of \(A\). This follows from the construction of the normalization and Lemma 035F. Alternatively, you can use Lemma 035P and the fact that normalization commutes with localization (Algebra, Lemma 0307). Since \(A_{red}\) has finitely many minimal primes (because these correspond exactly to the generic points of the irreducible components of \(X\) passing through \(x\)) we see that \(L\) is the total ring of fractions of \(A_{red}\) (Algebra, Lemma 02LX). Thus our ring is also the integral closure of \(A\) in the total ring of fractions of \(A_{red}\).

Lemma

Let \(X\) be a scheme such that every quasi-compact open has finitely many irreducible components.

The normalization \(X^\nu\) is a disjoint union of integral normal schemes.

The morphism \(\nu : X^\nu \to X\) is integral, surjective, and induces a bijection on irreducible components.

For any integral morphism \(\alpha : X' \to X\) such that for \(U \subset X\) quasi-compact open the inverse image \(\alpha^{-1}(U)\) has finitely many irreducible components and \(\alpha|_{\alpha^{-1}(U)} : \alpha^{-1}(U) \to U\) is birational¹⁸ there exists a factorization \(X^\nu \to X' \to X\) and \(X^\nu \to X'\) is the normalization of \(X'\).

For any morphism \(Z \to X\) with \(Z\) a normal scheme such that each irreducible component of \(Z\) dominates an irreducible component of \(X\) there exists a unique factorization \(Z \to X^\nu \to X\).

Proof

Let \(f : Y \to X\) be as in (035M). The scheme \(X^\nu\) is a disjoint union of normal integral schemes because \(Y\) is normal and every affine open of \(Y\) has finitely many irreducible components, see Lemma 035L. This proves (1). Alternatively one can deduce (1) from Lemmas 035O and 035P.

The morphism \(\nu\) is integral by Lemma 035I. By Lemma 035L the morphism \(Y \to X^\nu\) induces a bijection on irreducible components, and by construction of \(Y\) this implies that \(X^\nu \to X\) induces a bijection on irreducible components. By construction \(f : Y \to X\) is dominant, hence also \(\nu\) is dominant. Since an integral morphism is closed (Lemma 01WM) this implies that \(\nu\) is surjective. This proves (2).

Suppose that \(\alpha : X' \to X\) is as in (3). It is clear that \(X'\) satisfies the assumptions under which the normalization is defined. Let \(f' : Y' \to X'\) be the morphism (035M) constructed starting with \(X'\). As \(\alpha\) is locally birational it is clear that \(Y' = Y\) and \(f = \alpha \circ f'\). Hence the factorization \(X^\nu \to X' \to X\) exists and \(X^\nu \to X'\) is the normalization of \(X'\) by Lemma 035I. This proves (3).

Let \(g : Z \to X\) be a morphism whose domain is a normal scheme and such that every irreducible component dominates an irreducible component of \(X\). By Lemma 035O we have \(X^\nu = X_{red}^\nu\) and by Schemes, Lemma 0356 \(Z \to X\) factors through \(X_{red}\). Hence we may replace \(X\) by \(X_{red}\) and assume \(X\) is reduced. Moreover, as the factorization is unique it suffices to construct it locally on \(Z\). Let \(W \subset Z\) and \(U \subset X\) be affine opens such that \(g(W) \subset U\). Write \(U = \Spec(A)\) and \(W = \Spec(B)\), with \(g|_W\) given by \(\varphi : A \to B\). We will use the results of Lemma 035P freely. Let \(\mathfrak p_1, \ldots, \mathfrak p_t\) be the minimal primes of \(A\). As \(Z\) is normal, we see that \(B\) is a normal ring, in particular reduced. Moreover, by assumption any minimal prime \(\mathfrak q \subset B\) we have that \(\varphi^{-1}(\mathfrak q)\) is a minimal prime of \(A\). Hence if \(x \in A\) is a nonzerodivisor, i.e., \(x \not \in \bigcup \mathfrak p_i\), then \(\varphi(x)\) is a nonzerodivisor in \(B\). Thus we obtain a canonical ring map \(Q(A) \to Q(B)\). As \(B\) is normal it is equal to its integral closure in \(Q(B)\) (see Algebra, Lemma 034M). Hence we see that the integral closure \(A' \subset Q(A)\) of \(A\) maps into \(B\) via the canonical map \(Q(A) \to Q(B)\). Since \(\nu^{-1}(U) = \Spec(A')\) this gives the canonical factorization \(W \to \nu^{-1}(U) \to U\) of \(\nu|_W\). We omit the verification that it is unique.

Lemma

Let \(X\) be a scheme such that every quasi-compact open has finitely many irreducible components. Let \(Z_i \subset X\), \(i \in I\) be the irreducible components of \(X\) endowed with the reduced induced structure. Let \(Z_i^\nu \to Z_i\) be the normalization. Then \(\coprod_{i \in I} Z_i^\nu \to X\) is the normalization of \(X\).

Proof

We may assume \(X\) is reduced, see Lemma 035O. Then the lemma follows either from the local description in Lemma 035P or from Lemma 035Q part (3) because \(\coprod Z_i \to X\) is integral and locally birational (as \(X\) is reduced and has locally finitely many irreducible components).

Lemma

Let \(X\) be a reduced scheme with finitely many irreducible components. Then the normalization morphism \(X^\nu \to X\) is birational.

Proof

The normalization induces a bijection of irreducible components by Lemma 035Q. Let \(\eta \in X\) be a generic point of an irreducible component of \(X\) and let \(\eta^\nu \in X^\nu\) be the generic point of the corresponding irreducible component of \(X^\nu\). Then \(\eta^\nu \mapsto \eta\) and to finish the proof we have to show that \(\mathcal{O}_{X, \eta} \to \mathcal{O}_{X^\nu, \eta^\nu}\) is an isomorphism, see Definition 01RO. Because \(X\) and \(X^\nu\) are reduced, we see that both local rings are equal to their residue fields (Algebra, Lemma 00EU). On the other hand, by the construction of the normalization as the normalization of \(X\) in \(Y = \coprod \Spec(\kappa(\eta))\) we see that we have \(\kappa(\eta) \subset \kappa(\eta^\nu) \subset \kappa(\eta)\) and the proof is complete.

Lemma

A finite (or even integral) birational morphism \(f : X \to Y\) of integral schemes with \(Y\) normal is an isomorphism.

Proof

Let \(V \subset Y\) be an affine open with inverse image \(U \subset X\) which is an affine open too. Since \(f\) is a birational morphism of integral schemes, the homomorphism \(\mathcal{O}_Y(V) \to \mathcal{O}_X(U)\) is an injective map of domains which induces an isomorphism of fraction fields. As \(Y\) is normal, the ring \(\mathcal{O}_Y(V)\) is integrally closed in the fraction field. Since \(f\) is finite (or integral) every element of \(\mathcal{O}_X(U)\) is integral over \(\mathcal{O}_Y(V)\). We conclude that \(\mathcal{O}_Y(V) = \mathcal{O}_X(U)\). This proves that \(f\) is an isomorphism as desired.

Lemma

Let \(X\) be a scheme with locally finitely many irreducible components. The normalization morphism \(\nu : X^\nu \to X\) is an isomorphism if and only if \(X\) is normal.

Proof

If \(\nu\) is an isomorphism, then \(X\) is normal by Lemma 035Q. Conversely, suppose \(X\) is normal. By Lemma 035K and Properties, Lemma 0357 we may assume \(X\) is integral. By Lemma 035Q the morphism \(\nu\) is integral and \(X^\nu\) has a unique irreducible component (so it is an integral scheme). We conclude from Lemmas 0BXC and 0AB1.

Lemma

Let \(X\) be an integral, Japanese scheme. The normalization \(\nu : X^\nu \to X\) is a finite morphism.

Proof

Follows from the definition (Properties, Definition 033S) and Lemma 035P. Namely, in this case the lemma says that \(\nu^{-1}(\Spec(A))\) is the spectrum of the integral closure of \(A\) in its field of fractions.

Lemma

Let \(X\) be a Nagata scheme. The normalization \(\nu : X^\nu \to X\) is a finite morphism.

Proof

Note that a Nagata scheme is locally Noetherian, thus Definition 035N does apply. The lemma is now a special case of Lemma 0AVK but we can also prove it directly as follows. Write \(X^\nu \to X\) as the composition \(X^\nu \to X_{red} \to X\). As \(X_{red} \to X\) is a closed immersion it is finite. Hence it suffices to prove the lemma for a reduced Nagata scheme (by Lemma 01WK). Let \(\Spec(A) = U \subset X\) be an affine open. By Lemma 035P we have \(\nu^{-1}(U) = \Spec(\prod A_i')\) where \(A_i'\) is the integral closure of \(A/\mathfrak q_i\) in its fraction field. As \(A\) is a Nagata ring (see Properties, Lemma 033X) each of the ring extensions \(A/\mathfrak q_i \subset A'_i\) are finite. Hence \(A \to \prod A'_i\) is a finite ring map and we win.

Lemma

Let \(X\) be an irreducible, geometrically unibranch scheme. The normalization morphism \(\nu : X^\nu \to X\) is a universal homeomorphism.

Proof

We have to show that \(\nu\) is integral, universally injective, and surjective, see Lemma 04DF. By Lemma 035Q the morphism \(\nu\) is integral. Let \(x \in X\) and set \(A = \mathcal{O}_{X, x}\). Since \(X\) is irreducible we see that \(A\) has a single minimal prime \(\mathfrak p\) and \(A_{red} = A/\mathfrak p\). By Lemma 0C3B the stalk \(A' = (\nu_*\mathcal{O}_{X^\nu})_x\) is the integral closure of \(A\) in the fraction field of \(A_{red}\). By More on Algebra, Definition 0BPZ we see that \(A'\) has a single prime \(\mathfrak m'\) lying over \(\mathfrak m_x \subset A\) and \(\kappa(\mathfrak m')/\kappa(x)\) is purely inseparable. Hence \(\nu\) is bijective (hence surjective) and universally injective by Lemma 01S4.

Weak normalization

We will only define the weak normalization of a scheme when it locally has finitely many irreducible components; similar to the case of normalization.

Lemma

Let \(A \to B\) be a ring map inducing a dominant morphism \(\Spec(B) \to \Spec(A)\) of spectra. There exists an \(A\)-subalgebra \(B' \subset B\) such that

\(\Spec(B') \to \Spec(A)\) is a universal homeomorphism,

given a factorization \(A \to C \to B\) such that \(\Spec(C) \to \Spec(A)\) is a universal homeomorphism, the image of \(C \to B\) is contained in \(B'\).

Proof

We will use Lemma 054M without further mention. Consider the commutative diagram \[\xymatrix{ B \ar[r] & B_{red} \\ A \ar[u] \ar[r] & A_{red} \ar[u] }\] For any factorization \(A \to C \to B\) of \(A \to B\) as in (2), we see that \(A_{red} \to C_{red} \to B_{red}\) is a factorization of \(A_{red} \to B_{red}\) as in (2). It follows that if the lemma holds for \(A_{red} \to B_{red}\) and produces the \(A_{red}\)-subalgebra \(B'_{red} \subset B_{red}\), then setting \(B' \subset B\) equal to the inverse image of \(B'_{red}\) solves the lemma for \(A \to B\). This reduces us to the case discussed in the next paragraph.

Assume \(A\) and \(B\) are reduced. In this case \(A \subset B\) by Algebra, Lemma 00FL. Let \(A \to C \to B\) be a factorization as in (2). Then we may apply Proposition 0CNE to \(A \subset C\) to see that every element of \(C\) is contained in an extension \(A[c_1, \ldots, c_n] \subset C\) such that for \(i = 1, \ldots, n\) we have

\(c_i^2, c_i^3 \in A[c_1, \ldots, c_{i - 1}]\), or

there exists a prime number \(p\) with \(pc_i, c_i^p \in A[c_1, \ldots, c_{i - 1}]\).

Thus property (2) holds if we define \(B' \subset B\) to be the subset of elements \(b \in B\) which are contained in an extension \(A[b_1, \ldots, b_n] \subset B\) such that (*) holds: for \(i = 1, \ldots, n\) we have

\(b_i^2, b_i^3 \in A[b_1, \ldots, b_{i - 1}]\), or

there exists a prime number \(p\) with \(pb_i, b_i^p \in A[b_1, \ldots, b_{i - 1}]\).

There are only two things to check: (a) \(B'\) is an \(A\)-subalgebra, and (b) \(\Spec(B') \to \Spec(A)\) is a universal homeomorphism. Part (a) follows because given \(n \geq 0\) and \(b_1, \ldots, b_n \in B\) satisfying (*) and \(m \geq 0\) and \(b'_1, \ldots, b'_m \in B\) satisfying (*), the integer \(n + m\) and \(b_1, \ldots, b_n, b'_1, \ldots, b'_m \in B\) also satisfies (*). Finally, part (b) holds by Proposition 0CNE and our construction of \(B'\).

Lemma

Let \(A \to B\) be a ring map inducing a dominant morphism \(\Spec(B) \to \Spec(A)\) of spectra. Formation of the \(A\)-subalgebra \(B' \subset B\) in Lemma 0H3J commutes with localization (see proof for explanation).

Proof

Let \(S \subset A\) be a multiplicative subset. Then \(S^{-1}A \to S^{-1}B\) is a ring map which induces a dominant morphism \(\Spec(S^{-1}B) \to \Spec(S^{-1}A)\) as well (see Lemmas 0H3F and 03HV). Hence Lemma 0H3J produces an \(S^{-1}A\)-subalgebra \((S^{-1}B)' \subset S^{-1}B\). The statement means that \(S^{-1}B' = (S^{-1}B)'\) as \(S^{-1}A\)-subalgebras of \(S^{-1}B\).

To see this is true, we will use the construction of \(B'\) and \((S^{-1}B)'\) in the proof of Lemma 0H3J. In the first step, we see that \(B'\) is the inverse image of the \(A_{red}\)-subalgebra \(B'_{red} \subset B_{red}\) constructed for the ring map \(A_{red} \to B_{red}\) and similarly for \((S^{-1}B)'\). Noting that \(S^{-1}B_{red} = (S^{-1}B)_{red}\) this reduces us to the case discussed in the next paragraph.

If \(A\) and \(B\) are reduced, we have constructed \(B'\) as the union of the subalgebras \(A[b_1, \ldots, b_n]\) such that for \(i = 1, \ldots, n\) we have

\(b_i^2, b_i^3 \in A[b_1, \ldots, b_{i - 1}]\), or

there exists a prime number \(p\) with \(pb_i, b_i^p \in A[b_1, \ldots, b_{i - 1}]\).

Similarly for \((S^{-1}B)' \subset S^{-1}B\). Thus it is clear that the image of \(B' \to B \to S^{-1}B\) is contained in \((S^{-1}B)'\). To show that the corresponding map \(S^{-1}B' \to (S^{-1}B)'\) is surjective, one uses Lemma 0CN9 to clear denominators successively; we omit the details.

Lemma

Let \(A \to B\) be a ring map inducing a dominant morphism \(\Spec(B) \to \Spec(A)\) of spectra. There exists an \(A\)-subalgebra \(B' \subset B\) such that

\(\Spec(B') \to \Spec(A)\) is a universal homeomorphism inducing isomorphisms on residue fields,

given a factorization \(A \to C \to B\) such that \(\Spec(C) \to \Spec(A)\) is a universal homeomorphism inducing isomorphisms on residue fields, the image of \(C \to B\) is contained in \(B'\).

Proof

This proof is exactly the same as the proof of Lemma 0H3J except we use Proposition 0CND in stead of Proposition 0CNE

Lemma

Let \(A \to B\) be a ring map inducing a dominant morphism \(\Spec(B) \to \Spec(A)\) of spectra. Formation of the \(A\)-subalgebra \(B' \subset B\) in Lemma 0H3L commutes with localization (see proof for explanation).

Proof

The proof is the same as the proof of Lemma 0H3K.

Lemma

Let \(f : Y \to X\) be a quasi-compact, quasi-separated, and dominant morphism of schemes.

The category of factorizations \(Y \to X' \to X\) where \(X' \to X\) is a universal homeomorphism has an initial object \(Y \to X^{Y/wn} \to X\).

The category of factorizations \(Y \to X' \to X\) where \(X' \to X\) is a universal homeomorphism inducing isomorphisms on residue fields has an initial object \(Y \to X^{Y/sn} \to X\).

Moreover, formation of the factorization \(Y \to X^{Y/wn} \to X\) and \(Y \to X^{Y/sn} \to X\) commutes with base change to open subschemes of \(X\).

Proof

We will prove (1) and omit the proof of (2); also the final assertion will follow from the construction of the factorization. We will use Lemma 04DF without further mention. First, let \(Y \to X^{Y/n} \to X\) be the normalization of \(X\) in \(Y\), see Definition 035H. For \(Y \to X' \to X\) as in (1), we obtain a unique morphism \(X^{Y/n} \to X'\) compatible with the given morphisms, see Lemma 035I. Thus it suffices to prove the lemma with \(f\) replaced by \(X^{Y/n} \to X\). This reduces us to the case studied in the next paragraph.

Assume \(f\) is integral (the rest of the proof works more generally if \(f\) is affine). Let \(U = \Spec(A)\) be an affine open of \(X\) and let \(V = f^{-1}(U) = \Spec(B)\) be the inverse image in \(Y\). Then \(A \to B\) is a ring map which induces a dominant morphism on spectra. By Lemma 0H3J we obtain an \(A\)-subalgebra \(B' \subset B\) such that setting \(U^{V/wn} = \Spec(B')\) the factorization \(V \to U^{V/wn} \to U\) is initial in the category of factorizations \(V \to U' \to U\) where \(U' \to U\) is a universal homeomorphism.

If \(U_1 \subset U_2 \subset X\) are affine opens, then setting \(V_i = f^{-1}(U_i)\) we obtain a canonical morphism \[\rho_{U_1}^{U_2} : U_1^{V_1/wn} \to U_1 \times_{U_2} U_2^{V_2/wn}\] over \(U_1\) by the universal property of \(U_1^{V_1/wn}\). These morphisms satisfy a natural functoriality which we leave to the reader to formulate and prove. Furthermore, the morphism \(\rho_{U_1}^{U_2}\) is an isomorphism; this follows from Lemma 0H3K provided that \(U_1 \subset U_2\) is a standard open and in the general case can be reduced to this case by the functorial nature of these maps and Schemes, Lemma 01IW (details omitted). Thus by relative glueing (Constructions, Lemma 01LH) we obtain a morphism \(X^{Y/wn} \to X\) which restricts to \(U^{V/wn} \to U\) over \(U\) compatibly with the \(\rho_{U_1}^{U_2}\). Of course, the morphisms \(V \to U^{V/wn}\) glue to a morphism \(Y \to X^{Y/wn}\) (see Constructions, Remark 01LK) and we get our factorization \(Y \to X^{Y/wn} \to X\) where the second morphism is a universal homeomorphism.

Finally, let \(Y \to X' \to X\) be a factorization as in (1). With \(V \to U^{V/wn} \to U \subset X\) as above, we obtain a factorization \(V \to U \times_X X' \to U\) where the second arrow is a universal homeomorphism and we obtain a unique morphism \(g_U : U^{V/wn} \to U \times_X X'\) over \(U\) by the universal property of \(U^{V/wn}\). These \(g_U\) are compatible with the morphisms \(\rho_{U_1}^{U_2}\); details omitted. Hence there is a unique morphism \(g : X^{Y/wn} \to X'\) over \(X\) agreeing with \(g_U\) over \(U\), see Constructions, Remark 01LK. This proves that \(Y \to X^{Y/wn} \to X\) is initial in our category and the proof is complete.

Definition

Let \(f : Y \to X\) be a quasi-compact, quasi-separated, and dominant morphism of schemes.

The factorization \(Y \to X^{Y/sn} \to X\) constructed in Lemma 0H3N part (2) is the seminormalization of \(X\) in \(Y\).

The factorization \(Y \to X^{Y/wn} \to X\) constructed in Lemma 0H3N part (1) is the weak normalization of \(X\) in \(Y\).

Here is a way to reinterpret the seminormalization of a scheme which locally has finitely many irreducible components.

Lemma

Let \(X\) be a scheme such that every quasi-compact open has finitely many irreducible components. Let \(\nu : X^\nu \to X\) be the normalization of \(X\). Then the seminormalization of \(X\) in \(X^\nu\) is is the seminormalization of \(X\). In a formula: \(X^{sn} = X^{X^\nu/sn}\).

Proof

Let \(f : Y \to X\) be as in (035M) so that \(X^\nu\) is the normalization of \(X\) in \(Y\). The seminormalization \(X^{sn} \to X\) of \(X\) is the initial object in the category of universal homeomorphisms \(X' \to X\) inducing isomorphisms on residue fields. Since \(Y\) is the disjoint union of the spectra of the residue fields at the generic points of irreducible components of \(X\), we see that for any \(X' \to X\) in this category we obtain a canonical lift \(f' : Y \to X'\) of \(f\). Then by Lemma 035I we obtain a canonical morphism \(X^\nu \to X'\). Whence in turn a canonical morphism \(X^{X^\nu/sn} \to X'\) by the universal property of \(X^{X^\nu/sn}\). In this way we see that \(X^{X^\nu/sn}\) satisfies the same universal property that \(X^{sn}\) has and we conclude.

Lemma 0H3Q motivates the following definition. Since we have only constructed the normalization in case \(X\) locally has finitely many irreducible components, we will also restrict ourselves to that case for the weak normalization.

Definition

Let \(X\) be a scheme such that every quasi-compact open has finitely many irreducible components. We define the weak normalization of \(X\) as the weak normalization \[X^\nu \longrightarrow X^{wn} \longrightarrow X\] of \(X\) in the normalization \(X^\nu\) of \(X\) (Definition 035N). In a formula: \(X^{wn} = X^{X^\nu/wn}\).

Combined with Lemma 0H3Q we see that for a scheme \(X\) which locally has finitely many irreducible components there are canonical morphisms \[X^\nu \to X^{wn} \to X^{sn} \to X\] Having made this definition, we can say what it means for a scheme to be weakly normal (provided it has locally finitely many irreducible components).

Definition

Let \(X\) be a scheme such that every quasi-compact open has finitely many irreducible components. We say \(X\) is weakly normal if the weak normalization \(X^{wn} \to X\) is an isomorphism (Definition 0H3R).

It follows immediately from the definitions that for a scheme \(X\) such that every quasi-compact open has finitely many irreducible components we have \[X\text{ normal} \Rightarrow X\text{ weakly normal} \Rightarrow X\text{ seminormal}\] We can work out the meaning of weak normality in the affine case as follows.

Lemma

Let \(X = \Spec(A)\) be an affine scheme which has finitely many irreducible components. Then \(X\) is weakly normal if and only if

\(A\) is seminormal (Definition 0EUL),

for a prime number \(p\) and \(z, w \in A\) such that (a) \(z\) is a nonzerodivisor, (b) \(w^p\) is divisible by \(z^p\), and (c) \(pw\) is divisible by \(z\), then \(w\) is divisible by \(z\).

Proof

Assume \(X\) is weakly normal. Since a weakly normal scheme is seminormal, we see that (1) holds (by our definition of weakly normal schemes). In particular \(A\) is reduced. Let \(p, z, w\) be as in (2). Choose \(x, y \in A\) such that \(z^p x = w^p\) and \(zy = pw\). Then \(p^p x = y^p\). The ring map \(A \to C = A[t]/(t^p - x, pt - y)\) induces a universal homeomorphism on spectra. The normalization \(X^\nu\) of \(X\) is the spectrum of the integral closure \(A'\) of \(A\) in the total ring of fractions of \(A\), see Lemma 035P. Note that \(a = w/z \in A'\) because \(a^p = x\). Hence we have an \(A\)-algebra homomorphism \(A \to C \to A'\) sending \(t\) to \(a\). At this point the defining property \(X = X^{wn} = X^{X^\nu/wn}\) of being weakly normal tells us that \(C \to A'\) maps into \(A\). Thus we find \(a \in A\) as desired.

Conversely, assume (1) and (2). Let \(A'\) be as in the previous paragraph. We have to show that \(X^{X^\nu/wn} = X\). By construction in the proof of Lemma 0H3J, the scheme \(X^{X^\nu/wn}\) is the spectrum of the subring of \(A'\) which is the union of the subrings \(A[a_1, \ldots, a_n] \subset A'\) such that for \(i = 1, \ldots, n\) we have

\(a_i^2, a_i^3 \in A[a_1, \ldots, a_{i - 1}]\), or

there exists a prime number \(p\) with \(pa_i, a_i^p \in A[a_1, \ldots, a_{i - 1}]\).

Then we can use (1) and (2) to inductively see that \(a_1, \ldots, a_n \in A\); we omit the details. Consequently, we have \(X = X^{X^\nu/wn}\) and hence \(X\) is weakly normal.

Here is the obligatory lemma.

Lemma

Let \(X\) be a scheme such that every quasi-compact open has finitely many irreducible components. The following are equivalent:

The scheme \(X\) is weakly normal.

For every affine open \(U \subset X\) the ring \(\mathcal{O}_X(U)\) satisfies conditions (1) and (2) of Lemma 0H3T.

There exists an affine open covering \(X = \bigcup U_i\) such that each ring \(\mathcal{O}_X(U_i)\) satisfies conditions (1) and (2) of Lemma 0H3T.

There exists an open covering \(X = \bigcup X_j\) such that each open subscheme \(X_j\) is weakly normal.

Moreover, if \(X\) is weakly normal then every open subscheme is weakly normal.

Proof

The condition to \(X\) be weakly normal is that the morphism \(X^{wn} = X^{X^\nu/wn} \to X\) is an isomorphism. Since the construction of \(X^\nu \to X\) commutes with base change to open subschemes and since the construction of \(X^{X^\nu/wn}\) commutes with base change to open subschemes of \(X\) (Lemma 0H3N) the lemma is clear.

Zariski’s Main Theorem (algebraic version)

This is the version you can prove using purely algebraic methods. Before we can prove more powerful versions (for non-affine morphisms) we need to develop more tools. See Cohomology of Schemes, Section 02OF and More on Morphisms, Section 02LQ.

Theorem

Let \(f : Y \to X\) be an affine morphism of schemes. Assume \(f\) is of finite type. Let \(X'\) be the normalization of \(X\) in \(Y\). Picture: \[\xymatrix{ Y \ar[rd]_f \ar[rr]_{f'} & & X' \ar[ld]^\nu \\ & X & }\] Then there exists an open subscheme \(U' \subset X'\) such that

\((f')^{-1}(U') \to U'\) is an isomorphism, and

\((f')^{-1}(U') \subset Y\) is the set of points at which \(f\) is quasi-finite.

Proof

There is an immediate reduction to the case where \(X\) and hence \(Y\) are affine. Say \(X = \Spec(R)\) and \(Y = \Spec(A)\). Then \(X' = \Spec(A')\), where \(A'\) is the integral closure of \(R\) in \(A\), see Definitions 035G and 035H. By Algebra, Theorem 00Q9 for every \(y \in Y\) at which \(f\) is quasi-finite, there exists an open \(U'_y \subset X'\) such that \((f')^{-1}(U'_y) \to U'_y\) is an isomorphism. Set \(U' = \bigcup U'_y\) where \(y \in Y\) ranges over all points where \(f\) is quasi-finite. It remains to show that \(f\) is quasi-finite at all points of \((f')^{-1}(U')\). If \(y \in (f')^{-1}(U')\) with image \(x \in X\), then we see that \(Y_x \to X'_x\) is an isomorphism in a neighbourhood of \(y\). Hence there is no point of \(Y_x\) which specializes to \(y\), since this is true for \(f'(y)\) in \(X'_x\), see Lemma 02NT. By Lemma 01TH part (3) this implies \(f\) is quasi-finite at \(y\).

We can use the algebraic version of Zariski’s Main Theorem to show that the set of points where a morphism is quasi-finite is open.

Lemma

Let \(f : X \to S\) be a morphism of schemes. The set of points of \(X\) where \(f\) is quasi-finite is an open \(U \subset X\). The induced morphism \(U \to S\) is locally quasi-finite.

Proof

Suppose \(f\) is quasi-finite at \(x\). Let \(x \in U = \Spec(A) \subset X\), \(V = \Spec(R) \subset S\) be affine opens as in Definition 01TD. By either Theorem 03GT above or Algebra, Lemma 00QA, the set of primes \(\mathfrak q\) at which \(R \to A\) is quasi-finite is open in \(\Spec(A)\). Since these all correspond to points of \(X\) where \(f\) is quasi-finite we get the first statement. The second statement is obvious.

We will improve the following lemma to general quasi-finite separated morphisms later, see More on Morphisms, Lemma 05K0.

Lemma

Let \(f : Y \to X\) be a morphism of schemes. Assume

\(X\) and \(Y\) are affine, and

\(f\) is quasi-finite.

Then there exists a diagram \[\xymatrix{ Y \ar[rd]_f \ar[rr]_j & & Z \ar[ld]^\pi \\ & X & }\] with \(Z\) affine, \(\pi\) finite and \(j\) an open immersion.

Proof

This is Algebra, Lemma 00QB reformulated in the language of schemes.

Lemma

Let \(f : Y \to X\) be a quasi-finite morphism of schemes. Let \(T \subset Y\) be a closed nowhere dense subset of \(Y\). Then \(f(T) \subset X\) is a nowhere dense subset of \(X\).

Proof

As in the proof of Lemma 03HX this reduces immediately to the case where the base \(X\) is affine. In this case \(Y = \bigcup_{i = 1, \ldots, n} Y_i\) is a finite union of affine opens (as \(f\) is quasi-compact). Since each \(T \cap Y_i\) is nowhere dense, and since a finite union of nowhere dense sets is nowhere dense (see Topology, Lemma 03HO), it suffices to prove that the image \(f(T \cap Y_i)\) is nowhere dense in \(X\). This reduces us to the case where both \(X\) and \(Y\) are affine. At this point we apply Lemma 03GU above to get a diagram \[\xymatrix{ Y \ar[rd]_f \ar[rr]_j & & Z \ar[ld]^\pi \\ & X & }\] with \(Z\) affine, \(\pi\) finite and \(j\) an open immersion. Set \(\overline{T} = \overline{j(T)} \subset Z\). By Topology, Lemma 03J0 we see \(\overline{T}\) is nowhere dense in \(Z\). Since \(f(T) \subset \pi(\overline{T})\) the lemma follows from the corresponding result in the finite case, see Lemma 03HX.

Universally bounded fibres

Let \(X\) be a scheme over a field \(k\). If \(X\) is finite over \(k\), then \(X = \Spec(A)\) where \(A\) is a finite \(k\)-algebra. Another way to say this is that \(X\) is finite locally free over \(\Spec(k)\), see Definition 02KA. Hence \(X \to \Spec(k)\) has a degree which is an integer \(d \geq 0\), namely \(d = \dim_k(A)\). We sometime call this the degree of the (finite) scheme \(X\) over \(k\).

Definition

Let \(f : X \to Y\) be a morphism of schemes.

We say the integer \(n\) bounds the degrees of the fibres of \(f\) if for all \(y \in Y\) the fibre \(X_y\) is a finite scheme over \(\kappa(y)\) whose degree over \(\kappa(y)\) is \(\leq n\).

We say the fibres of \(f\) are universally bounded¹⁹ if there exists an integer \(n\) which bounds the degrees of the fibres of \(f\).

Note that in particular the number of points in a fibre is bounded by \(n\) as well. (The converse does not hold, even if all fibres are finite reduced schemes.)

Lemma

Let \(f : X \to Y\) be a morphism of schemes. Let \(n \geq 0\). The following are equivalent:

the integer \(n\) bounds the degrees of the fibres of \(f\), and

for every morphism \(\Spec(k) \to Y\), where \(k\) is a field, the fibre product \(X_k = \Spec(k) \times_Y X\) is finite over \(k\) of degree \(\leq n\).

In this case the fibres of \(f\) are universally bounded and the schemes \(X_k\) have at most \(n\) points. More precisely, if \(X_k = \{x_1, \ldots, x_t\}\), then we have \[n \geq \sum\nolimits_{i = 1, \ldots, t} [\kappa(x_i) : k]\]

Proof

The implication (2) \(\Rightarrow\) (1) is trivial. The other implication holds because if the image of \(\Spec(k) \to Y\) is \(y\), then \(X_k = \Spec(k) \times_{\Spec(\kappa(y))} X_y\). By definition the fibres of \(f\) being universally bounded means that some \(n\) exists. Finally, suppose that \(X_k = \Spec(A)\). Then \(\dim_k A = n\). Hence \(A\) is Artinian, all prime ideals are maximal ideals \(\mathfrak m_i\), and \(A\) is the product of the localizations at these maximal ideals. See Algebra, Lemmas 00J6 and 00JB. Then \(\mathfrak m_i\) corresponds to \(x_i\), we have \(A_{\mathfrak m_i} = \mathcal{O}_{X_k, x_i}\) and hence there is a surjection \(A \to \bigoplus \kappa(\mathfrak m_i) = \bigoplus \kappa(x_i)\) which implies the inequality in the statement of the lemma by linear algebra.

Lemma

If \(f\) is a finite locally free morphism of degree \(d\), then \(d\) bounds the degree of the fibres of \(f\).

Proof

This is true because any base change of \(f\) is finite locally free of degree \(d\) (Lemma 02KD) and hence the fibres of \(f\) all have degree \(d\).

Lemma

A composition of morphisms with universally bounded fibres is a morphism with universally bounded fibres. More precisely, assume that \(n\) bounds the degrees of the fibres of \(f : X \to Y\) and \(m\) bounds the degrees of \(g : Y \to Z\). Then \(nm\) bounds the degrees of the fibres of \(g \circ f : X \to Z\).

Proof

Let \(f : X \to Y\) and \(g : Y \to Z\) have universally bounded fibres. Say that \(\deg(X_y/\kappa(y)) \leq n\) for all \(y \in Y\), and that \(\deg(Y_z/\kappa(z)) \leq m\) for all \(z \in Z\). Let \(z \in Z\) be a point. By assumption the scheme \(Y_z\) is finite over \(\Spec(\kappa(z))\). In particular, the underlying topological space of \(Y_z\) is a finite discrete set. The fibres of the morphism \(f_z : X_z \to Y_z\) are the fibres of \(f\) at the corresponding points of \(Y\), which are finite discrete sets by the reasoning above. Hence we conclude that the underlying topological space of \(X_z\) is a finite discrete set as well. Thus \(X_z\) is an affine scheme (this is a nice exercise; it also follows for example from Properties, Lemma 01ZV applied to the set of all points of \(X_z\)). Write \(X_z = \Spec(A)\), \(Y_z = \Spec(B)\), and \(k = \kappa(z)\). Then \(k \to B \to A\) and we know that (a) \(\dim_k(B) \leq m\), and (b) for every maximal ideal \(\mathfrak m \subset B\) we have \(\dim_{\kappa(\mathfrak m)}(A/\mathfrak mA) \leq n\). We claim this implies that \(\dim_k(A) \leq nm\). Note that \(B\) is the product of its localizations \(B_{\mathfrak m}\), for example because \(Y_z\) is a disjoint union of \(1\)-point schemes, or by Algebra, Lemmas 00J6 and 00JB. So we see that \(\dim_k(B) = \sum_{\mathfrak m} \dim_k(B_{\mathfrak m})\) and \(\dim_k(A) = \sum_{\mathfrak m} \dim_k(A_{\mathfrak m})\) where in both cases \(\mathfrak m\) runs over the maximal ideals of \(B\) (not of \(A\)). By the above, and Nakayama’s Lemma (Algebra, Lemma 00DV) we see that each \(A_{\mathfrak m}\) is a quotient of \(B_{\mathfrak m}^{\oplus n}\) as a \(B_{\mathfrak m}\)-module. Hence \(\dim_k(A_{\mathfrak m}) \leq n \dim_k(B_{\mathfrak m})\). Putting everything together we see that \[\dim_k(A) = \sum\nolimits_{\mathfrak m} \dim_ta (A_{\mathfrak m}) \leq \sum\nolimits_{\mathfrak m} n \dim_k(B_{\mathfrak m}) = n \dim_k(B) \leq nm\] as desired.

Lemma

A base change of a morphism with universally bounded fibres is a morphism with universally bounded fibres. More precisely, if \(n\) bounds the degrees of the fibres of \(f : X \to Y\) and \(Y' \to Y\) is any morphism, then the degrees of the fibres of the base change \(f' : Y' \times_Y X \to Y'\) is also bounded by \(n\).

Proof

This is clear from the result of Lemma 03J5.

Lemma

Let \(f : X \to Y\) be a morphism of schemes. Let \(Y' \to Y\) be a morphism of schemes, and let \(f' : X' = X_{Y'} \to Y'\) be the base change of \(f\). If \(Y' \to Y\) is surjective and \(f'\) has universally bounded fibres, then \(f\) has universally bounded fibres. More precisely, if \(n\) bounds the degree of the fibres of \(f'\), then also \(n\) bounds the degrees of the fibres of \(f\).

Proof

Let \(n \geq 0\) be an integer bounding the degrees of the fibres of \(f'\). We claim that \(n\) works for \(f\) also. Namely, if \(y \in Y\) is a point, then choose a point \(y' \in Y'\) lying over \(y\) and observe that \[X'_{y'} = \Spec(\kappa(y')) \times_{\Spec(\kappa(y))} X_y.\] Since \(X'_{y'}\) is assumed finite of degree \(\leq n\) over \(\kappa(y')\) it follows that also \(X_y\) is finite of degree \(\leq n\) over \(\kappa(y)\). (Some details omitted.)

Lemma

An immersion has universally bounded fibres.

Proof

The integer \(n = 1\) works in the definition.

Lemma

Let \(f : X \to Y\) be an étale morphism of schemes. Let \(n \geq 0\). The following are equivalent

the integer \(n\) bounds the degrees of the fibres,

for every field \(k\) and morphism \(\Spec(k) \to Y\) the base change \(X_k = \Spec(k) \times_Y X\) has at most \(n\) points, and

for every \(y \in Y\) and every separable algebraic closure \(\kappa(y) \subset \kappa(y)^{sep}\) the scheme \(X_{\kappa(y)^{sep}}\) has at most \(n\) points.

Proof

This follows from Lemma 03J5 and the fact that the fibres \(X_y\) are disjoint unions of spectra of finite separable field extensions of \(\kappa(y)\), see Lemma 02GL.

Having universally bounded fibres is an absolute notion and not a relative notion. This is why the condition in the following lemma is that \(X\) is quasi-compact, and not that \(f\) is quasi-compact.

Lemma

Let \(f : X \to Y\) be a morphism of schemes. Assume that

\(f\) is locally quasi-finite, and

\(X\) is quasi-compact.

Then \(f\) has universally bounded fibres.

Proof

Since \(X\) is quasi-compact, there exists a finite affine open covering \(X = \bigcup_{i = 1, \ldots, n} U_i\) and affine opens \(V_i \subset Y\), \(i = 1, \ldots, n\) such that \(f(U_i) \subset V_i\). Because of the local nature of “local quasi-finiteness” (see Lemma 01TH part (4)) we see that the morphisms \(f|_{U_i} : U_i \to V_i\) are locally quasi-finite morphisms of affines, hence quasi-finite, see Lemma 01TJ. For \(y \in Y\) it is clear that \(X_y = \bigcup_{y \in V_i} (U_i)_y\) is an open covering. Hence it suffices to prove the lemma for a quasi-finite morphism of affines (namely, if \(n_i\) works for the morphism \(f|_{U_i} : U_i \to V_i\), then \(\sum n_i\) works for \(f\)).

Assume \(f : X \to Y\) is a quasi-finite morphism of affines. By Lemma 03GU we can find a diagram \[\xymatrix{ X \ar[rd]_f \ar[rr]_j & & Z \ar[ld]^\pi \\ & Y & }\] with \(Z\) affine, \(\pi\) finite and \(j\) an open immersion. Since \(j\) has universally bounded fibres (Lemma 03J9) this reduces us to showing that \(\pi\) has universally bounded fibres (Lemma 03J6).

This reduces us to a morphism of the form \(\Spec(B) \to \Spec(A)\) where \(A \to B\) is finite. Say \(B\) is generated by \(x_1, \ldots, x_n\) as an \(A\)-module, so that \[A^{\oplus n} \longrightarrow B, \quad (a_1, \ldots, a_n) \longmapsto \sum a_i x_i\] is a surjective \(A\)-module map. For any prime \(\mathfrak p \subset A\) this induces a surjective map \(\kappa(\mathfrak p)\)-vector spaces \[\kappa(\mathfrak p)^{\oplus n} \longrightarrow B \otimes_A \kappa(\mathfrak p)\] In other words, the integer \(n\) works in the definition of a morphism with universally bounded fibres.

Lemma

Consider a commutative diagram of morphisms of schemes \[\xymatrix{ X \ar[rd]_g \ar[rr]_f & & Y \ar[ld]^h \\ & Z & }\] If \(g\) has universally bounded fibres, and \(f\) is surjective and flat, then also \(h\) has universally bounded fibres. More precisely, if \(n\) bounds the degree of the fibres of \(g\), then also \(n\) bounds the degree of the fibres of \(h\).

Proof

Assume \(g\) has universally bounded fibres, and \(f\) is surjective and flat. Say the degree of the fibres of \(g\) is bounded by \(n \in \mathbf{N}\). We claim \(n\) also works for \(h\). Let \(z \in Z\). Consider the morphism of schemes \(X_z \to Y_z\). It is flat and surjective. By assumption \(X_z\) is a finite scheme over \(\kappa(z)\), in particular it is the spectrum of an Artinian ring (by Algebra, Lemma 00J6). By Lemma 01SI the morphism \(X_z \to Y_z\) is affine in particular quasi-compact. It follows from Lemma 02JY that \(Y_z\) is a finite discrete as this holds for \(X_z\). Hence \(Y_z\) is an affine scheme (this is a nice exercise; it also follows for example from Properties, Lemma 01ZV applied to the set of all points of \(Y_z\)). Write \(Y_z = \Spec(B)\) and \(X_z = \Spec(A)\). Then \(A\) is faithfully flat over \(B\), so \(B \subset A\). Hence \(\dim_k(B) \leq \dim_k(A) \leq n\) as desired.

Miscellany

Results which do not fit elsewhere.

Lemma

Let \(f : Y \to X\) be a morphism of schemes. Let \(x \in X\) be a point. Assume that \(Y\) is reduced and \(f(Y)\) is set-theoretically contained in \(\{x\}\). Then \(f\) factors through the canonical morphism \(x = \Spec(\kappa(x)) \to X\).

Proof

Omitted. Hints: working affine locally one reduces to a commutative algebra lemma. Given a ring map \(A \to B\) with \(B\) reduced such that there exists a unique prime ideal \(\mathfrak p \subset A\) in the image of \(\Spec(B) \to \Spec(A)\), then \(A \to B\) factors through \(\kappa(\mathfrak p)\). This is a nice exercise.

Lemma

Let \(f : Y \to X\) be a morphism of schemes. Let \(E \subset X\). Assume \(X\) is locally Noetherian, there are no nontrivial specializations among the elements of \(E\), \(Y\) is reduced, and \(f(Y) \subset E\). Then \(f\) factors through \(\coprod_{x \in E} x \to X\).

Proof

When \(E\) is a singleton this follows from Lemma 0H1M. If \(E\) is finite, then \(E\) (with the induced topology of \(X\)) is a finite discrete space by our assumption on specializations. Hence this case reduces to the singleton case. In general, there is a reduction to the case where \(X\) and \(Y\) are affine schemes. Say \(f : Y \to X\) corresponds to the ring map \(\varphi : A \to B\). Denote \(A' \subset B\) the image of \(\varphi\). Let \(E' \subset \Spec(A') \subset \Spec(A)\) be the set of minimal primes of \(A'\). By Algebra, Lemma 00FK the set \(E'\) is contained in the image of \(\Spec(B) \to \Spec(A') \subset \Spec(A)\). We conclude that \(E' \subset E\). Since \(A'\) is Noetherian we have \(E'\) is finite by Algebra, Lemma 00FR. Since any other point in the image of \(\Spec(B) \to \Spec(A)\) is a specialization of an element of \(E'\) and in \(E\), we conclude that the image is contained in \(E'\) (by our assumption on specializations between points of \(E\)). Thus we reduce to the case where \(E\) is finite which we dealt with above.

This was proved in a more general situation in the proof of Modules, Lemma 08KS.↩︎
This is likely nonstandard notation.↩︎
By Lemma 01R8 set-theoretically \(Z\) agrees with the closure of \(f(X)\) in \(Y\).↩︎
But only those properties that are not already dealt with separately elsewhere.↩︎
But only those properties that are not already dealt with separately elsewhere.↩︎
But only those properties that are not already dealt with separately elsewhere.↩︎
This is very different from the notion of a submersion of differential manifolds.↩︎
In fact, if \(f\) is surjective, flat, and locally of finite presentation and \(p\) is syntomic, then both \(q\) and \(f\) are syntomic, see Descent, Lemma 05B7.↩︎
Namely, the local section \(\text{d}_{X/S}(f) = 1 \otimes f - f \otimes 1\) of the ideal sheaf of \(\Delta\) maps via \(\text{d}_{X \times_S X/X}\) to the local section \(1 \otimes 1 \otimes 1 \otimes f - 1 \otimes f \otimes 1 \otimes 1 -1 \otimes 1 \otimes f \otimes 1 + f \otimes 1 \otimes 1 \otimes 1 = \text{pr}_2^*\text{d}_{X/S}(f) - \text{pr}_1^*\text{d}_{X/S}(f)\).↩︎
In fact this is implied by (1) and (2), see Descent, Lemma 02KL. Moreover, it suffices to assume \(f\) is surjective, flat and locally of finite presentation, see Descent, Lemma 05B5.↩︎
In fact this is implied by (1) and (2), see Descent, Lemma 02KL. Moreover, it suffices to assume that \(f\) is surjective, flat and locally of finite presentation, see Descent, Lemma 05B5.↩︎
This follows if \(g\) is quasi-separated by Schemes, Lemma 03GI.↩︎
This is not exactly the same as the definition in Hartshorne. Namely, the definition in Hartshorne (8th corrected printing, 1997) is that \(f\) should be the composition of an open immersion followed by a H-projective morphism (see Definition 01W8), which does not imply \(f\) is quasi-compact. See Lemma 01WA for the implication in the other direction.↩︎
This follows if \(g\) is quasi-separated by Schemes, Lemma 03GI.↩︎
Let \(A\) be a ring such that for all \(x, y \in A\) with \(x^3 = y^2\) there is an \(a \in A\) with \(x = a^2\) and \(y = a^3\). Then \(A\) is reduced: if \(x^2 = 0\), then \(x^2 = x^3\) and hence there exists an \(a\) such that \(x = a^3\) and \(x = a^2\). Then \(x = a^3 = ax = a^4 = x^2 = 0\). Finally, if \(a_1^2 = a_2^2\) and \(a_1^3 = a_2^3\) for \(a_1, a_2\) in a reduced ring, then \((a_1 - a_2)^3 = a_1^3 - 3a_1^2a_2 +3a_1a_2^2 - a_2^3 = (1 - 3 + 3 - 1)a_1^3 = 0\) and hence \(a_1 = a_2\).↩︎
For example if \(f\) is proper, see Definition 01W1.↩︎
The scheme \(X'\) need not be normal, for example if \(Y = X\) and \(f = \text{id}_X\), then \(X' = X\).↩︎
This awkward formulation is necessary as we’ve only defined what it means for a morphism to be birational if the source and target have finitely many irreducible components. It suffices if \(X'_{red} \to X_{red}\) satisfies the condition.↩︎
This is probably nonstandard notation.↩︎