More on Algebra

Introduction

In this chapter we prove some results in commutative algebra which are less elementary than those in the first chapter on commutative algebra, see Algebra, Section 00AP. A reference is [MatCA].

Advice for the reader

More than in the chapter on commutative algebra, each of the sections in this chapter stands on its own. Starting with Section 0914 we freely use the (unbounded) derived category of modules over rings and all the machinery that comes with it.

Stably free modules

Here is what seems to be the generally accepted definition.

Definition

Let \(R\) be a ring.

1. Two modules \(M\), \(N\) over \(R\) are said to be stably isomorphic if there exist \(n, m \geq 0\) such that \(M \oplus R^{\oplus m} \cong N \oplus R^{\oplus n}\) as \(R\)-modules.

2. A module \(M\) is stably free if it is stably isomorphic to a free module.

Observe that a stably free module is projective.

Lemma

Let \(R\) be a ring. Let \(0 \to P' \to P \to P'' \to 0\) be a short exact sequence of finite projective \(R\)-modules. If \(2\) out of \(3\) of these modules are stably free, then so is the third.

Proof

Since the modules are projective, the sequence is split. Thus we can choose an isomorphism \(P = P' \oplus P''\). If \(P' \oplus R^{\oplus n}\) and \(P'' \oplus R^{\oplus m}\) are free, then we see that \(P \oplus R^{\oplus n + m}\) is free. Suppose that \(P'\) and \(P\) are stably free, say \(P \oplus R^{\oplus n}\) is free and \(P' \oplus R^{\oplus m}\) is free. Then \[P'' \oplus (P' \oplus R^{\oplus m}) \oplus R^{\oplus n} = (P'' \oplus P') \oplus R^{\oplus m} \oplus R^{\oplus n} = (P \oplus R^{\oplus n}) \oplus R^{\oplus m}\] is free. Thus \(P''\) is stably free. By symmetry we get the last of the three cases.

Lemma

Let \(R\) be a ring. Let \(I \subset R\) be an ideal. Assume that every element of \(1 + I\) is a unit (in other words \(I\) is contained in the Jacobson radical of \(R\)). For every finite stably free \(R/I\)-module \(E\) there exists a finite stably free \(R\)-module \(M\) such that \(M/IM \cong E\).

Proof

Choose a \(n\) and \(m\) and an isomorphism \(E \oplus (R/I)^{\oplus n} \cong (R/I)^{\oplus m}\). Choose \(R\)-linear maps \(\varphi : R^{\oplus m} \to R^{\oplus n}\) and \(\psi : R^{\oplus n} \to R^{\oplus m}\) lifting the projection \((R/I)^{\oplus m} \to (R/I)^{\oplus n}\) and injection \((R/I)^{\oplus n} \to (R/I)^{\oplus m}\). Then \(\varphi \circ \psi : R^{\oplus n} \to R^{\oplus n}\) reduces to the identity modulo \(I\). Thus the determinant of this map is invertible by our assumption on \(I\). Hence \(P = \Ker(\varphi)\) is stably free and lifts \(E\).

Lemma

Let \(R\) be a ring. Let \(I \subset R\) be an ideal. Assume that every element of \(1 + I\) is a unit (in other words \(I\) is contained in the Jacobson radical of \(R\)). Let \(M\) be a finite flat \(R\)-module such that \(M/IM\) is a projective \(R/I\)-module. Then \(M\) is a finite projective \(R\)-module.

Proof

By Algebra, Lemma 00NZ we see that \(M_\mathfrak p\) is finite free for all prime ideals \(\mathfrak p \subset R\). By Algebra, Lemma 00NX it suffices to show that the function \(\rho_M : \mathfrak p \mapsto \dim_{\kappa(\mathfrak p)} M \otimes_R \kappa(\mathfrak p)\) is locally constant on \(\Spec(R)\). Because \(M/IM\) is finite projective, this is true on \(V(I) \subset \Spec(R)\). Since every closed point of \(\Spec(R)\) is in \(V(I)\) and since \(\rho_M(\mathfrak p) = \rho_M(\mathfrak q)\) whenever \(\mathfrak p \subset \mathfrak q \subset R\) are prime ideals, we conclude by an elementary argument on topological spaces which we omit.

The lift of Lemma 0BC5 is unique up to isomorphism by the following lemma.

Lemma

Let \(R\) be a ring. Let \(I \subset R\) be an ideal. Assume that every element of \(1 + I\) is a unit (in other words \(I\) is contained in the Jacobson radical of \(R\)). If \(P\) and \(P'\) are finite projective \(R\)-modules, then

1. if \(\varphi : P \to P'\) is an \(R\)-module map inducing an isomorphism \(\overline{\varphi} : P/IP \to P'/IP'\), then \(\varphi\) is an isomorphism,

2. if \(P/IP \cong P'/IP'\), then \(P \cong P'\).

Proof

Proof of (1). As \(P'\) is projective as an \(R\)-module we may choose a lift \(\psi : P' \to P\) of the map \(P' \to P'/IP' \xrightarrow{\overline{\varphi}^{-1}} P/IP\). By Nakayama’s lemma (Algebra, Lemma 00DV) \(\psi \circ \varphi\) and \(\varphi \circ \psi\) are surjective. Hence these maps are isomorphisms (Algebra, Lemma 05G8). Thus \(\varphi\) is an isomorphism.

Proof of (2). Choose an isomorphism \(P/IP \cong P'/IP'\). Since \(P\) is projective we can choose a lift \(\varphi : P \to P'\) of the map \(P \to P/IP \to P'/IP'\). Then \(\varphi\) is an isomorphism by (1).

A comment on the Artin-Rees property

Some of this material is taken from [conrad-dejong]. A general discussion with additional references can be found in [Eis, Section 1].

Let \(A\) be a Noetherian ring and let \(I \subset A\) be an ideal. Given a homomorphism \(f : M \to N\) of finite \(A\)-modules there exists a \(c \geq 0\) such that \[f(M) \cap I^nN \subset f(I^{n - c}M)\] for all \(n \geq c\), see Algebra, Lemma 00IO. In this situation we will say \(c\) works for \(f\) in the Artin-Rees lemma.

Lemma

Let \(A\) be a Noetherian ring. Let \(I \subset A\) be an ideal contained in the Jacobson radical of \(A\). Let \[S : L \xrightarrow{f} M \xrightarrow{g} N \quad\text{and}\quad S' : L \xrightarrow{f'} M \xrightarrow{g'} N\] be two complexes of finite \(A\)-modules as shown. Assume that

1. \(c\) works in the Artin-Rees lemma for \(f\) and \(g\),

2. the complex \(S\) is exact, and

3. \(f' = f \bmod I^{c + 1}M\) and \(g' = g \bmod I^{c + 1}N\).

Then \(c\) works in the Artin-Rees lemma for \(g'\) and the complex \(S'\) is exact.

Proof

We first show that \(g'(M) \cap I^nN \subset g'(I^{n - c}M)\) for \(n \geq c\). Let \(a\) be an element of \(M\) such that \(g'(a) \in I^nN\). We want to adjust \(a\) by an element of \(f'(L)\), i.e, without changing \(g'(a)\), so that \(a \in I^{n-c}M\). Assume that \(a \in I^rM\), where \(r < n - c\). Then \[g(a) = g'(a) + (g - g')(a) \in I^n N + I^{r + c + 1}N = I^{r + c + 1}N.\] By Artin-Rees for \(g\) we have \(g(a) \in g(I^{r + 1}M)\). Say \(g(a) = g(a_1)\) with \(a_1 \in I^{r + 1}M\). Since the sequence \(S\) is exact, \(a - a_1 \in f(L)\). Accordingly, we write \(a = f(b) + a_1\) for some \(b \in L\). Then \(f(b) = a - a_1 \in I^rM\). Artin-Rees for \(f\) shows that if \(r \geq c\), we may replace \(b\) by an element of \(I^{r - c}L\). Then in all cases, \(a = f'(b) + a_2\), where \(a_2 = (f - f')(b) + a_1 \in I^{r + 1}M\). (Namely, either \(c \geq r\) and \((f - f')(b) \in I^{r + 1}M\) by assumption, or \(c < r\) and \(b \in I^{r - c}\), whence again \((f - f')(b) \in I^{c + 1} I^{r - c} M = I^{r + 1}M\).) So we can adjust \(a\) by the element \(f'(b) \in f'(L)\) to increase \(r\) by \(1\).

In fact, the argument above shows that \((g')^{-1}(I^nN) \subset f'(L) + I^{n - c}M\) for all \(n \geq c\). Hence \(S'\) is exact because \[(g')^{-1}(0) = (g')^{-1}(\bigcap I^nN) \subset \bigcap f'(L) + I^{n - c}M = f'(L)\] as \(I\) is contained in the Jacobson radical of \(A\), see Algebra, Lemma 00IQ.

Given an ideal \(I \subset A\) of a ring \(A\) and an \(A\)-module \(M\) we set \[\text{Gr}_I(M) = \bigoplus I^nM/I^{n + 1}M.\] We think of this as a graded \(\text{Gr}_I(A)\)-module.

Lemma

Assumptions as in Lemma 07VE. Let \(Q = \Coker(g)\) and \(Q' = \Coker(g')\). Then \(\text{Gr}_I(Q) \cong \text{Gr}_I(Q')\) as graded \(\text{Gr}_I(A)\)-modules.

Proof

In degree \(n\) we have \(\text{Gr}_I(Q)_n = I^nN/(I^{n + 1}N + g(M) \cap I^nN)\) and similarly for \(Q'\). We claim that \[g(M) \cap I^nN \subset I^{n + 1}N + g'(M) \cap I^nN.\] By symmetry (the proof of the claim will only use that \(c\) works for \(g\) which also holds for \(g'\) by the lemma) this will imply that \[I^{n + 1}N + g(M) \cap I^nN = I^{n + 1}N + g'(M) \cap I^nN\] whence \(\text{Gr}_I(Q)_n\) and \(\text{Gr}_I(Q')_n\) agree as subquotients of \(N\), implying the lemma. Observe that the claim is clear for \(n \leq c\) as \(g = g' \bmod I^{c + 1}N\). If \(n > c\), then suppose \(b \in g(M) \cap I^nN\). Write \(b = g(a)\) for \(a \in I^{n - c}M\). Set \(b' = g'(a)\). We have \(b - b' = (g - g')(a) \in I^{n + 1}N\) as desired.

Lemma

Let \(A \to B\) be a flat map of Noetherian rings. Let \(I \subset A\) be an ideal. Let \(f : M \to N\) be a homomorphism of finite \(A\)-modules. Assume that \(c\) works for \(f\) in the Artin-Rees lemma. Then \(c\) works for \(f \otimes 1 : M \otimes_A B \to N \otimes_A B\) in the Artin-Rees lemma for the ideal \(IB\).

Proof

Note that \[(f \otimes 1)(M) \cap I^n N \otimes_A B = (f \otimes 1)\left((f \otimes 1)^{-1}(I^n N \otimes_A B)\right)\] On the other hand, \[\begin{align*} (f \otimes 1)^{-1}(I^n N \otimes_A B) & = \Ker(M \otimes_A B \to N \otimes_A B/(I^n N \otimes_A B)) \\ & = \Ker(M \otimes_A B \to (N/I^nN) \otimes_A B) \end{align*}\] As \(A \to B\) is flat taking kernels and cokernels commutes with tensoring with \(B\), whence this is equal to \(f^{-1}(I^nN) \otimes_A B\). By assumption \(f^{-1}(I^nN)\) is contained in \(\Ker(f) + I^{n - c}M\). Thus the lemma holds.

Fibre products of rings, I

Fibre products of rings have to do with pushouts of schemes. Some cases of pushouts of schemes are discussed in More on Morphisms, Section 07RS.

Lemma

Let \(R\) be a ring. Let \(A \to B\) and \(C \to B\) be \(R\)-algebra maps. Assume

1. \(R\) is Noetherian,

2. \(A\), \(B\), \(C\) are of finite type over \(R\),

3. \(A \to B\) is surjective, and

4. \(B\) is finite over \(C\).

Then \(A \times_B C\) is of finite type over \(R\).

Proof

Set \(D = A \times_B C\). There is a commutative diagram \[\xymatrix{ 0 & B \ar[l] & A \ar[l] & I \ar[l] & 0 \ar[l] \\ 0 & C \ar[l] \ar[u] & D \ar[l] \ar[u] & I \ar[l] \ar[u] & 0 \ar[l] }\] with exact rows. Choose \(y_1, \ldots, y_n \in B\) which are generators for \(B\) as a \(C\)-module. Choose \(x_i \in A\) mapping to \(y_i\). Then \(1, x_1, \ldots, x_n\) are generators for \(A\) as a \(D\)-module. The map \(D \to A \times C\) is injective, and the ring \(A \times C\) is finite as a \(D\)-module (because it is the direct sum of the finite \(D\)-modules \(A\) and \(C\)). Hence the lemma follows from the Artin-Tate lemma (Algebra, Lemma 00IS).

Lemma

Let \(R\) be a Noetherian ring. Let \(I\) be a finite set. Suppose given a cartesian diagram \[\xymatrix{ \prod B_i & \prod A_i \ar[l]^{\prod \varphi_i} \\ Q \ar[u]^{\prod \psi_i} & P \ar[u] \ar[l] }\] with \(\psi_i\) and \(\varphi_i\) surjective, and \(Q\), \(A_i\), \(B_i\) of finite type over \(R\). Then \(P\) is of finite type over \(R\).

Proof

Follows from Lemma 00IT and induction on the size of \(I\). Namely, let \(I = I' \amalg \{i_0\}\). Let \(P'\) be the ring defined by the diagram of the lemma using \(I'\). Then \(P'\) is of finite type by induction hypothesis. Finally, \(P\) sits in a fibre product diagram \[\xymatrix{ B_{i_0} & A_{i_0} \ar[l] \\ P' \ar[u] & P \ar[u] \ar[l] & }\] to which the lemma applies.

Lemma

Suppose given a cartesian diagram of rings \[\xymatrix{ R & R' \ar[l]^t \\ B \ar[u]_s & B'\ar[u] \ar[l] }\] i.e., \(B' = B \times_R R'\). If \(h \in B'\) corresponds to \(g \in B\) and \(f \in R'\) such that \(s(g) = t(f)\), then the diagram \[\xymatrix{ R_{s(g)} = R_{t(f)} & (R')_f \ar[l]^-t \\ B_g \ar[u]_s & (B')_h \ar[u] \ar[l] }\] is cartesian too.

Proof

The equality \(B' = B \times_R R'\) tells us that \[0 \to B' \to B \oplus R' \xrightarrow{s, -t} R\] is an exact sequence of \(B'\)-modules. We have \(B_g = B_h\), \(R'_f = R'_h\), and \(R_{s(g)} = R_{t(f)} = R_h\) as \(B'\)-modules. By exactness of localization (Algebra, Proposition 00CS) we find that \[0 \to B'_h \to B_g \oplus R'_f \xrightarrow{s, -t} R_{s(g)} = R_{t(f)}\] is an exact sequence. This proves the lemma.

Consider a commutative diagram of rings \[\xymatrix{ R & R' \ar[l] \\ B \ar[u] & B' \ar[u] \ar[l] }\] Consider the functor (where the fibre product of categories is as constructed in Categories, Example 003R) [0D2E]\[\begin{equation} \text{Mod}_{B'} \longrightarrow \text{Mod}_B \times_{\text{Mod}_R} \text{Mod}_{R'},\quad L' \longmapsto (L' \otimes_{B'} B, L' \otimes_{B'} R', can) \end{equation}\] where \(can\) is the canonical identification \(L' \otimes_{B'} B \otimes_B R = L' \otimes_{B'} R' \otimes_{R'} R\). In the following we will write \((N, M', \varphi)\) for an object of the right hand side, i.e., \(N\) is a \(B\)-module, \(M'\) is an \(R'\)-module and \(\varphi : N \otimes_B R \to M' \otimes_{R'} R\) is an isomorphism.

Lemma

Given a commutative diagram of rings \[\xymatrix{ R & R' \ar[l] \\ B \ar[u] & B' \ar[u] \ar[l] }\] the functor (0D2E) has a right adjoint, namely the functor \[F : (N, M', \varphi) \longmapsto N \times_\varphi M'\] (see proof for elucidation).

Proof

Given an object \((N, M', \varphi)\) of the category \(\text{Mod}_B \times_{\text{Mod}_R} \text{Mod}_{R'}\) we set \[N \times_\varphi M' = \{(n, m') \in N \times M' \mid \varphi(n \otimes 1) = m' \otimes 1\text{ in }M' \otimes_{R'} R\}\] viewed as a \(B'\)-module. The adjointness statement is that for a \(B'\)-module \(L'\) and a triple \((N, M', \varphi)\) we have \[\Hom_{B'}(L', N \times_\varphi M') = \Hom_B(L' \otimes_{B'} B, N) \times_{\Hom_R(L' \otimes_{B'} R, M' \otimes_{R'} R)} \Hom_{R'}(L' \otimes_{B'} R', M')\] By Algebra, Lemma 05DQ the right hand side is equal to \[\Hom_{B'}(L', N) \times_{\Hom_{B'}(L', M' \otimes_{R'} R)} \Hom_{B'}(L', M')\] Thus it is clear that for a pair \((g, f')\) of elements of this fibre product we get an \(B'\)-linear map \(L' \to N \times_\varphi M'\), \(l' \mapsto (g(l'), f'(l'))\). Conversely, given a \(B'\) linear map \(g' : L' \to N \times_\varphi M'\) we can set \(g\) equal to the composition \(L' \to N \times_\varphi M' \to N\) and \(f'\) equal to the composition \(L' \to N \times_\varphi M' \to M'\). These constructions are mutually inverse to each other and define the desired isomorphism.

Fibre products of rings, II

In this section we discuss fibre products in the following situation.

Situation

In the following we will consider ring maps \[\xymatrix{ B \ar[r] & A & A' \ar[l] }\] where we assume \(A' \to A\) is surjective with kernel \(I\). In this situation we set \(B' = B \times_A A'\) to obtain a cartesian square \[\xymatrix{ A & A' \ar[l] \\ B \ar[u] & B' \ar[l] \ar[u] }\]

Lemma

In Situation 08KH we have \[\Spec(B') = \Spec(B) \amalg_{\Spec(A)} \Spec(A')\] as topological spaces.

Proof

Since \(B' = B \times_A A'\) we obtain a commutative square of spectra, which induces a continuous map \[can : \Spec(B) \amalg_{\Spec(A)} \Spec(A') \longrightarrow \Spec(B')\] as the source is a pushout in the category of topological spaces (which exists by Topology, Section 0B1W).

To show the map \(can\) is surjective, let \(\mathfrak q' \subset B'\) be a prime ideal. If \(I \subset \mathfrak q'\) (here and below we take the liberty of considering \(I\) as an ideal of \(B'\) as well as an ideal of \(A'\)), then \(\mathfrak q'\) corresponds to a prime ideal of \(B\) and is in the image. If not, then pick \(h \in I\), \(h \not \in \mathfrak q'\). In this case \(B_h = A_h = 0\) and the ring map \(B'_h \to A'_h\) is an isomorphism, see Lemma 01Z8. Thus we see that \(\mathfrak q'\) corresponds to a unique prime ideal \(\mathfrak p' \subset A'\) which does not contain \(I\).

Since \(B' \to B\) is surjective, we see that \(can\) is injective on the summand \(\Spec(B)\). We have seen above that \(\Spec(A') \to \Spec(B')\) is injective on the complement of \(V(I) \subset \Spec(A')\). Since \(V(I) \subset \Spec(A')\) is exactly the image of \(\Spec(A) \to \Spec(A')\) a trivial set theoretic argument shows that \(can\) is injective.

To finish the proof we have to show that \(can\) is open. To do this, observe that an open of the pushout is of the form \(V \amalg U'\) where \(V \subset \Spec(B)\) and \(U' \subset \Spec(A')\) are opens whose inverse images in \(\Spec(A)\) agree. Let \(v \in V\). We can find a \(g \in B\) such that \(v \in D(g) \subset V\). Let \(f \in A\) be the image. Pick \(f' \in A'\) mapping to \(f\). Then \(D(f') \cap U' \cap V(I) = D(f') \cap V(I)\). Hence \(V(I) \cap D(f')\) and \(D(f') \cap (U')^c\) are disjoint closed subsets of \(D(f') = \Spec(A'_{f'})\). Write \((U')^c = V(J)\) for some ideal \(J \subset A'\). Since \(A'_{f'} \to A'_{f'}/IA'_{f'} \times A'_{f'}/JA'_{f'}\) is surjective by the disjointness just shown, we can find an \(a'' \in A'_{f'}\) mapping to \(1\) in \(A'_{f'}/IA'_{f'}\) and mapping to zero in \(A'_{f'}/JA'_{f'}\). Clearing denominators, we find an element \(a' \in J\) mapping to \(f^n\) in \(A\). Then \(D(a'f') \subset U'\). Let \(h' = (g^{n + 1}, a'f') \in B'\). Since \(B'_{h'} = B_{g^{n + 1}} \times_{A_{f^{n + 1}}} A'_{a'f'}\) by a previously cited lemma, we see that \(D(h')\) pulls back to an open neighbourhood of \(v\) in the pushout, i.e., the image of \(V \amalg U'\) contains an open neighbourhood of the image of \(v\). We omit the (easier) proof that the same thing is true for \(u' \in U'\) with \(u' \not \in V(I)\).

Lemma

In Situation 08KH if \(B \to A\) is integral, then \(B' \to A'\) is integral.

Proof

Let \(a' \in A'\) with image \(a \in A\). Let \(x^d + b_1 x^{d - 1} + \ldots + b_d\) be a monical polynomial with coefficients in \(B\) satisfied by \(a\). Choose \(b'_i \in B'\) mapping to \(b_i \in B\) (possible). Then \((a')^d + b'_1 (a')^{d - 1} + \ldots + b'_d\) is in the kernel of \(A' \to A\). Since \(\Ker(B' \to B) = \Ker(A' \to A)\) we can modify our choice of \(b'_d\) to get \((a')^d + b'_1 (a')^{d - 1} + \ldots + b'_d = 0\) as desired.

In Situation 08KH we’d like to understand \(B'\)-modules in terms of modules over \(A'\), \(A\), and \(B\). In order to do this we consider the functor (where the fibre product of categories as constructed in Categories, Example 003R) [08KI]\[\begin{equation} \text{Mod}_{B'} \longrightarrow \text{Mod}_B \times_{\text{Mod}_A} \text{Mod}_{A'},\quad L' \longmapsto (L' \otimes_{B'} B, L' \otimes_{B'} A', can) \end{equation}\] where \(can\) is the canonical identification \(L' \otimes_{B'} B \otimes_B A = L' \otimes_{B'} A' \otimes_{A'} A\). In the following we will write \((N, M', \varphi)\) for an object of the right hand side, i.e., \(N\) is a \(B\)-module, \(M'\) is an \(A'\)-module and \(\varphi : N \otimes_B A \to M' \otimes_{A'} A\) is an isomorphism. However, it is often more convenient think of \(\varphi\) as a \(B\)-linear map \(\varphi : N \to M'/IM'\) which induces an isomorphism \(N \otimes_B A \to M' \otimes_{A'} A = M'/IM'\).

Lemma

In Situation 08KH the functor (08KI) has a right adjoint, namely the functor \[F : (N, M', \varphi) \longmapsto N \times_{\varphi, M} M'\] where \(M = M'/IM'\). Moreover, the composition of \(F\) with (08KI) is the identity functor on \(\text{Mod}_B \times_{\text{Mod}_A} \text{Mod}_{A'}\). In other words, setting \(N' = N \times_{\varphi, M} M'\) we have \(N' \otimes_{B'} B = N\) and \(N' \otimes_{B'} A' = M'\).

Proof

The adjointness statement follows from the more general Lemma 0D2F. To prove the final assertion, recall that \(B' = B \times_A A'\) and \(N' = N \times_{\varphi, M} M'\) and extend these equalities to \[\vcenter{ \xymatrix{ A & A' \ar[l] & I \ar[l] \\ B \ar[u] & B' \ar[l] \ar[u] & J \ar[l] \ar[u] } } \quad\text{and}\quad \vcenter{ \xymatrix{ M & M' \ar[l] & K \ar[l] \\ N \ar[u]_\varphi & N' \ar[l] \ar[u] & L \ar[l] \ar[u] } }\] where \(I, J, K, L\) are the kernels of the horizontal maps of the original diagrams. We present the proof as a sequence of observations:

1. \(K = IM'\) (see statement lemma),

2. \(B' \to B\) is surjective with kernel \(J\) and \(J \to I\) is bijective,

3. \(N' \to N\) is surjective with kernel \(L\) and \(L \to K\) is bijective,

4. \(JN' \subset L\),

5. \(\Im(N \to M)\) generates \(M\) as an \(A\)-module (because \(N \otimes_B A = M\)),

6. \(\Im(N' \to M')\) generates \(M'\) as an \(A'\)-module (because it holds modulo \(K\) and \(L\) maps isomorphically to \(K\)),

7. \(JN' = L\) (because \(L \cong K = I M'\) is generated by images of elements \(x n'\) with \(x \in I\) and \(n' \in N'\) by the previous statement),

8. \(N' \otimes_{B'} B = N\) (because \(N = N'/L\), \(B = B'/J\), and the previous statement),

9. there is a map \(\gamma : N' \otimes_{B'} A' \to M'\),

10. \(\gamma\) is surjective (see above),

11. the kernel of the composition \(N' \otimes_{B'} A' \to M' \to M\) is generated by elements \(l \otimes 1\) and \(n' \otimes x\) with \(l \in K\), \(n' \in N'\), \(x \in I\) (because \(M = N \otimes_B A\) by assumption and because \(N' \to N\) and \(A' \to A\) are surjective with kernels \(L\) and \(I\)),

12. any element of \(N' \otimes_{B'} A'\) in the submodule generated by the elements \(l \otimes 1\) and \(n' \otimes x\) with \(l \in L\), \(n' \in N'\), \(x \in I\) can be written as \(l \otimes 1\) for some \(l \in L\) (because \(J\) maps isomorphically to \(I\) we see that \(n' \otimes x = n'x \otimes 1\) in \(N' \otimes_{B'} A'\); similarly \(x n' \otimes a' = n' \otimes xa' = n'(xa') \otimes 1\) in \(N' \otimes_{B'} A'\) when \(n' \in N'\), \(x \in J\) and \(a' \in A'\); since we have seen that \(JN' = L\) this proves the assertion),

13. the kernel of \(\gamma\) is zero (because by (10) and (11) any element of the kernel is of the form \(l \otimes 1\) with \(l \in L\) which is mapped to \(l \in K \subset M'\) by \(\gamma\)).

This finishes the proof.

Lemma

In the situation of Lemma 07RU for a \(B'\)-module \(L'\) the adjunction map \[L' \longrightarrow (L' \otimes_{B'} B) \times_{(L' \otimes_{B'} A)} (L' \otimes_{B'} A')\] is surjective but in general not injective.

Proof

As in the proof of Lemma 07RU let \(J \subset B'\) be the kernel of the map \(B' \to B\). Then \(L' \otimes_{B'} B = L'/JL'\). Hence to prove surjectivity it suffices to show that elements of the form \((0, z)\) of the fibre product are in the image of the map of the lemma. The kernel of the map \(L' \otimes_{B'} A' \to L' \otimes_{B'} A\) is the image of \(L' \otimes_{B'} I \to L' \otimes_{B'} A'\). Since the map \(J \to I\) induced by \(B' \to A'\) is an isomorphism the composition \[L' \otimes_{B'} J \to L' \to (L' \otimes_{B'} B) \times_{(L' \otimes_{B'} A)} (L' \otimes_{B'} A')\] induces a surjection of \(L' \otimes_{B'} J\) onto the set of elements of the form \((0, z)\). To see the map is not injective in general we present a simple example. Namely, take a field \(k\), set \(B' = k[x, y]/(xy)\), \(A' = B'/(x)\), \(B = B'/(y)\), \(A = B'/(x, y)\) and \(L' = B'/(x - y)\). In that case the class of \(x\) in \(L'\) is nonzero but is mapped to zero under the displayed arrow.

Lemma

In Situation 08KH let \((N_1, M'_1, \varphi_1) \to (N_2, M'_2, \varphi_2)\) be a morphism of \(\text{Mod}_B \times_{\text{Mod}_A} \text{Mod}_{A'}\) with \(N_1 \to N_2\) and \(M'_1 \to M'_2\) surjective. Then \[N_1 \times_{\varphi_1, M_1} M'_1 \to N_2 \times_{\varphi_2, M_2} M'_2\] where \(M_1 = M'_1/IM'_1\) and \(M_2 = M'_2/IM'_2\) is surjective.

Proof

Pick \((x_2, y_2) \in N_2 \times_{\varphi_2, M_2} M'_2\). Choose \(x_1 \in N_1\) mapping to \(x_2\). Since \(M'_1 \to M_1\) is surjective we can find \(y_1 \in M'_1\) mapping to \(\varphi_1(x_1)\). Then \((x_1, y_1)\) maps to \((x_2, y'_2)\) in \(N_2 \times_{\varphi_2, M_2} M'_2\). Thus it suffices to show that elements of the form \((0, y_2)\) are in the image of the map. Here we see that \(y_2 \in IM'_2\). Write \(y_2 = \sum t_i y_{2, i}\) with \(t_i \in I\). Choose \(y_{1, i} \in M'_1\) mapping to \(y_{2, i}\). Then \(y_1 = \sum t_iy_{1, i} \in IM'_1\) and the element \((0, y_1)\) does the job.

Lemma

Let \(A, A', B, B', I, M, M', N, \varphi\) be as in Lemma 07RU. If \(N\) finite over \(B\) and \(M'\) finite over \(A'\), then \(N' = N \times_{\varphi, M} M'\) is finite over \(B'\).

Proof

We will use the results of Lemma 07RU without further mention. Choose generators \(y_1, \ldots, y_r\) of \(N\) over \(B\) and generators \(x_1, \ldots, x_s\) of \(M'\) over \(A'\). Using that \(N = N' \otimes_{B'} B\) and \(B' \to B\) is surjective we can find \(u_1, \ldots, u_r \in N'\) mapping to \(y_1, \ldots, y_r\) in \(N\). Using that \(M' = N' \otimes_{B'} A'\) we can find \(v_1, \ldots, v_t \in N'\) such that \(x_i = \sum v_j \otimes a'_{ij}\) for some \(a'_{ij} \in A'\). In particular we see that the images \(\overline{v}_j \in M'\) of the \(v_j\) generate \(M'\) over \(A'\). We claim that \(u_1, \ldots, u_r, v_1, \ldots, v_t\) generate \(N'\) as a \(B'\)-module. Namely, pick \(\xi \in N'\). We first choose \(b'_1, \ldots, b'_r \in B'\) such that \(\xi\) and \(\sum b'_i u_i\) map to the same element of \(N\). This is possible because \(B' \to B\) is surjective and \(y_1, \ldots, y_r\) generate \(N\) over \(B\). The difference \(\xi - \sum b'_i u_i\) is of the form \((0, \theta)\) for some \(\theta\) in \(IM'\). Say \(\theta\) is \(\sum t_j\overline{v}_j\) with \(t_j \in I\). As \(J = \Ker(B' \to B)\) maps isomorphically to \(I\) we can choose \(s_j \in J \subset B'\) mapping to \(t_j\). Because \(N' = N \times_{\varphi, M} M'\) it follows that \(\xi = \sum b'_i u_i + \sum s_j v_j\) as desired.

Lemma

With \(A, A', B, B', I\) as in Situation 08KH.

1. Let \((N, M', \varphi)\) be an object of \(\text{Mod}_B \times_{\text{Mod}_A} \text{Mod}_{A'}\). If \(M'\) is flat over \(A'\) and \(N\) is flat over \(B\), then \(N' = N \times_{\varphi, M} M'\) is flat over \(B'\).

2. If \(L'\) is a flat \(B'\)-module, then \(L' = (L \otimes_{B'} B) \times_{(L \otimes_{B'} A)} (L \otimes_{B'} A')\).

3. The category of flat \(B'\)-modules is equivalent to the full subcategory of \(\text{Mod}_B \times_{\text{Mod}_A} \text{Mod}_{A'}\) consisting of triples \((N, M', \varphi)\) with \(N\) flat over \(B\) and \(M'\) flat over \(A'\).

Proof

In the proof we will use Lemma 07RU without further mention.

Proof of (1). Set \(J = \Ker(B' \to B)\). This is an ideal of \(B'\) mapping isomorphically to \(I = \Ker(A' \to A)\). Let \(\mathfrak b' \subset B'\) be an ideal. We have to show that \(\mathfrak b' \otimes_{B'} N' \to N'\) is injective, see Algebra, Lemma 00HD. We know that \[\mathfrak b'/(\mathfrak b' \cap J) \otimes_{B'} N' = \mathfrak b'/(\mathfrak b' \cap J) \otimes_B N \to N\] is injective as \(N\) is flat over \(B\). As \(\mathfrak b' \cap J \to \mathfrak b' \to \mathfrak b'/(\mathfrak b' \cap J) \to 0\) is exact, we conclude that it suffices to show that \((\mathfrak b' \cap J) \otimes_{B'} N' \to N'\) is injective. Thus we may assume that \(\mathfrak b' \subset J\). Next, since \(J \to I\) is an isomorphism we have \[J \otimes_{B'} N' = I \otimes_{A'} A' \otimes_{B'} N' = I \otimes_{A'} M'\] which maps injectively into \(M'\) as \(M'\) is a flat \(A'\)-module. Hence \(J \otimes_{B'} N' \to N'\) is injective and we conclude that \(\text{Tor}_1^{B'}(B'/J, N') = 0\), see Algebra, Remark 00M6. Thus we may apply Algebra, Lemma 051C to \(N'\) over \(B'\) and the ideal \(J\). Going back to our ideal \(\mathfrak b' \subset J\), let \(\mathfrak b' \subset \mathfrak b'' \subset J\) be the smallest ideal whose image in \(I\) is an \(A'\)-submodule of \(I\). In other words, we have \(\mathfrak b'' = A' \mathfrak b'\) if we view \(J = I\) as \(A'\)-module. Then \(\mathfrak b''/\mathfrak b'\) is killed by \(J\) and we get a short exact sequence \[0 \to \mathfrak b' \otimes_{B'} N' \to \mathfrak b'' \otimes_{B'} N' \to \mathfrak b''/\mathfrak b' \otimes_{B'} N' \to 0\] by the vanishing of \(\text{Tor}_1^{B'}(\mathfrak b''/\mathfrak b', N')\) we get from the application of the lemma. Thus we may replace \(\mathfrak b'\) by \(\mathfrak b''\). In particular we may assume \(\mathfrak b'\) is an \(A'\)-module and maps to an ideal of \(A'\). Then \[\mathfrak b' \otimes_{B'} N' = \mathfrak b' \otimes_{A'} A' \otimes_{B'} N' = \mathfrak b' \otimes_{A'} M'\] This tensor product maps injectively into \(M'\) by our assumption that \(M'\) is flat over \(A'\). We conclude that \(\mathfrak b' \otimes_{B'} N' \to N' \to M'\) is injective and hence the first map is injective as desired.

Proof of (2). This follows by tensoring the short exact sequence \(0 \to B' \to B \oplus A' \to A \to 0\) with \(L'\) over \(B'\).

Proof of (3). Immediate consequence of (1) and (2).

Lemma

Let \(A, A', B, B', I\) be as in Situation 08KH. The category of finite projective \(B'\)-modules is equivalent to the full subcategory of \(\text{Mod}_B \times_{\text{Mod}_A} \text{Mod}_{A'}\) consisting of triples \((N, M', \varphi)\) with \(N\) finite projective over \(B\) and \(M'\) finite projective over \(A'\).

Proof

Recall that a module is finite projective if and only if it is finitely presented and flat, see Algebra, Lemma 00NX. Using Lemmas 0D2I and 0D2H we reduce to showing that \(N' = N \times_{\varphi, M} M'\) is a \(B'\)-module of finite presentation if \(N\) finite projective over \(B\) and \(M'\) finite projective over \(A'\).

By Lemma 0D2H the module \(N'\) is finite over \(B'\). Choose a surjection \((B')^{\oplus n} \to N'\) with kernel \(K'\). By base change we obtain maps \(B^{\oplus n} \to N\), \((A')^{\oplus n} \to M'\), and \(A^{\oplus n} \to M\) with kernels \(K_B\), \(K_{A'}\), and \(K_A\). There is a canonical map \[K' \longrightarrow K_B \times_{K_A} K_{A'}\] On the other hand, since \(N' = N \times_{\varphi, M} M'\) and \(B' = B \times_A A'\) there is also a canonical map \(K_B \times_{K_A} K_{A'} \to K'\) inverse to the displayed arrow. Hence the displayed map is an isomorphism. By Algebra, Lemma 0519 the modules \(K_B\) and \(K_{A'}\) are finite. We conclude from Lemma 0D2H that \(K'\) is a finite \(B'\)-module provided that \(K_B \to K_A\) and \(K_{A'} \to K_A\) induce isomorphisms \(K_B \otimes_B A = K_A = K_{A'} \otimes_{A'} A\). This is true because the flatness assumptions implies the sequences \[0 \to K_B \to B^{\oplus n} \to N \to 0 \quad\text{and}\quad 0 \to K_{A'} \to (A')^{\oplus n} \to M' \to 0\] stay exact upon tensoring, see Algebra, Lemma 00HL.

Fibre products of rings, III

In this section we discuss fibre products in the following situation.

Situation

Let \(A, A', B, B', I\) be as in Situation 08KH. Let \(B' \to D'\) be a ring map. Set \(D = D' \otimes_{B'} B\), \(C' = D' \otimes_{B'} A'\), and \(C = D' \otimes_{B'} A\). This leads to a big commutative diagram \[\xymatrix{ C & & & C' \ar[lll] \\ & A \ar[ul] & A' \ar[l] \ar[ru] \\ & B \ar[u] \ar[ld] & B' \ar[l] \ar[u] \ar[rd] \\ D \ar[uuu] & & & D' \ar[lll] \ar[uuu] }\] of rings. Observe that we do not assume that the map \(D' \to D \times_C C'\) is an isomorphism¹. In this situation we have the functor [08KL]\[\begin{equation} \text{Mod}_{D'} \longrightarrow \text{Mod}_D \times_{\text{Mod}_C} \text{Mod}_{C'},\quad L' \longmapsto (L' \otimes_{D'} D, L' \otimes_{D'} C', can) \end{equation}\] analogous to (08KI). Note that \(L' \otimes_{D'} D = L \otimes_{D'} (D' \otimes_{B'} B) = L \otimes_{B'} B\) and similarly \(L' \otimes_{D'} C' = L \otimes_{D'} (D' \otimes_{B'} A') = L \otimes_{B'} A'\) hence the diagram \[\xymatrix{ \text{Mod}_{D'} \ar[r] \ar[d] & \text{Mod}_D \times_{\text{Mod}_C} \text{Mod}_{C'} \ar[d] \\ \text{Mod}_{B'} \ar[r] & \text{Mod}_B \times_{\text{Mod}_A} \text{Mod}_{A'} }\] is commutative. In the following we will write \((N, M', \varphi)\) for an object of \(\text{Mod}_D \times_{\text{Mod}_C} \text{Mod}_{C'}\), i.e., \(N\) is a \(D\)-module, \(M'\) is an \(C'\)-module and \(\varphi : N \otimes_B A \to M' \otimes_{A'} A\) is an isomorphism of \(C\)-modules. However, it is often more convenient think of \(\varphi\) as a \(D\)-linear map \(\varphi : N \to M'/IM'\) which induces an isomorphism \(N \otimes_B A \to M' \otimes_{A'} A = M'/IM'\).

Lemma

In Situation 08KK the functor (08KL) has a right adjoint, namely the functor \[F : (N, M', \varphi) \longmapsto N \times_{\varphi, M} M'\] where \(M = M'/IM'\). Moreover, the composition of \(F\) with (08KL) is the identity functor on \(\text{Mod}_D \times_{\text{Mod}_C} \text{Mod}_{C'}\). In other words, setting \(N' = N \times_{\varphi, M} M'\) we have \(N' \otimes_{D'} D = N\) and \(N' \otimes_{D'} C' = M'\).

Proof

The adjointness statement follows from the more general Lemma 0D2F. The final assertion follows from the corresponding assertion of Lemma 07RU because \(N' \otimes_{D'} D = N' \otimes_{D'} D' \otimes_{B'} B = N' \otimes_{B'} B\) and \(N' \otimes_{D'} C' = N' \otimes_{D'} D' \otimes_{B'} A' = N' \otimes_{B'} A'\).

Lemma

In Situation 08KK the map \(JD' \to IC'\) is surjective where \(J = \Ker(B' \to B)\).

Proof

Since \(C' = D' \otimes_{B'} A'\) we have that \(IC'\) is the image of \(D' \otimes_{B'} I = C' \otimes_{A'} I \to C'\). As the ring map \(B' \to A'\) induces an isomorphism \(J \to I\) the lemma follows.

Lemma

Let \(A, A', B, B', C, C', D, D', I, M', M, N, \varphi\) be as in Lemma 08KM. If \(N\) finite over \(D\) and \(M'\) finite over \(C'\), then \(N' = N \times_{\varphi, M} M'\) is finite over \(D'\).

Proof

Recall that \(D' \to D \times_C C'\) is surjective by Lemma 08IG. Observe that \(N' = N \times_{\varphi, M} M'\) is a module over \(D \times_C C'\). We can apply Lemma 0D2H to the data \(C, C', D, D', IC', M', M, N, \varphi\) to see that \(N' = N \times_{\varphi, M} M'\) is finite over \(D \times_C C'\). Thus it is finite over \(D'\).

Lemma

With \(A, A', B, B', C, C', D, D', I\) as in Situation 08KK.

1. Let \((N, M', \varphi)\) be an object of \(\text{Mod}_D \times_{\text{Mod}_C} \text{Mod}_{C'}\). If \(M'\) is flat over \(A'\) and \(N\) is flat over \(B\), then \(N' = N \times_{\varphi, M} M'\) is flat over \(B'\).

2. If \(L'\) is a \(D'\)-module flat over \(B'\), then \(L' = (L \otimes_{D'} D) \times_{(L \otimes_{D'} C)} (L \otimes_{D'} C')\).

3. The category of \(D'\)-modules flat over \(B'\) is equivalent to the categories of objects \((N, M', \varphi)\) of \(\text{Mod}_D \times_{\text{Mod}_C} \text{Mod}_{C'}\) with \(N\) flat over \(B\) and \(M'\) flat over \(A'\).

Proof

Part (1) follows from part (1) of Lemma 0D2I.

Part (2) follows from part (2) of Lemma 0D2I using that \(L' \otimes_{D'} D = L' \otimes_{B'} B\), \(L' \otimes_{D'} C' = L' \otimes_{B'} A'\), and \(L' \otimes_{D'} C = L' \otimes_{B'} A\), see discussion in Situation 08KK.

Part (3) is an immediate consequence of (1) and (2).

The following lemma is a good deal more interesting than its counter part in the absolute case (Lemma 0D2J), although the proof is essentially the same.

Lemma

Let \(A, A', B, B', C, C', D, D', I, M', M, N, \varphi\) be as in Lemma 08KM. If

1. \(N\) is finitely presented over \(D\) and flat over \(B\),

2. \(M'\) finitely presented over \(C'\) and flat over \(A'\), and

3. the ring map \(B' \to D'\) factors as \(B' \to D'' \to D'\) with \(B' \to D''\) flat and \(D'' \to D'\) of finite presentation,

then \(N' = N \times_M M'\) is finitely presented over \(D'\).

Proof

Choose a surjection \(D''' = D''[x_1, \ldots, x_n] \to D'\) with finitely generated kernel \(J\). By Algebra, Lemma 0564 it suffices to show that \(N'\) is finitely presented as a \(D'''\)-module. Moreover, \(D''' \otimes_{B'} B \to D' \otimes_{B'} B = D\) and \(D''' \otimes_{B'} A' \to D' \otimes_{B'} A' = C'\) are surjections whose kernels are generated by the image of \(J\), hence \(N\) is a finitely presented \(D''' \otimes_{B'} B\)-module and \(M'\) is a finitely presented \(D''' \otimes_{B'} A'\)-module by Algebra, Lemma 0564 again. Thus we may replace \(D'\) by \(D'''\) and \(D\) by \(D''' \otimes_{B'} B\), etc. Since \(D'''\) is flat over \(B'\), it follows that we may assume that \(B' \to D'\) is flat.

Assume \(B' \to D'\) is flat. By Lemma 08IH the module \(N'\) is finite over \(D'\). Choose a surjection \((D')^{\oplus n} \to N'\) with kernel \(K'\). By base change we obtain maps \(D^{\oplus n} \to N\), \((C')^{\oplus n} \to M'\), and \(C^{\oplus n} \to M\) with kernels \(K_D\), \(K_{C'}\), and \(K_C\). There is a canonical map \[K' \longrightarrow K_D \times_{K_C} K_{C'}\] On the other hand, since \(N' = N \times_M M'\) and \(D' = D \times_C C'\) (by Lemma 0D2I; applied to the flat \(B'\)-module \(D'\)) there is also a canonical map \(K_D \times_{K_C} K_{C'} \to K'\) inverse to the displayed arrow. Hence the displayed map is an isomorphism. By Algebra, Lemma 0519 the modules \(K_D\) and \(K_{C'}\) are finite. We conclude from Lemma 08IH that \(K'\) is a finite \(D'\)-module provided that \(K_D \to K_C\) and \(K_{C'} \to K_C\) induce isomorphisms \(K_D \otimes_B A = K_C = K_{C'} \otimes_{A'} A\). This is true because the flatness assumptions implies the sequences \[0 \to K_D \to D^{\oplus n} \to N \to 0 \quad\text{and}\quad 0 \to K_{C'} \to (C')^{\oplus n} \to M' \to 0\] stay exact upon tensoring, see Algebra, Lemma 00HL.

Lemma

Let \(A, A', B, B', I\) be as in Situation 08KH. Let \((D, C', \varphi)\) be a system consisting of an \(B\)-algebra \(D\), a \(A'\)-algebra \(C'\) and an isomorphism \(D \otimes_B A \to C'/IC' = C\). Set \(D' = D \times_C C'\) (as in Lemma 07RU). Then

1. \(B' \to D'\) is finite type if and only if \(B \to D\) and \(A' \to C'\) are finite type,

2. \(B' \to D'\) is flat if and only if \(B \to D\) and \(A' \to C'\) are flat,

3. \(B' \to D'\) is flat and of finite presentation if and only if \(B \to D\) and \(A' \to C'\) are flat and of finite presentation,

4. \(B' \to D'\) is smooth if and only if \(B \to D\) and \(A' \to C'\) are smooth,

5. \(B' \to D'\) is étale if and only if \(B \to D\) and \(A' \to C'\) are étale.

Moreover, if \(D'\) is a flat \(B'\)-algebra, then \(D' \to (D' \otimes_{B'} B) \times_{(D' \otimes_{B'} A)} (D' \otimes_{B'} A')\) is an isomorphism. In this way the category of flat \(B'\)-algebras is equivalent to the categories of systems \((D, C', \varphi)\) as above with \(D\) flat over \(B\) and \(C'\) flat over \(A'\).

Proof

The implication “\(\Rightarrow\)” follows from Algebra, Lemmas 05G5, 00HI, 00T4, and 00U2 because we have \(D' \otimes_{B'} B = D\) and \(D' \otimes_{B'} A' = C'\) by Lemma 07RU. Thus it suffices to prove the implications in the other direction.

Ad (1). Assume \(D\) of finite type over \(B\) and \(C'\) of finite type over \(A'\). We will use the results of Lemma 07RU without further mention. Choose generators \(x_1, \ldots, x_r\) of \(D\) over \(B\) and generators \(y_1, \ldots, y_s\) of \(C'\) over \(A'\). Using that \(D = D' \otimes_{B'} B\) and \(B' \to B\) is surjective we can find \(u_1, \ldots, u_r \in D'\) mapping to \(x_1, \ldots, x_r\) in \(D\). Using that \(C' = D' \otimes_{B'} A'\) we can find \(v_1, \ldots, v_t \in D'\) such that \(y_i = \sum v_j \otimes a'_{ij}\) for some \(a'_{ij} \in A'\). In particular, the images of \(v_j\) in \(C'\) generate \(C'\) as an \(A'\)-algebra. Set \(N = r + t\) and consider the cube of rings \[\xymatrix{ A[x_1, \ldots, x_N] & & A'[x_1, \ldots, x_N] \ar[ll] \\ & A \ar[lu] & & A' \ar[ll] \ar[lu] \\ B[x_1, \ldots, x_N] \ar[uu] & & B'[x_1, \ldots, x_N] \ar[uu] \ar[ll] \\ & B \ar[uu] \ar[lu] & & B' \ar[ll] \ar[uu] \ar[lu] }\] Observe that the back square is cartesian as well. Consider the ring map \[B'[x_1, \ldots, x_N] \to D',\quad x_i \mapsto u_i \quad\text{and}\quad x_{r + j} \mapsto v_j.\] Then we see that the induced maps \(B[x_1, \ldots, x_N] \to D\) and \(A'[x_1, \ldots, x_N] \to C'\) are surjective, in particular finite. We conclude from Lemma 08IH that \(B'[x_1, \ldots, x_N] \to D'\) is finite, which implies that \(D'\) is of finite type over \(B'\) for example by Algebra, Lemma 00F4.

Ad (2). The implication “\(\Leftarrow\)” follows from Lemma 07RW. Moreover, the final statement follows from the final statement of Lemma 07RW.

Ad (3). Assume \(B \to D\) and \(A' \to C'\) are flat and of finite presentation. The flatness of \(B' \to D'\) we’ve seen in (2). We know \(B' \to D'\) is of finite type by (1). Choose a surjection \(B'[x_1, \ldots, x_N] \to D'\). By Algebra, Lemma 00R2 the ring \(D\) is of finite presentation as a \(B[x_1, \ldots, x_N]\)-module and the ring \(C'\) is of finite presentation as a \(A'[x_1, \ldots, x_N]\)-module. By Lemma 08KP we see that \(D'\) is of finite presentation as a \(B'[x_1, \ldots, x_N]\)-module, i.e., \(B' \to D'\) is of finite presentation.

Ad (4). Assume \(B \to D\) and \(A' \to C'\) smooth. By (3) we see that \(B' \to D'\) is flat and of finite presentation. By Algebra, Lemma 00TF it suffices to check that \(D' \otimes_{B'} k\) is smooth for any field \(k\) over \(B'\). If the composition \(J \to B' \to k\) is zero, then \(B' \to k\) factors as \(B' \to B \to k\) and we see that \[D' \otimes_{B'} k = D' \otimes_{B'} B \otimes_B k = D \otimes_B k\] is smooth as \(B \to D\) is smooth. If the composition \(J \to B' \to k\) is nonzero, then there exists an \(h \in J\) which does not map to zero in \(k\). Then \(B' \to k\) factors as \(B' \to B'_h \to k\). Observe that \(h\) maps to zero in \(B\), hence \(B_h = 0\). Thus by Lemma 01Z8 we have \(B'_h = A'_h\) and we get \[D' \otimes_{B'} k = D' \otimes_{B'} B'_h \otimes_{B'_h} k = C'_h \otimes_{A'_h} k\] is smooth as \(A' \to C'\) is smooth.

Ad (5). Assume \(B \to D\) and \(A' \to C'\) are étale. By (4) we see that \(B' \to D'\) is smooth. As we can read off whether or not a smooth map is étale from the dimension of fibres we see that (5) holds (argue as in the proof of (4) to identify fibres – some details omitted).

Remark

In Situation 08KK. Assume \(B' \to D'\) is of finite presentation and suppose we are given a \(D'\)-module \(L'\). We claim there is a bijective correspondence between

1. surjections of \(D'\)-modules \(L' \to Q'\) with \(Q'\) of finite presentation over \(D'\) and flat over \(B'\), and

2. pairs of surjections of modules \((L' \otimes_{D'} D \to Q_1, L' \otimes_{D'} C' \to Q_2)\) with

   1. \(Q_1\) of finite presentation over \(D\) and flat over \(B\),

   2. \(Q_2\) of finite presentation over \(C'\) and flat over \(A'\),

   3. \(Q_1 \otimes_D C = Q_2 \otimes_{C'} C\) as quotients of \(L' \otimes_{D'} C\).

The correspondence between these is given by \(Q \mapsto (Q_1, Q_2)\) with \(Q_1 = Q \otimes_{D'} D\) and \(Q_2 = Q \otimes_{D'} C'\). And for the converse we use \(Q = Q_1 \times_{Q_{12}} Q_2\) where \(Q_{12}\) the common quotient \(Q_1 \otimes_D C = Q_2 \otimes_{C'} C\) of \(L' \otimes_{D'} C\). As quotient map we use \[L' \longrightarrow (L' \otimes_{D'} D) \times_{(L' \otimes_{D'} C)} (L' \otimes_{D'} C') \longrightarrow Q_1 \times_{Q_{12}} Q_2 = Q\] where the first arrow is surjective by Lemma 08IG and the second by Lemma 08KJ. The claim follows by Lemmas 07RW and 08KP.

Fitting ideals

The Fitting ideals of a finite module are the ideals determined by the construction of Lemma 07Z8.

Lemma

Let \(R\) be a ring. Let \(A\) be an \(n \times m\) matrix with coefficients in \(R\). Let \(I_r(A)\) be the ideal generated by the \(r \times r\)-minors of \(A\) with the convention that \(I_0(A) = R\) and \(I_r(A) = 0\) if \(r > \min(n, m)\). Then

1. \(I_0(A) \supset I_1(A) \supset I_2(A) \supset \ldots\),

2. if \(B\) is an \((n + n') \times m\) matrix, and \(A\) is the first \(n\) rows of \(B\), then \(I_{r + n'}(B) \subset I_r(A)\),

3. if \(C\) is an \(n \times n\) matrix then \(I_r(CA) \subset I_r(A)\).

4. If \(A\) is a block matrix \[\left( \begin{matrix} A_1 & 0 \\ 0 & A_2 \end{matrix} \right)\] then \(I_r(A) = \sum_{r_1 + r_2 = r} I_{r_1}(A_1) I_{r_2}(A_2)\).

5. Add more here.

Proof

Omitted. (Hint: Use that a determinant can be computed by expanding along a column or a row.)

Lemma

Let \(R\) be a ring. Let \(M\) be a finite \(R\)-module. Choose a presentation \[\bigoplus\nolimits_{j \in J} R \longrightarrow R^{\oplus n} \longrightarrow M \longrightarrow 0.\] of \(M\). Let \(A = (a_{ij})_{i = 1, \ldots, n, j \in J}\) be the matrix of the map \(\bigoplus_{j \in J} R \to R^{\oplus n}\). The ideal \(\text{Fit}_k(M)\) generated by the \((n - k) \times (n - k)\) minors of \(A\) is independent of the choice of the presentation.

Proof

Let \(K \subset R^{\oplus n}\) be the kernel of the surjection \(R^{\oplus n} \to M\). Pick \(z_1, \ldots, z_{n - k} \in K\) and write \(z_j = (z_{1j}, \ldots, z_{nj})\). Another description of the ideal \(\text{Fit}_k(M)\) is that it is the ideal generated by the \((n - k) \times (n - k)\) minors of all the matrices \((z_{ij})\) we obtain in this way.

Suppose we change the surjection into the surjection \(R^{\oplus n + n'} \to M\) with kernel \(K'\) where we use the original map on the first \(n\) standard basis elements of \(R^{\oplus n + n'}\) and \(0\) on the last \(n'\) basis vectors. Then the corresponding ideals are the same. Namely, if \(z_1, \ldots, z_{n - k} \in K\) as above, let \(z'_j = (z_{1j}, \ldots, z_{nj}, 0, \ldots, 0) \in K'\) for \(j = 1, \ldots, n - k\) and \(z'_{n + j'} = (0, \ldots, 0, 1, 0, \ldots, 0) \in K'\). Then we see that the ideal of \((n - k) \times (n - k)\) minors of \((z_{ij})\) agrees with the ideal of \((n + n' - k) \times (n + n' - k)\) minors of \((z'_{ij})\). This gives one of the inclusions. Conversely, given \(z'_1, \ldots, z'_{n + n' - k}\) in \(K'\) we can project these to \(R^{\oplus n}\) to get \(z_1, \ldots, z_{n + n' - k}\) in \(K\). By Lemma 07Z7 we see that the ideal generated by the \((n + n' - k) \times (n + n' - k)\) minors of \((z'_{ij})\) is contained in the ideal generated by the \((n - k) \times (n - k)\) minors of \((z_{ij})\). This gives the other inclusion.

Let \(R^{\oplus m} \to M\) be another surjection with kernel \(L\). By Schanuel’s lemma (Algebra, Lemma 00O3) and the results of the previous paragraph, we may assume \(m = n\) and that there is an isomorphism \(R^{\oplus n} \to R^{\oplus m}\) commuting with the surjections to \(M\). Let \(C = (c_{li})\) be the (invertible) matrix of this map (it is a square matrix as \(n = m\)). Then given \(z'_1, \ldots, z'_{n - k} \in L\) as above we can find \(z_1, \ldots, z_{n - k} \in K\) with \(z_1' = Cz_1, \ldots, z'_{n - k} = Cz_{n - k}\). By Lemma 07Z7 we get one of the inclusions. By symmetry we get the other.

Definition

Let \(R\) be a ring. Let \(M\) be a finite \(R\)-module. Let \(k \geq 0\). The \(k\)th Fitting ideal of \(M\) is the ideal \(\text{Fit}_k(M)\) constructed in Lemma 07Z8. Set \(\text{Fit}_{-1}(M) = 0\).

Since the Fitting ideals are the ideals of minors of a big matrix (numbered in reverse ordering from the ordering in Lemma 07Z7) we see that \[0 = \text{Fit}_{-1}(M) \subset \text{Fit}_0(M) \subset \text{Fit}_1(M) \subset \ldots \subset \text{Fit}_t(M) = R\] for some \(t \gg 0\). Here are some basic properties of Fitting ideals.

Lemma

Let \(R\) be a ring. Let \(M\) be a finite \(R\)-module.

1. If \(M\) can be generated by \(n\) elements, then \(\text{Fit}_n(M) = R\).

2. Given a second finite \(R\)-module \(M'\) we have \(\text{Fit}_0(M \oplus M') = \text{Fit}_0(M)\text{Fit}_0(M')\) and more generally \[\text{Fit}_l(M \oplus M') = \sum\nolimits_{k + k' = l} \text{Fit}_k(M)\text{Fit}_{k'}(M')\]

3. If \(R \to R'\) is a ring map, then \(\text{Fit}_k(M \otimes_R R')\) is the ideal of \(R'\) generated by the image of \(\text{Fit}_k(M)\).

4. If \(M\) is of finite presentation, then \(\text{Fit}_k(M)\) is a finitely generated ideal.

5. If \(M \to M'\) is a surjection, then \(\text{Fit}_k(M) \subset \text{Fit}_k(M')\).

6. We have \(\text{Fit}_0(M) \subset \text{Ann}_R(M)\).

7. We have \(V(\text{Fit}_0(M)) = \text{Supp}(M)\).

8. If \(I\) is an ideal of \(R\), then \(\text{Fit}_0(R/I) = I\).

9. Add more here.

Proof

Part (1) follows from the fact that \(I_0(A) = R\) in Lemma 07Z7.

Part (2) follows form the corresponding statement in Lemma 07Z7.

Part (3) follows from the fact that \(\otimes_R R'\) is right exact, so the base change of a presentation of \(M\) is a presentation of \(M \otimes_R R'\).

Proof of (4). Let \(R^{\oplus m} \xrightarrow{A} R^{\oplus n} \to M \to 0\) be a presentation. Then \(\text{Fit}_k(M)\) is the ideal generated by the \((n - k) \times (n - k)\) minors of the matrix \(A\).

Part (5) is immediate from the definition.

Proof of (6). Choose a presentation of \(M\) with matrix \(A\) as in Lemma 07Z8. Let \(J' \subset J\) be a subset of cardinality \(n\). It suffices to show that \(f = \det(a_{ij})_{i = 1, \ldots, n, j \in J'}\) annihilates \(M\). This is clear because the cokernel of \[R^{\oplus n} \xrightarrow{A' = (a_{ij})_{i = 1, \ldots, n, j \in J'}} R^{\oplus n} \to M \to 0\] is killed by \(f\) as there is a matrix \(B\) with \(A' B = f1_{n \times n}\).

Proof of (7). Choose a presentation of \(M\) with matrix \(A\) as in Lemma 07Z8. By Nakayama’s lemma (Algebra, Lemma 00DV) we have \[M_\mathfrak p \not = 0 \Leftrightarrow M \otimes_R \kappa(\mathfrak p) \not = 0 \Leftrightarrow \text{rank}(\text{image }A\text{ in }\kappa(\mathfrak p)) < n\] Clearly \(\text{Fit}_0(M)\) exactly cuts out the set of primes with this property.

Example

Let \(R\) be a ring. The Fitting ideals of the finite free module \(M = R^{\oplus n}\) are \(\text{Fit}_k(M) = 0\) for \(k < n\) and \(\text{Fit}_k(M) = R\) for \(k \geq n\).

Example

Let \(R\) be a ring and let \(I, J\subset R\) be ideals. Then \(\text{Fit}_0(R/I) = I\), \(\text{Fit}_0(R/I \oplus R/J) = IJ\), \(\text{Fit}_1(R/I\oplus R/J) = I+J\).

Lemma

Let \(R\) be a ring. Let \(M\) be a finite \(R\)-module. Let \(k \geq 0\). Let \(\mathfrak p \subset R\) be a prime ideal. The following are equivalent

1. \(\text{Fit}_k(M) \not \subset \mathfrak p\),

2. \(\dim_{\kappa(\mathfrak p)} M \otimes_R \kappa(\mathfrak p) \leq k\),

3. \(M_\mathfrak p\) can be generated by \(k\) elements over \(R_\mathfrak p\), and

4. \(M_f\) can be generated by \(k\) elements over \(R_f\) for some \(f \in R\), \(f \not \in \mathfrak p\).

Proof

By Nakayama’s lemma (Algebra, Lemma 00DV) we see that \(M_f\) can be generated by \(k\) elements over \(R_f\) for some \(f \in R\), \(f \not \in \mathfrak p\) if \(M \otimes_R \kappa(\mathfrak p)\) can be generated by \(k\) elements. Hence (2), (3), and (4) are equivalent. Using Lemma 07ZA part (3) this reduces the problem to the case where \(R\) is a field and \(\mathfrak p = (0)\). In this case the result follows from Example 07ZB.

Lemma

Let \(R\) be a ring. Let \(M\) be a finite \(R\)-module. Let \(r \geq 0\). The following are equivalent

1. \(M\) is finite locally free of rank \(r\) (Algebra, Definition 00NW),

2. \(\text{Fit}_{r - 1}(M) = 0\) and \(\text{Fit}_r(M) = R\), and

3. \(\text{Fit}_k(M) = 0\) for \(k < r\) and \(\text{Fit}_k(M) = R\) for \(k \geq r\).

Proof

It is immediate that (2) is equivalent to (3) because the Fitting ideals form an increasing sequence of ideals. Since the formation of \(\text{Fit}_k(M)\) commutes with base change (Lemma 07ZA) we see that (1) implies (2) by Example 07ZB and glueing results (Algebra, Section 00EN). Conversely, assume (2). By Lemma 07ZC we may assume that \(M\) is generated by \(r\) elements. Thus a presentation \(\bigoplus_{j \in J} R \to R^{\oplus r} \to M \to 0\). But now the assumption that \(\text{Fit}_{r - 1}(M) = 0\) implies that all entries of the matrix of the map \(\bigoplus_{j \in J} R \to R^{\oplus r}\) are zero. Thus \(M\) is free.

Lemma

Let \(R\) be a local ring. Let \(M\) be a finite \(R\)-module. Let \(k \geq 0\). Assume that \(\text{Fit}_k(M) = (f)\) for some \(f \in R\). Let \(M'\) be the quotient of \(M\) by \(\{x \in M \mid fx = 0\}\). Then \(M'\) can be generated by \(k\) elements.

Proof

Choose generators \(x_1, \ldots, x_n \in M\) corresponding to the surjection \(R^{\oplus n} \to M\). Since \(R\) is local if a set of elements \(E \subset (f)\) generates \((f)\), then some \(e \in E\) generates \((f)\), see Algebra, Lemma 00DV. Hence we may pick \(z_1, \ldots, z_{n - k}\) in the kernel of \(R^{\oplus n} \to M\) such that some \((n - k) \times (n - k)\) minor of the \(n \times (n - k)\) matrix \(A = (z_{ij})\) generates \((f)\). After renumbering the \(x_i\) we may assume the first minor \(\det(z_{ij})_{1 \leq i, j \leq n - k}\) generates \((f)\), i.e., \(\det(z_{ij})_{1 \leq i, j \leq n - k} = uf\) for some unit \(u \in R\). Every other minor is a multiple of \(f\). By Algebra, Lemma 080R there exists a \(n - k \times n - k\) matrix \(B\) such that \[AB = f \left( \begin{matrix} u 1_{n - k \times n - k} \\ C \end{matrix} \right)\] for some matrix \(C\) with coefficients in \(R\). This implies that for every \(i \leq n - k\) the element \(y_i = ux_i + \sum_j c_{ji}x_j\) is annihilated by \(f\). Since \(M/\sum Ry_i\) is generated by the images of \(x_{n - k + 1}, \ldots, x_n\) we win.

Lemma

Let \(R\) be a ring. Let \(M\) be a finitely presented \(R\)-module. Let \(k \geq 0\). Assume that \(\text{Fit}_k(M) = (f)\) for some nonzerodivisor \(f \in R\) and \(\text{Fit}_{k - 1}(M) = 0\). Then

1. \(M\) has projective dimension \(\leq 1\),

2. \(M' = \Ker(f : M \to M)\) is the \(f\)-power torsion submodule of \(M\),

3. \(M'\) has projective dimension \(\leq 1\),

4. \(M/M'\) is finite locally free of rank \(k\), and

5. \(M \cong M/M' \oplus M'\).

Proof

Choose a presentation \[R^{\oplus m} \xrightarrow{A} R^{\oplus n} \to M \to 0\] for some matrix \(A\) with coefficients in \(R\).

We first prove the lemma when \(R\) is local. Set \(M' = \{x \in M \mid fx = 0\}\) as in the statement. By Lemma 080Z we can choose \(x_1, \ldots, x_k \in M\) which generate \(M/M'\). Then \(x_1, \ldots, x_k\) generate \(M_f = (M/M')_f\). Hence, if there is a relation \(\sum a_ix_i = 0\) in \(M\), then we see that \(a_1, \ldots, a_k\) map to zero in \(R_f\) since otherwise \(\text{Fit}_{k - 1}(M) R_f = \text{Fit}_{k - 1}(M_f)\) would be nonzero. Since \(f\) is a nonzerodivisor, we conclude \(a_1 = \ldots = a_k = 0\). Thus \(M \cong R^{\oplus k} \oplus M'\). After a change of basis in our presentation above, we may assume the first \(n - k\) basis vectors of \(R^{\oplus n}\) map into the summand \(M'\) of \(M\) and the last \(k\)-basis vectors of \(R^{\oplus n}\) map to basis elements of the summand \(R^{\oplus k}\) of \(M\). Having done so, the last \(k\) rows of the matrix \(A\) vanish. In this way we see that, replacing \(M\) by \(M'\), \(k\) by \(0\), \(n\) by \(n - k\), and \(A\) by the submatrix where we delete the last \(k\) rows, we reduce to the case discussed in the next paragraph.

Assume \(R\) is local, \(k = 0\), and \(M\) annihilated by \(f\). Now the \(0\)th Fitting ideal of \(M\) is \((f)\) and is generated by the \(n \times n\) minors of the matrix \(A\) of size \(n \times m\). (This in particular implies \(m \geq n\).) Since \(R\) is local, some \(n \times n\) minor of \(A\) is \(uf\) for a unit \(u \in R\). After renumbering we may assume this minor is the first one. Moreover, we know all other \(n \times n\) minors of \(A\) are divisible by \(f\). Write \(A = (A_1 A_2)\) in block form where \(A_1\) is an \(n \times n\) matrix and \(A_2\) is an \(n \times (m - n)\) matrix. By Algebra, Lemma 080R applied to the transpose of \(A\) (!) we find there exists an \(n \times n\) matrix \(B\) such that \[BA = B(A_1 A_2) = f \left( \begin{matrix} u 1_{n \times n} & C \end{matrix} \right)\] for some \(n \times (m - n)\) matrix \(C\) with coefficients in \(R\). Then we first conclude \(BA_1 = fu 1_{n \times n}\). Thus \[BA_2 = fC = u^{-1}fuC = u^{-1}BA_1C\] Since the determinant of \(B\) is a nonzerodivisor we conclude that \(A_2 = u^{-1}A_1C\). Therefore the image of \(A\) is equal to the image of \(A_1\) which is isomorphic to \(R^{\oplus n}\) because the determinant of \(A_1\) is a nonzerodivisor. Hence \(M\) has projective dimension \(\leq 1\).

We return to the case of a general ring \(R\). By the local case we see that \(M/M'\) is a finite locally free module of rank \(k\), see Algebra, Lemma 00NX. Hence the extension \(0 \to M' \to M \to M/M' \to 0\) splits. It follows that \(M'\) is a finitely presented module. Choose a short exact sequence \(0 \to K \to R^{\oplus a} \to M' \to 0\). Then \(K\) is a finite \(R\)-module, see Algebra, Lemma 0519. By the local case we see that \(K_\mathfrak p \cong R_\mathfrak p^{\oplus a}\) for all primes. Hence by Algebra, Lemma 00NX again we see that \(K\) is finite locally free of rank \(a\). It follows that \(M'\) has projective dimension \(\leq 1\) and the lemma is proved.

Lifting

In this section we collection some lemmas concerning lifting statements of the following kind: If \(A\) is a ring and \(I \subset A\) is an ideal, and \(\overline{\xi}\) is some kind of structure over \(A/I\), then we can lift \(\overline{\xi}\) to a similar kind of structure \(\xi\) over \(A\) or over some étale extension of \(A\). Here are some types of structure for which we have already proved some results:

1. idempotents, see Algebra, Lemmas 00J9 and 05BU,

2. projective modules, see Algebra, Lemmas 07LV and 0D47,

3. finite stably free modules, see Lemma 0BC5,

4. basis elements, see Algebra, Lemmas 051F and 051H,

5. ring maps, i.e., proving certain algebras are formally smooth, see Algebra, Lemma 00TK, Proposition 00TN, and Lemma 07K4,

6. syntomic ring maps, see Algebra, Lemma 00T0,

7. smooth ring maps, see Algebra, Lemma 04B1,

8. étale ring maps, see Algebra, Lemma 04D1,

9. factoring polynomials, see Algebra, Lemma 00UH, and

10. Algebra, Section 04GE discusses henselian local rings.

The interested reader will find more results of this nature in Smoothing Ring Maps, Section 07CD in particular Smoothing Ring Maps, Proposition 07M8.

Let \(A\) be a ring and let \(I \subset A\) be an ideal. Let \(\overline{\xi}\) be some kind of structure over \(A/I\). In the following lemmas we look for étale ring maps \(A \to A'\) which induce isomorphisms \(A/I \to A'/IA'\) and objects \(\xi'\) over \(A'\) lifting \(\overline{\xi}\). A general remark is that given étale ring maps \(A \to A' \to A''\) such that \(A/I \cong A'/IA'\) and \(A'/IA' \cong A''/IA''\) the composition \(A \to A''\) is also étale (Algebra, Lemma 00U2) and also satisfies \(A/I \cong A''/IA''\). We will frequently use this in the following lemmas without further mention. Here is a trivial example of the type of result we are looking for.

Lemma

Let \(A\) be a ring, let \(I \subset A\) be an ideal, let \(\overline{u} \in A/I\) be an invertible element. There exists an étale ring map \(A \to A'\) which induces an isomorphism \(A/I \to A'/IA'\) and an invertible element \(u' \in A'\) lifting \(\overline{u}\).

Proof

Choose any lift \(f \in A\) of \(\overline{u}\) and set \(A' = A_f\) and \(u\) the image of \(f\) in \(A'\).

Lemma

Let \(A\) be a ring, let \(I \subset A\) be an ideal, let \(\overline{e} \in A/I\) be an idempotent. There exists an étale ring map \(A \to A'\) which induces an isomorphism \(A/I \to A'/IA'\) and an idempotent \(e' \in A'\) lifting \(\overline{e}\).

Proof

Choose any lift \(x \in A\) of \(\overline{e}\). Set \[A' = A[t]/(t^2 - t)\left[\frac{1}{t - 1 + x}\right].\] The ring map \(A \to A'\) is étale because \((2t - 1)\text{d}t = 0\) and \((2t - 1)(2t - 1) = 1\) which is invertible. We have \(A'/IA' = A/I[t]/(t^2 - t)[\frac{1}{t - 1 + \overline{e}}] \cong A/I\) the last map sending \(t\) to \(\overline{e}\) which works as \(\overline{e}\) is a root of \(t^2 - t\). This also shows that setting \(e'\) equal to the class of \(t\) in \(A'\) works.

Lemma

Let \(A\) be a ring, let \(I \subset A\) be an ideal. Let \(\Spec(A/I) = \coprod_{j \in J} \overline{U}_j\) be a finite disjoint open covering. Then there exists an étale ring map \(A \to A'\) which induces an isomorphism \(A/I \to A'/IA'\) and a finite disjoint open covering \(\Spec(A') = \coprod_{j \in J} U'_j\) lifting the given covering.

Proof

This follows from Lemma 07LY and the fact that open and closed subsets of Spectra correspond to idempotents, see Algebra, Lemma 00EE.

Lemma

Let \(A \to B\) be a ring map and \(J \subset B\) an ideal. If \(A \to B\) is étale at every prime of \(V(J)\), then there exists a \(g \in B\) mapping to an invertible element of \(B/J\) such that \(A' = B_g\) is étale over \(A\).

Proof

The set of points of \(\Spec(B)\) where \(A \to B\) is not étale is a closed subset of \(\Spec(B)\), see Algebra, Definition 00U1. Write this as \(V(J')\) for some ideal \(J' \subset B\). Then \(V(J') \cap V(J) = \emptyset\) hence \(J + J' = B\) by Algebra, Lemma 00E0. Write \(1 = f + g\) with \(f \in J\) and \(g \in J'\). Then \(g\) works.

Next we have three lemmas saying we can lift factorizations of polynomials.

Lemma

Let \(A\) be a ring, let \(I \subset A\) be an ideal. Let \(f \in A[x]\) be a monic polynomial. Let \(\overline{f} = \overline{g} \overline{h}\) be a factorization of \(f\) in \(A/I[x]\) such that \(\overline{g}\) and \(\overline{h}\) are monic and generate the unit ideal in \(A/I[x]\). Then there exists an étale ring map \(A \to A'\) which induces an isomorphism \(A/I \to A'/IA'\) and a factorization \(f = g' h'\) in \(A'[x]\) with \(g'\), \(h'\) monic lifting the given factorization over \(A/I\).

Proof

We will deduce this from results on the universal factorization proved earlier; however, we encourage the reader to find their own proof not using this trick. Say \(\deg(\overline{g}) = n\) and \(\deg(\overline{h}) = m\) so that \(\deg(f) = n + m\). Write \(f = x^{n + m} + \sum \alpha_i x^{n + m - i}\) for some \(\alpha_1, \ldots, \alpha_{n + m} \in A\). Consider the ring map \[R = \mathbf{Z}[a_1, \ldots, a_{n + m}] \longrightarrow S = \mathbf{Z}[b_1, \ldots, b_n, c_1, \ldots, c_m]\] of Algebra, Example 00UA. Let \(R \to A\) be the ring map which sends \(a_i\) to \(\alpha_i\). Set \[B = A \otimes_R S\] By construction the image \(f_B\) of \(f\) in \(B[x]\) factors, say \(f_B = g_B h_B\) with \(g_B = x^n + \sum (1 \otimes b_i) x^{n - i}\) and similarly for \(h_B\). Write \(\overline{g} = x^n + \sum \overline{\beta}_i x^{n - i}\) and \(\overline{h} = x^m + \sum \overline{\gamma}_i x^{m - i}\). The \(A\)-algebra map \[B \longrightarrow A/I, \quad 1 \otimes b_i \mapsto \overline{\beta}_i, \quad 1 \otimes c_i \mapsto \overline{\gamma}_i\] maps \(g_B\) and \(h_B\) to \(\overline{g}\) and \(\overline{h}\) in \(A/I[x]\). The displayed map is surjective; denote \(J \subset B\) its kernel. From the discussion in Algebra, Example 00UA it is clear that \(A \to B\) is etale at all points of \(V(J) \subset \Spec(B)\). Choose \(g \in B\) as in Lemma 07M0 and consider the \(A\)-algebra \(B_g\). Since \(g\) maps to a unit in \(B/J = A/I\) we obtain also a map \(B_g/I B_g \to A/I\) of \(A/I\)-algebras. Since \(A/I \to B_g/I B_g\) is étale, also \(B_g/IB_g \to A/I\) is étale (Algebra, Lemma 00U7). Hence there exists an idempotent \(e \in B_g/I B_g\) such that \(A/I = (B_g/I B_g)_e\) (Algebra, Lemma 00U8). Choose a lift \(h \in B_g\) of \(e\). Then \(A \to A' = (B_g)_h\) with factorization given by the image of the factorization \(f_B = g_B h_B\) in \(A'\) is a solution to the problem posed by the lemma.

Lemma

Let \(A\) be a ring, let \(I \subset A\) be an ideal. Let \(f \in A[x]\) be a monic polynomial. Let \(\overline{f} = \overline{g} \overline{h}\) be a factorization of \(f\) in \(A/I[x]\) and assume

1. the leading coefficient of \(\overline{g}\) is an invertible element of \(A/I\), and

2. \(\overline{g}\), \(\overline{h}\) generate the unit ideal in \(A/I[x]\).

Then there exists an étale ring map \(A \to A'\) which induces an isomorphism \(A/I \to A'/IA'\) and a factorization \(f = g' h'\) in \(A'[x]\) lifting the given factorization over \(A/I\).

Proof

Applying Lemma 07LX we may assume that the leading coefficient of \(\overline{g}\) is the reduction of an invertible element \(u \in A\). Then we may replace \(\overline{g}\) by \(\overline{u}^{-1}\overline{g}\) and \(\overline{h}\) by \(\overline{u}\overline{h}\). Thus we may assume that \(\overline{g}\) is monic. Since \(f\) is monic we conclude that \(\overline{h}\) is monic too. In this case the result follows from Lemma 0ALH.

Example

This example shows that one cannot drop the condition on the leading coefficient in Lemma 07M1. Namely, let \(A = \mathbf{Z}\), \(I = 4\mathbf{Z}\), \(f = 1\) and \(\bar{g} = \bar{h} = 2x^2 + 2x + 1\) in \(A/I[x]\). If \(A \to A'\) is étale and \(A'/I A' = A/I\), then the \(2\)-adic completion of \(A'\) is isomorphic to \(\mathbf{Z}_2\). Now any polynomial \(g'\) in \(\mathbf{Z}_2[x]\) congruent to \(2x^2 + 2x + 1\) mod \(4\) will not have a polynomial multiplicative inverse in \(\mathbf{Z}_2[x]\), whence the conclusion of the lemma does not hold.

Lemma

Let \(R \to S\) be a ring map. Let \(I \subset R\) be an ideal of \(R\) and let \(J \subset S\) be an ideal of \(S\). If the closure of the image of \(V(J)\) in \(\Spec(R)\) is disjoint from \(V(I)\), then there exists an element \(f \in R\) which maps to \(1\) in \(R/I\) and to an element of \(J\) in \(S\).

Proof

Let \(I' \subset R\) be an ideal such that \(V(I')\) is the closure of the image of \(V(J)\). Then \(V(I) \cap V(I') = \emptyset\) by assumption and hence \(I + I' = R\) by Algebra, Lemma 00E0. Write \(1 = g + f\) with \(g \in I\) and \(f \in I'\). We have \(V(f') \supset V(J)\) where \(f'\) is the image of \(f\) in \(S\). Hence \((f')^n \in J\) for some \(n\), see Algebra, Lemma 00E0. Replacing \(f\) by \(f^n\) we win.

Lemma

Let \(I\) be an ideal of a ring \(A\). Let \(A \to B\) be an integral ring map. Let \(b \in B\) map to an idempotent in \(B/IB\). Then there exists a monic \(f \in A[x]\) with \(f(b) = 0\) and \(f \bmod I = x^d(x - 1)^d\) for some \(d \geq 1\).

Proof

Observe that \(z = b^2 - b\) is an element of \(IB\). By Algebra, Lemma 00H5 there exist a monic polynomial \(g(x) = x^d + \sum a_j x^j\) of degree \(d\) with \(a_j \in I\) such that \(g(z) = 0\) in \(B\). Hence \(f(x) = g(x^2 - x) \in A[x]\) is a monic polynomial such that \(f(x) \equiv x^d(x - 1)^d \bmod I\) and such that \(f(b) = 0\) in \(B\).

Lemma

Let \(A\) be a ring, let \(I \subset A\) be an ideal. Let \(A \to B\) be an integral ring map. Let \(\overline{e} \in B/IB\) be an idempotent. Then there exists an étale ring map \(A \to A'\) which induces an isomorphism \(A/I \to A'/IA'\) and an idempotent \(e' \in B \otimes_A A'\) lifting \(\overline{e}\).

Proof

Choose an element \(y \in B\) lifting \(\overline{e}\). Choose \(f \in A[x]\) as in Lemma 09XG for \(y\). By Lemma 07M1 we can find an étale ring map \(A \to A'\) which induces an isomorphism \(A/I \to A'/IA'\) and such that \(f = gh\) in \(A[x]\) with \(g(x) = x^d \bmod IA'\) and \(h(x) = (x - 1)^d \bmod IA'\). After replacing \(A\) by \(A'\) we may assume that the factorization is defined over \(A\). In that case we see that \(b_1 = g(y) \in B\) is a lift of \(\overline{e}^d = \overline{e}\) and \(b_2 = h(y) \in B\) is a lift of \((\overline{e} - 1)^d = (-1)^d (1 - \overline{e})^d = (-1)^d(1 - \overline{e})\) and moreover \(b_1b_2 = 0\). Thus \((b_1, b_2)B/IB = B/IB\) and \(V(b_1, b_2) \subset \Spec(B)\) is disjoint from \(V(IB)\). Since \(\Spec(B) \to \Spec(A)\) is closed (see Algebra, Lemmas 00GU and 00HZ) we can find an \(a \in A\) which maps to an invertible element of \(A/I\) whose image in \(B\) lies in \((b_1, b_2)\), see Lemma 07M3. After replacing \(A\) by the localization \(A_a\) we get that \((b_1, b_2) = B\). Then \(\Spec(B) = D(b_1) \amalg D(b_2)\); disjoint union because \(b_1b_2 = 0\) and covers \(\Spec(B)\) because \((b_1, b_2) = B\). Let \(e \in B\) be the idempotent corresponding to the open and closed subset \(D(b_1)\), see Algebra, Lemma 00EE. Since \(b_1\) is a lift of \(\overline{e}\) and \(b_2\) is a lift of \(\pm (1 - \overline{e})\) we conclude that \(e\) is a lift of \(\overline{e}\) by the uniqueness statement in Algebra, Lemma 00EE.

Lemma

Let \(A\) be a ring, let \(I \subset A\) be an ideal. Let \(\overline{P}\) be a finite projective \(A/I\)-module. Then there exists an étale ring map \(A \to A'\) which induces an isomorphism \(A/I \to A'/IA'\) and a finite projective \(A'\)-module \(P'\) lifting \(\overline{P}\).

Proof

We can choose an integer \(n\) and a direct sum decomposition \((A/I)^{\oplus n} = \overline{P} \oplus \overline{K}\) for some \(R/I\)-module \(\overline{K}\). Choose a lift \(\varphi : A^{\oplus n} \to A^{\oplus n}\) of the projector \(\overline{p}\) associated to the direct summand \(\overline{P}\). Let \(f \in A[x]\) be the characteristic polynomial of \(\varphi\). Set \(B = A[x]/(f)\). By Cayley-Hamilton (Algebra, Lemma 00DX) there is a map \(B \to \text{End}_A(A^{\oplus n})\) mapping \(x\) to \(\varphi\). For every prime \(\mathfrak p \supset I\) the image of \(f\) in \(\kappa(\mathfrak p)\) is \((x - 1)^rx^{n - r}\) where \(r\) is the dimension of \(\overline{P} \otimes_{A/I} \kappa(\mathfrak p)\). Hence \((x - 1)^nx^n\) maps to zero in \(B \otimes_A \kappa(\mathfrak p)\) for all \(\mathfrak p \supset I\). Thus \(x(1 - x)\) is contained in every prime ideal of \(B/IB\). Hence \(x^N(1 - x)^N\) is contained in \(IB\) for some \(N \geq 1\). It follows that \(x^N + (1 - x)^N\) is a unit in \(B/IB\) and that \[\overline{e} = \text{image of }\frac{x^N}{x^N + (1 - x)^N}\text{ in }B/IB\] is an idempotent as both assertions hold in \(\mathbf{Z}[x]/(x^N(x - 1)^N)\). The image of \(\overline{e}\) in \(\text{End}_{A/I}((A/I)^{\oplus n})\) is \[\frac{\overline{p}^N}{\overline{p}^N + (1 - \overline{p})^N} = \overline{p}\] as \(\overline{p}\) is an idempotent. After replacing \(A\) by an étale extension \(A'\) as in the lemma, we may assume there exists an idempotent \(e \in B\) which maps to \(\overline{e}\) in \(B/IB\), see Lemma 07M4. Then the image of \(e\) under the map \[B = A[x]/(f) \longrightarrow \text{End}_A(A^{\oplus n}).\] is an idempotent element \(p\) which lifts \(\overline{p}\). Setting \(P = \Im(p)\) we win.

Lemma

Let \(A\) be a ring. Let \(0 \to K \to A^{\oplus m} \to M \to 0\) be a sequence of \(A\)-modules. Consider the \(A\)-algebra \(C = \text{Sym}^*_A(M)\) with its presentation \(\alpha : A[y_1, \ldots, y_m] \to C\) coming from the surjection \(A^{\oplus m} \to M\). Then \[\NL(\alpha) = (K \otimes_A C \to \bigoplus\nolimits_{j = 1, \ldots, m} C \text{d}y_j)\] (see Algebra, Section 00S0) in particular \(\Omega_{C/A} = M \otimes_A C\).

Proof

Let \(J = \Ker(\alpha)\). The lemma asserts that \(J/J^2 \cong K \otimes_A C\). Note that \(\alpha\) is a homomorphism of graded algebras. We will prove that in degree \(d\) we have \((J/J^2)_d = K \otimes_A C_{d - 1}\). Note that \[J_d = \Ker(\text{Sym}^d_A(A^{\oplus m}) \to \text{Sym}^d_A(M)) = \Im(K \otimes_A \text{Sym}^{d - 1}_A(A^{\oplus m}) \to \text{Sym}^d_A(A^{\oplus m})),\] see Algebra, Lemma 00DO. It follows that \((J^2)_d = \sum_{a + b = d} J_a \cdot J_b\) is the image of \[K \otimes_A K \otimes_A \text{Sym}^{d - 2}_A(A^{\otimes m}) \to \text{Sym}^d_A(A^{\oplus m}).\] The cokernel of the map \(K \otimes_A \text{Sym}^{d - 2}_A(A^{\otimes m}) \to \text{Sym}^{d - 1}_A(A^{\oplus m})\) is \(\text{Sym}^{d - 1}_A(M)\) by the lemma referenced above. Hence it is clear that \((J/J^2)_d = J_d/(J^2)_d\) is equal to \[\begin{align*} \Coker( K \otimes_A K \otimes_A \text{Sym}^{d - 2}_A(A^{\otimes m}) \to K \otimes_A \text{Sym}^{d - 1}_A(A^{\otimes m})) & = K \otimes_A \text{Sym}^{d - 1}_A(M) \\ & = K \otimes_A C_{d -1} \end{align*}\] as desired.

Lemma

Let \(A\) be a ring. Let \(M\) be an \(A\)-module. Then \(C = \text{Sym}_A^*(M)\) is smooth over \(A\) if and only if \(M\) is a finite projective \(A\)-module.

Proof

Let \(\sigma : C \to A\) be the projection onto the degree \(0\) part of \(C\). Then \(J = \Ker(\sigma)\) is the part of degree \(> 0\) and we see that \(J/J^2 = M\) as an \(A\)-module. Hence if \(A \to C\) is smooth then \(M\) is a finite projective \(A\)-module by Algebra, Lemma 05D5.

Conversely, assume that \(M\) is finite projective and choose a surjection \(A^{\oplus n} \to M\) with kernel \(K\). Of course the sequence \(0 \to K \to A^{\oplus n} \to M \to 0\) is split as \(M\) is projective. In particular we see that \(K\) is a finite \(A\)-module and hence \(C\) is of finite presentation over \(A\) as \(C\) is a quotient of \(A[x_1, \ldots, x_n]\) by the ideal generated by \(K \subset \bigoplus Ax_i\). The computation of Lemma 07EV shows that \(\NL_{C/A}\) is homotopy equivalent to \((K \to M) \otimes_A C\). Hence \(\NL_{C/A}\) is quasi-isomorphic to \(C \otimes_A M\) placed in degree \(0\) which means that \(C\) is smooth over \(A\) by Algebra, Definition 00T2.

Lemma

Let \(A\) be a ring, let \(I \subset A\) be an ideal. Consider a commutative diagram \[\xymatrix{ B \ar[rd] \\ A \ar[u] \ar[r] & A/I }\] where \(B\) is a smooth \(A\)-algebra. Then there exists an étale ring map \(A \to A'\) which induces an isomorphism \(A/I \to A'/IA'\) and an \(A\)-algebra map \(B \to A'\) lifting the ring map \(B \to A/I\).

Proof

Let \(J \subset B\) be the kernel of \(B \to A/I\) so that \(B/J = A/I\). By Algebra, Lemma 06A9 the sequence \[0 \to I/I^2 \to J/J^2 \to \Omega_{B/A} \otimes_B B/J \to 0\] is split exact. Thus \(\overline{P} = J/(J^2 + IB) = \Omega_{B/A} \otimes_B B/J\) is a finite projective \(A/I\)-module. Choose an integer \(n\) and a direct sum decomposition \(A/I^{\oplus n} = \overline{P} \oplus \overline{K}\). By Lemma 07M5 we can find an étale ring map \(A \to A'\) which induces an isomorphism \(A/I \to A'/IA'\) and a finite projective \(A\)-module \(K\) which lifts \(\overline{K}\). We may and do replace \(A\) by \(A'\). Set \(B' = B \otimes_A \text{Sym}_A^*(K)\). Since \(A \to \text{Sym}_A^*(K)\) is smooth by Lemma 07M6 we see that \(B \to B'\) is smooth which in turn implies that \(A \to B'\) is smooth (see Algebra, Lemmas 00T4 and 00TC). Moreover the section \(\text{Sym}^*_A(K) \to A\) determines a section \(B' \to B\) and we let \(B' \to A/I\) be the composition \(B' \to B \to A/I\). Let \(J' \subset B'\) be the kernel of \(B' \to A/I\). We have \(JB' \subset J'\) and \(B \otimes_A K \subset J'\). These maps combine to give an isomorphism \[(A/I)^{\oplus n} \cong J/(J^2 + IB) \oplus \overline{K} \longrightarrow J'/((J')^2 + IB')\] Thus, after replacing \(B\) by \(B'\) we may assume that \(J/(J^2 + IB) = \Omega_{B/A} \otimes_B B/J\) is a free \(A/I\)-module of rank \(n\).

In this case, choose \(f_1, \ldots, f_n \in J\) which map to a basis of \(J/(J^2 + IB)\). Consider the finitely presented \(A\)-algebra \(C = B/(f_1, \ldots, f_n)\). Note that we have an exact sequence \[0 \to H_1(L_{C/A}) \to (f_1, \ldots, f_n)/(f_1, \ldots, f_n)^2 \to \Omega_{B/A} \otimes_B C \to \Omega_{C/A} \to 0\] see Algebra, Lemma 00S2 (note that \(H_1(L_{B/A}) = 0\) and that \(\Omega_{B/A}\) is finite projective, in particular flat so the Tor group vanishes). For any prime \(\mathfrak q \supset J\) of \(B\) the module \(\Omega_{B/A, \mathfrak q}\) is free of rank \(n\) because \(\Omega_{B/A}\) is finite projective and because \(\Omega_{B/A} \otimes_B B/J\) is free of rank \(n\) (see Algebra, Lemma 00NX). By our choice of \(f_1, \ldots, f_n\) the map \[\left((f_1, \ldots, f_n)/(f_1, \ldots, f_n)^2\right)_{\mathfrak q} \to \Omega_{B/A, \mathfrak q}\] is surjective modulo \(J\). Hence we see that this map of modules over the local ring \(C_{\mathfrak q}\) has to be an isomorphism (by Algebra, Lemma 00DV the map is surjective and for example by Algebra, Lemma 05G8 because \(((f_1, \ldots, f_n)/(f_1, \ldots, f_n)^2)_{\mathfrak q}\) is generated by \(n\) elements the map is injective). Thus \(H_1(L_{C/A})_{\mathfrak q} = 0\) and \(\Omega_{C/A, \mathfrak q} = 0\). By Algebra, Lemma 07BU we see that \(A \to C\) is smooth at the prime \(\overline{\mathfrak q}\) of \(C\) corresponding to \(\mathfrak q\). Since \(\Omega_{C/A, \mathfrak q} = 0\) it is actually étale at \(\overline{\mathfrak q}\). Thus \(A \to C\) is étale at all primes of \(C\) containing \(JC\). By Lemma 07M0 we can find an \(f \in C\) mapping to an invertible element of \(C/JC\) such that \(A \to C_f\) is étale. By our choice of \(f\) it is still true that \(C_f/JC_f = A/I\). The map \(C_f/IC_f \to A/I\) is surjective and étale by Algebra, Lemma 00U7. Hence \(A/I\) is isomorphic to the localization of \(C_f/IC_f\) at some element \(g \in C\), see Algebra, Lemma 00U8. Set \(A' = C_{fg}\) to conclude the proof.

Zariski pairs

In this section and the next a pair is a pair \((A, I)\) where \(A\) is a ring and \(I \subset A\) is an ideal. A morphism of pairs \((A, I) \to (B, J)\) is a ring map \(\varphi : A \to B\) with \(\varphi(I) \subset J\).

Definition

A Zariski pair is a pair \((A, I)\) such that \(I\) is contained in the Jacobson radical of \(A\).

Lemma

Let \((A, I)\) be a Zariski pair. Then the map from idempotents of \(A\) to idempotents of \(A/I\) is injective.

Proof

An idempotent of a local ring is either \(0\) or \(1\). Thus an idempotent is determined by the set of maximal ideals where it vanishes, by Algebra, Lemma 00HN.

Lemma

Let \((A, I)\) be a Zariski pair. Let \(A \to B\) be a flat, integral, finitely presented ring map such that \(A/I \to B/IB\) is an isomorphism. Then \(A \to B\) is an isomorphism.

Proof

The ring map \(A \to B\) is finite by Algebra, Lemma 02JJ. Hence \(B\) is finitely presented as an \(A\)-module by Algebra, Lemma 0564. Hence \(B\) is a finite locally free \(A\)-module by Algebra, Lemma 00NX. Since the module \(B\) has rank \(1\) along \(V(I)\) (see rank function described in Algebra, Lemma 00NX), and as \((A, I)\) is a Zariski pair, we conclude that the rank is \(1\) everywhere. It follows that \(A \to B\) is an isomorphism: it is a pleasant exercise to show that a ring map \(R \to S\) such that \(S\) is a locally free \(R\)-module of rank \(1\) is an isomorphism (hint: look at local rings).

Lemma

Let \((A, I)\) be a Zariski pair. Let \(A \to B\) be a finite ring map. Assume

1. \(B/IB = B_1 \times B_2\) is a product of \(A/I\)-algebras

2. \(A/I \to B_1/IB_1\) is surjective,

3. \(b \in B\) maps to \((1, 0)\) in the product.

Then there exists a monic \(f \in A[x]\) with \(f(b) = 0\) and \(f \bmod I = (x - 1)x^d\) for some \(d \geq 1\).

Proof

By Lemma 07M4 we can find an étale ring map \(A \to A'\) inducing an isomorphism \(A/I \to A'/IA'\) such that \(B' = B \otimes_A A'\) contains an idempotent \(e'\) lifting the image of \(b\) in \(B'/IB'\). Consider the corresponding \(A'\)-algebra decomposition \[B' = B'_1 \times B'_2\] which is compatible with the one given in the lemma upon reduction modulo \(I\). The map \(A' \to B'_1\) is surjective modulo \(IA'\). By Nakayama’s lemma (Algebra, Lemma 00DV) we can find \(i \in IA'\) such that after replacing \(A'\) by \(A'_{1 + i}\) the map \(A' \to B'_1\) is surjective. Observe that the image \(b'_1 \in B'_1\) of \(b\) satisfies \(b'_1 - 1 \in IB'_1\). Thus we may pick \(a' \in IA'\) mapping to \(b'_1 - 1\). On the other hand, the image \(b'_2 \in B'_2\) of \(b\) is in \(IB'_2\). By Algebra, Lemma 00H5 there exist a monic polynomial \(g(x) = x^d + \sum a'_j x^j\) of degree \(d\) with \(a'_j \in IA'\) such that \(g(b'_2) = 0\) in \(B'_2\). Thus the image \(b' = (b'_1, b'_2) \in B'\) of \(b\) is a root of the polynomial \((x - 1 - a')g(x)\). We conclude that \[(b' - 1)(b')^d \in \sum\nolimits_{j = 0, \ldots, d} IA' \cdot (b')^j\] We claim that this implies \[(b - 1)b^d \in \sum\nolimits_{j = 0, \ldots, d} I \cdot b^j\] in \(B\). For this it is enough to see that the ring map \(A \to A'\) is faithfully flat, because the condition is that the image of \((b - 1)b^d\) is zero in \(B/\sum_{j = 0, \ldots, d} Ib^j\) (use Algebra, Lemma 05CK). The map \(A \to A'\) flat because it is étale (Algebra, Lemma 00U2). On the other hand, the induced map on spectra is open (see Algebra, Proposition 00I1 and use previous lemma referenced) and the image contains \(V(I)\). Since \(I\) is contained in the Jacobson radical of \(A\) the image is all of \(\Spec(A)\) by Algebra, Lemma algebra-lemma-open-containing-vanishing-jacobson-radical. Thus \(A \to A'\) is faithfully flat by Algebra, Lemma 00HQ, and we conclude.

Lemma

Let \((A, I)\) be a Zariski pair with \(A\) Noetherian. Let \(f \in I\). Then \(A_f\) is a Jacobson ring.

Proof

We will use the criterion of Algebra, Lemma 00KX. Let \(\mathfrak p \subset A\) be a prime ideal such that \(\mathfrak p_f = \mathfrak p A_f\) is prime and not maximal. We have to show that \(A_f/\mathfrak p_f = (A/\mathfrak p)_f\) has infinitely many prime ideals. After replacing \(A\) by \(A/\mathfrak p\) we may assume \(A\) is a domain, \(\dim A_f > 0\), and our goal is to show that \(\Spec(A_f)\) is infinite. Since \(\dim A_f > 0\) we can find a nonzero prime ideal \(\mathfrak q \subset A\) not containing \(f\). Choose a maximal ideal \(\mathfrak m \subset A\) containing \(\mathfrak q\). Since \((A, I)\) is a Zariski pair, we see \(I \subset \mathfrak m\). Hence \(\mathfrak m \not = \mathfrak q\) and \(\dim(A_\mathfrak m) > 1\). Hence \(\Spec((A_\mathfrak m)_f) \subset \Spec(A_f)\) is infinite by Algebra, Lemma 02IG and we win.

Henselian pairs

Some of the results of Section 07LW may be viewed as results about henselian pairs. In this section a pair is a pair \((A, I)\) where \(A\) is a ring and \(I \subset A\) is an ideal. A morphism of pairs \((A, I) \to (B, J)\) is a ring map \(\varphi : A \to B\) with \(\varphi(I) \subset J\). As in Section 07LW given an object \(\xi\) over \(A\) we denote \(\overline{\xi}\) the “base change” of \(\xi\) to an object over \(A/I\) (provided this makes sense).

Definition

A henselian pair is a pair \((A, I)\) satisfying

1. \(I\) is contained in the Jacobson radical of \(A\), and

2. for any monic polynomial \(f \in A[T]\) and factorization \(\overline{f} = g_0h_0\) with \(g_0, h_0 \in A/I[T]\) monic generating the unit ideal in \(A/I[T]\), there exists a factorization \(f = gh\) in \(A[T]\) with \(g, h\) monic and \(g_0 = \overline{g}\) and \(h_0 = \overline{h}\).

Observe that if \(A\) is a local ring and \(I = \mathfrak m\) is the maximal ideal, then \((A, I)\) is a henselian pair if and only if \(A\) is a henselian local ring, see Algebra, Lemma 04GG. In Lemma 09XI we give a number of equivalent characterizations of henselian pairs (and we will add more as time goes on).

Lemma

Let \((A, I)\) be a pair with \(I\) locally nilpotent. Then the functor \(B \mapsto B/IB\) induces an equivalence between the category of étale algebras over \(A\) and the category of étale algebras over \(A/I\). Moreover, the pair is henselian.

Proof

Essential surjectivity holds by Algebra, Lemma 04D1. If \(B\), \(B'\) are étale over \(A\) and \(B/IB \to B'/IB'\) is a morphism of \(A/I\)-algebras, then we can lift this by Algebra, Lemma 07K4. Finally, suppose that \(f, g : B \to B'\) are two \(A\)-algebra maps with \(f \bmod I = g \bmod I\). Choose an idempotent \(e \in B \otimes_A B\) generating the kernel of the multiplication map \(B \otimes_A B \to B\), see Algebra, Lemmas 02FL and 00UV (to see that étale is unramified). Then \((f \otimes g)(e) \in IB'\). Since \(IB'\) is locally nilpotent (Algebra, Lemma 0544) this implies \((f \otimes g)(e) = 0\) by Algebra, Lemma 00J9. Thus \(f = g\).

It is clear that \(I\) is contained in the Jacobson radical of \(A\). Let \(f \in A[T]\) be a monic polynomial and let \(\overline{f} = g_0h_0\) be a factorization of \(\overline{f} = f \bmod I\) with \(g_0, h_0 \in A/I[T]\) monic generating the unit ideal in \(A/I[T]\). By Lemma 0ALH there exists an étale ring map \(A \to A'\) which induces an isomorphism \(A/I \to A'/IA'\) such that the factorization lifts to a factorization into monic polynomials over \(A'\). By the above we have \(A = A'\) and the factorization is over \(A\).

Lemma

Let \(A = \lim A_n\) where \((A_n)\) is an inverse system of rings whose transition maps are surjective and have locally nilpotent kernels. Then \((A, I_n)\) is a henselian pair, where \(I_n = \Ker(A \to A_n)\).

Proof

Fix \(n\). Let \(a \in A\) be an element which maps to \(1\) in \(A_n\). By Algebra, Lemma 0AMG we see that \(a\) maps to a unit in \(A_m\) for all \(m \geq n\). Hence \(a\) is a unit in \(A\). Thus by Algebra, Lemma 0AME the ideal \(I_n\) is contained in the Jacobson radical of \(A\). Let \(f \in A[T]\) be a monic polynomial and let \(\overline{f} = g_nh_n\) be a factorization of \(\overline{f} = f \bmod I_n\) with \(g_n, h_n \in A_n[T]\) monic generating the unit ideal in \(A_n[T]\). By Lemma 0ALI we can successively lift this factorization to \(f \bmod I_m = g_m h_m\) with \(g_m, h_m\) monic in \(A_m[T]\) for all \(m \geq n\). At each step we have to verify that our lifts \(g_m, h_m\) generate the unit ideal in \(A_n[T]\); this follows from the corresponding fact for \(g_n, h_n\) and the fact that \(\Spec(A_n[T]) = \Spec(A_m[T])\) because the kernel of \(A_m \to A_n\) is locally nilpotent. As \(A = \lim A_m\) this finishes the proof.

Lemma

Let \((A, I)\) be a pair. If \(A\) is \(I\)-adically complete, then the pair is henselian.

Proof

By Algebra, Lemma 05GI the ideal \(I\) is contained in the Jacobson radical of \(A\). Let \(f \in A[T]\) be a monic polynomial and let \(\overline{f} = g_0h_0\) be a factorization of \(\overline{f} = f \bmod I\) with \(g_0, h_0 \in A/I[T]\) monic generating the unit ideal in \(A/I[T]\). By Lemma 0ALI we can successively lift this factorization to \(f \bmod I^n = g_n h_n\) with \(g_n, h_n\) monic in \(A/I^n[T]\) for all \(n \geq 1\). As \(A = \lim A/I^n\) this finishes the proof.

Lemma

Let \((A, I)\) be a pair. Let \(A \to B\) be a finite type ring map such that \(B/IB = C_1 \times C_2\) with \(A/I \to C_1\) finite. Let \(B'\) be the integral closure of \(A\) in \(B\). Then we can write \(B'/IB' = C_1 \times C'_2\) such that the map \(B'/IB' \to B/IB\) preserves product decompositions and there exists a \(g \in B'\) mapping to \((1, 0)\) in \(C_1 \times C'_2\) with \(B'_g \to B_g\) an isomorphism.

Proof

Observe that \(A \to B\) is quasi-finite at every prime of the closed subset \(T = \Spec(C_1) \subset \Spec(B)\) (this follows by looking at fibre rings, see Algebra, Definition 00PL). Consider the diagram of topological spaces \[\xymatrix{ \Spec(B) \ar[rr]_\phi \ar[rd]_\psi & & \Spec(B') \ar[ld]^{\psi'} \\ & \Spec(A) }\] By Algebra, Theorem 00Q9 for every \(\mathfrak p \in T\) there is a \(h_\mathfrak p \in B'\), \(h_\mathfrak p \not \in \mathfrak p\) such that \(B'_h \to B_h\) is an isomorphism. The union \(U = \bigcup D(h_\mathfrak p)\) gives an open \(U \subset \Spec(B')\) such that \(\phi^{-1}(U) \to U\) is a homeomorphism and \(T \subset \phi^{-1}(U)\). Since \(T\) is open in \(\psi^{-1}(V(I))\) we conclude that \(\phi(T)\) is open in \(U \cap (\psi')^{-1}(V(I))\). Thus \(\phi(T)\) is open in \((\psi')^{-1}(V(I))\). On the other hand, since \(C_1\) is finite over \(A/I\) it is finite over \(B'\). Hence \(\phi(T)\) is a closed subset of \(\Spec(B')\) by Algebra, Lemmas 00HZ and 00GU. We conclude that \(\Spec(B'/IB') \supset \phi(T)\) is open and closed. By Algebra, Lemma 00EM we get a corresponding product decomposition \(B'/IB' = C'_1 \times C'_2\). The map \(B'/IB' \to B/IB\) maps \(C'_1\) into \(C_1\) and \(C'_2\) into \(C_2\) as one sees by looking at what happens on spectra (hint: the inverse image of \(\phi(T)\) is exactly \(T\); some details omitted). Pick a \(g \in B'\) mapping to \((1, 0)\) in \(C'_1 \times C'_2\) such that \(D(g) \subset U\); this is possible because \(\Spec(C'_1)\) and \(\Spec(C'_2)\) are disjoint and closed in \(\Spec(B')\) and \(\Spec(C'_1)\) is contained in \(U\). Then \(B'_g \to B_g\) defines a homeomorphism on spectra and an isomorphism on local rings (by our choice of \(U\) above). Hence it is an isomorphism, as follows for example from Algebra, Lemma 00HN. Finally, it follows that \(C'_1 = C_1\) and the proof is complete.

Lemma

Let \((A, I)\) be a pair. The following are equivalent

1. \((A, I)\) is a henselian pair,

2. given an étale ring map \(A \to A'\) and an \(A\)-algebra map \(\sigma : A' \to A/I\), there exists an \(A\)-algebra map \(A' \to A\) lifting \(\sigma\),

3. for any finite \(A\)-algebra \(B\) the map \(B \to B/IB\) induces a bijection on idempotents,

4. for any integral \(A\)-algebra \(B\) the map \(B \to B/IB\) induces a bijection on idempotents, and

5. (Gabber) \(I\) is contained in the Jacobson radical of \(A\) and every monic polynomial \(f(T) \in A[T]\) of the form \[f(T) = T^n(T - 1) + a_n T^n + \ldots + a_1 T + a_0\] with \(a_n, \ldots, a_0 \in I\) and \(n \ge 1\) has a root \(\alpha \in 1 + I\).

Moreover, in part (5) the root is unique.

Proof

Assume (2) holds. Then \(I\) is contained in the Jacobson radical of \(A\), since otherwise there would be a nonunit \(f \in A\) congruent to \(1\) modulo \(I\) and the map \(A \to A_f\) would contradict (2). Hence \(IB \subset B\) is contained in the Jacobson radical of \(B\) for \(B\) integral over \(A\) because \(\Spec(B) \to \Spec(A)\) is closed by Algebra, Lemmas 00HZ and 00GU. Thus the map from idempotents of \(B\) to idempotents of \(B/IB\) is injective by Lemma 09XF. On the other hand, since (2) holds, every idempotent of \(B/IB\) lifts to an idempotent of \(B\) by Lemma 07M4. In this way we see that (2) implies (4).

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

Assume (3). Let \(\mathfrak m\) be a maximal ideal and consider the finite map \(A \to B = A/(I \cap \mathfrak m)\). The condition that \(B \to B/IB\) induces a bijection on idempotents implies that \(I \subset \mathfrak m\) (if not, then \(B = A/I \times A/\mathfrak m\) and \(B/IB = A/I\)). Thus we see that \(I\) is contained in the Jacobson radical of \(A\). Let \(f \in A[T]\) be monic and suppose given a factorization \(\overline{f} = g_0h_0\) with \(g_0, h_0 \in A/I[T]\) monic generating the unit ideal in \(A/I[T]\). Set \(B = A[T]/(f)\). Let \(\overline{e}\) be the idempotent of \(B/IB\) corresponding to the decomposition \[B/IB = A/I[T]/(g_0) \times A/I[T]/(h_0)\] of \(A\)-algebras. Let \(e \in B\) be an idempotent lifting \(\overline{e}\) which exists as we assumed (3). This gives a product decomposition \[B = eB \times (1 - e)B\] Note that \(B\) is free of rank \(\deg(f)\) as an \(A\)-module. Hence \(eB\) and \((1 - e)B\) are finite locally free \(A\)-modules. However, since \(eB\) and \((1 - e)B\) have constant rank \(\deg(g_0)\) and \(\deg(h_0)\) over \(A/I\) we find that the same is true over \(\Spec(A)\). We conclude that \[\begin{align*} f & = \text{CharPol}_A(T : B \to B) \\ & = \text{CharPol}_A(T : eB \to eB) \text{CharPol}_A(T : (1 - e)B \to (1 - e)B) \end{align*}\] is a factorization into monic polynomials reducing to the given factorization modulo \(I\). Here \(\text{CharPol}_A\) denotes the characteristic polynomial of an endomorphism of a finite locally free module over \(A\). If the module is free the \(\text{CharPol}_A\) is defined as the characteristic polynomial of the corresponding matrix and in general one uses Algebra, Lemma 00EJ to glue. Details omitted. Thus (3) implies (1).

Assume (1). Let \(f\) be as in (5). The factorization of \(f \bmod I\) as \(T^n\) times \(T - 1\) lifts to a factorization \(f = gh\) with \(g\) and \(h\) monic by Definition 09XE. Then \(h\) has to have degree \(1\) and we see that \(f\) has a root reducing to \(1\) modulo \(I\). Finally, \(I\) is contained in the Jacobson radical by the definition of a henselian pair. Thus (1) implies (5).

Before we give the proof of the last step, let us show that the root \(\alpha\) in (5), if it exists, is unique. Namely, due to the explicit shape of \(f(T)\), we have \(f'(\alpha) \in 1 + I\) where \(f'\) is the derivative of \(f\) with respect to \(T\). An elementary argument shows that \[f(T) = f(\alpha + T - \alpha) = f(\alpha) + f'(\alpha) \cdot (T - \alpha) \bmod (T - \alpha)^2 A[T]\] This shows that any other root \(\alpha' \in 1 + I\) of \(f(T)\) satisfies \(0 = f(\alpha') - f(\alpha) = (\alpha' - \alpha)(1 + i)\) for some \(i \in I\), so that, since \(1 + i\) is a unit in \(A\), we have \(\alpha = \alpha'\).

Assume (5). We will show that (2) holds, in other words, that for every étale map \(A \to A'\), every section \(\sigma : A' \to A/I\) modulo \(I\) lifts to a section \(A' \to A\). Since \(A \to A'\) is étale, the section \(\sigma\) determines a decomposition [0EM1]\[\begin{equation} A'/IA' \cong A/I \times C \end{equation}\] of \(A/I\)-algebras. Namely, the surjective ring map \(A'/IA' \to A/I\) is étale by Algebra, Lemma 00U7 and then we get the desired idempotent by Algebra, Lemma 00U8. We will show that this decomposition lifts to a decomposition [0EM2]\[\begin{equation} A' \cong A'_1 \times A'_2 \end{equation}\] of \(A\)-algebras with \(A'_1\) integral over \(A\). Then \(A \to A'_1\) is integral and étale and \(A/I \to A'_1/IA'_1\) is an isomorphism, thus \(A \to A'_1\) is an isomorphism by Lemma 0ELZ (here we also use that an étale ring map is flat and of finite presentation, see Algebra, Lemma 00U2).

Let \(B'\) be the integral closure of \(A\) in \(A'\). By Lemma 09XH we may decompose [0EM3]\[\begin{equation} B'/IB' \cong A/I \times C' \end{equation}\] as \(A/I\)-algebras compatibly with (0EM1) and we may find \(b \in B'\) that lifts \((1, 0)\) such that \(B'_b \to A'_b\) is an isomorphism. If the decomposition (0EM3) lifts to a decomposition [0EM4]\[\begin{equation} B' \cong B'_1 \times B'_2 \end{equation}\] of \(A\)-algebras, then the induced decomposition \(A' = A'_1 \times A'_2\) will give the desired (0EM2): indeed, since \(b\) is a unit in \(B'_1\) (details omitted), we will have \(B'_1 \cong A'_1\), so that \(A'_1\) will be integral over \(A\).

Choose a finite \(A\)-subalgebra \(B'' \subset B'\) containing \(b\) (observe that any finitely generated \(A\)-subalgebra of \(B'\) is finite over \(A\)). After enlarging \(B''\) we may assume \(b\) maps to an idempotent in \(B''/IB''\) producing [0EM5]\[\begin{equation} B''/IB'' \cong C''_1 \times C''_2 \end{equation}\] Since \(B'_b \cong A'_b\) we see that \(B'_b\) is of finite type over \(A\). Say \(B'_b\) is generated by \(b_1/b^n, \ldots, b_t/b^n\) over \(A\) and enlarge \(B''\) so that \(b_1, \ldots, b_t \in B''\). Then \(B''_b \to B'_b\) is surjective as well as injective, hence an isomorphism. In particular, we see that \(C''_1 = A/I\)! Therefore \(A/I \to C''_1\) is an isomorphism, in particular surjective. By Lemma 0EM0 we can find an \(f(T) \in A[T]\) of the form \[f(T) = T^n(T - 1) + a_n T^n + \ldots + a_1 T + a_0\] with \(a_n, \ldots, a_0 \in I\) and \(n \ge 1\) such that \(f(b) = 0\). In particular, we find that \(B'\) is a \(A[T]/(f)\)-algebra. By (5) we deduce there is a root \(a \in 1 + I\) of \(f\). This produces a product decomposition \(A[T]/(f) = A[T]/(T - a) \times D\) compatible with the splitting (0EM3) of \(B'/IB'\). The induced splitting of \(B'\) is then a desired (0EM4).

Lemma

Let \(A\) be a ring. Let \(I, J \subset A\) be ideals with \(V(I) = V(J)\). Then \((A, I)\) is henselian if and only if \((A, J)\) is henselian.

Proof

For any integral ring map \(A \to B\) we see that \(V(IB) = V(JB)\). Hence idempotents of \(B/IB\) and \(B/JB\) are in bijective correspondence (Algebra, Lemma 00EE). It follows that \(B \to B/IB\) induces a bijection on sets of idempotents if and only if \(B \to B/JB\) induces a bijection on sets of idempotents. Thus we conclude by Lemma 09XI.

Lemma

Let \((A, I)\) be a henselian pair and let \(A \to B\) be an integral ring map. Then \((B, IB)\) is a henselian pair.

Proof

Immediate from the fourth characterization of henselian pairs in Lemma 09XI and the fact that the composition of integral ring maps is integral.

Lemma

Let \(I \subset J \subset A\) be ideals of a ring \(A\). The following are equivalent

1. \((A, I)\) and \((A/I, J/I)\) are henselian pairs, and

2. \((A, J)\) is an henselian pair.

Proof

Assume (1). Let \(B\) be an integral \(A\)-algebra. Consider the ring maps \[B \to B/IB \to B/JB\] By Lemma 09XI we find that both arrows induce bijections on idempotents. Hence so does the composition. Whence \((A, J)\) is a henselian pair by Lemma 09XI.

Conversely, assume (2) holds. Then \((A/I, J/I)\) is a henselian pair by Lemma 09XK. Let \(B\) be an integral \(A\)-algebra. Consider the ring maps \[B \to B/IB \to B/JB\] By Lemma 09XI we find that the composition and the second arrow induce bijections on idempotents. Hence so does the first arrow. It follows that \((A, I)\) is a henselian pair (by the lemma again).

Lemma

Let \(A\) be a ring and let \((A, I)\) and \((A, I')\) be henselian pairs. Then \((A, I + I')\) is an henselian pair.

Proof

By Lemma 09XK the pair \((A/I, (I' + I)/I)\) is henselian. Thus we get the conclusion from Lemma 0DYD.

Lemma

Let \(J\) be a set and let \(\{ (A_j, I_j)\}_{j \in J}\) be a collection of pairs. Then \((\prod_{j \in J} A_j, \prod_{j\in J} I_j)\) is Henselian if and only if so is each \((A_j, I_j)\).

Proof

For every \(j \in J\), the projection \(\prod_{j \in J} A_j \rightarrow A_j\) is an integral ring map, so Lemma 09XK proves that each \((A_j, I_j)\) is Henselian if \((\prod_{j \in J} A_j, \prod_{j\in J} I_j)\) is Henselian.

Conversely, suppose that each \((A_j, I_j)\) is a Henselian pair. Then every \(1 + x\) with \(x \in \prod_{j \in J} I_j\) is a unit in \(\prod_{j \in J} A_j\) because it is so componentwise by Algebra, Lemma 0AME and Definition 09XE. Thus, by Algebra, Lemma 0AME again, \(\prod_{j \in J} I_j\) is contained in the Jacobson radical of \(\prod_{j \in J} A_j\). Continuing to work componentwise, it likewise follows that for every monic \(f \in (\prod_{j \in J} A_j)[T]\) and every factorization \(\overline{f} = g_0h_0\) with monic \(g_0, h_0 \in (\prod_{j \in J} A_j / \prod_{j \in J} I_j)[T] = (\prod_{j \in J} A_j/I_j)[T]\) that generate the unit ideal in \((\prod_{j \in J} A_j / \prod_{j \in J} I_j)[T]\), there exists a factorization \(f = gh\) in \((\prod_{j \in J} A_j)[T]\) with \(g\), \(h\) monic and reducing to \(g_0\), \(h_0\). In conclusion, according to Definition 09XE \((\prod_{j \in J} A_j, \prod_{j\in J} I_j)\) is a Henselian pair.

Lemma

The property of being Henselian is preserved under limits of pairs. More precisely, let \(J\) be a preordered set and let \((A_j, I_j)\) be an inverse system of henselian pairs over \(J\). Then \(A = \lim A_j\) equipped with the ideal \(I = \lim I_j\) is a henselian pair \((A, I)\).

Proof

By Categories, Lemma 002N, we only need to consider products and equalizers. For products, the claim follows from Lemma 0ATD. Thus, consider an equalizer diagram \[\xymatrix{ (A, I) \ar[r] & (A', I') \ar@<1ex>[r]^{\varphi} \ar@<-1ex>[r]_{\psi} & (A'', I'') }\] in which the pairs \((A', I')\) and \((A'', I'')\) are henselian. To check that the pair \((A, I)\) is also henselian, we will use the Gabber’s criterion in Lemma 09XI. Every element of \(1 + I\) is a unit in \(A\) because, due to the uniqueness of the inverses of units, this may be checked in \((A', I')\). Thus \(I\) is contained in the Jacobson radical of \(A\), see Algebra, Lemma 0AME. Thus, let \[f(T) = T^{N - 1}(T - 1) + a_{N - 1} T^{N - 1} + \dotsb + a_1 T + a_0\] be a polynomial in \(A[T]\) with \(a_{N - 1}, \dotsc, a_0 \in I\) and \(N \ge 1\). The image of \(f(T)\) in \(A'[T]\) has a unique root \(\alpha' \in 1 + I'\) and likewise for the further image in \(A''[T]\). Thus, due to the uniqueness, \(\varphi(\alpha') = \psi(\alpha')\), to the effect that \(\alpha'\) defines a root of \(f(T)\) in \(1 + I\), as desired.

Lemma

The property of being Henselian is preserved under filtered colimits of pairs. More precisely, let \(J\) be a directed set and let \((A_j, I_j)\) be a system of henselian pairs over \(J\). Then \(A = \colim A_j\) equipped with the ideal \(I = \colim I_j\) is a henselian pair \((A, I)\).

Proof

If \(u \in 1 + I\) then for some \(j \in J\) we see that \(u\) is the image of some \(u_j \in 1 + I_j\). Then \(u_j\) is invertible in \(A_j\) by Algebra, Lemma 0AME and the assumption that \(I_j\) is contained in the Jacobson radical of \(A_j\). Hence \(u\) is invertible in \(A\). Thus \(I\) is contained in the Jacobson radical of \(A\) (by the lemma).

Let \(f \in A[T]\) be a monic polynomial and let \(\overline{f} = g_0 h_0\) be a factorization with \(g_0, h_0 \in A/I[T]\) monic generating the unit ideal in \(A/I[T]\). Write \(1 = g_0 g'_0 + h_0 h'_0\) for some \(g'_0, h'_0 \in A/I[T]\). Since \(A = \colim A_j\) and \(A/I = \colim A_j/I_j\) are filtered colimits we can find a \(j \in J\) and \(f_j \in A_j\) and a factorization \(\overline{f}_j = g_{j, 0} h_{j, 0}\) with \(g_{j, 0}, h_{j, 0} \in A_j/I_j[T]\) monic and \(1 = g_{j, 0} g'_{j, 0} + h_{j, 0} h'_{j, 0}\) for some \(g'_{j, 0}, h'_{j, 0} \in A_j/I_j[T]\) with \(f_j, g_{j, 0}, h_{j, 0}, g'_{j, 0}, h'_{j, 0}\) mapping to \(f, g_0, h_0, g'_0, h'_0\). Since \((A_j, I_j)\) is a henselian pair, we can lift \(\overline{f}_j = g_{j, 0} h_{j, 0}\) to a factorization over \(A_j\) and taking the image in \(A\) we obtain a corresponding factorization in \(A\). Hence \((A, I)\) is henselian.

Example

Lemma 0FWT is wrong if the colimit isn’t filtered. For example, if we take the coproduct of the henselian pairs \((\mathbf{Z}_p, (p))\) and \((\mathbf{Z}_p, (p))\), then we obtain \((A, pA)\) with \(A = \mathbf{Z}_p \otimes_\mathbf{Z} \mathbf{Z}_p\). This isn’t a henselian pair: \(A/pA = \mathbf{F}_p\) hence if \((A, pA)\) where henselian, then \(A\) would have to be local. However, \(\Spec(A)\) is disconnected; for example for odd primes \(p\) we have the nontrivial idempotent \[(1/2 \otimes 1) \left(1 \otimes 1 - (1 + p)^{-1}u \otimes u\right)\] where \(u \in \mathbf{Z}_p\) is a square root of \(1 + p\). Some details omitted.

Lemma

Let \(A\) be a ring. There exists a largest ideal \(I \subset A\) such that \((A, I)\) is a henselian pair.

Proof

Combine Lemmas 0DYD, 0G1R, and 0FWT.

Lemma

Let \((A, I)\) be a henselian pair. Let \(\mathfrak p \subset A\) be a prime ideal. Then \(V(\mathfrak p + I)\) is connected.

Proof

By Lemma 09XK we see that \((A/\mathfrak p, I + \mathfrak p/\mathfrak p)\) is a henselian pair. Thus it suffices to prove: If \((A, I)\) is a henselian pair and \(A\) is a domain, then \(\Spec(A/I) = V(I)\) is connected. If not, then \(A/I\) has a nontrivial idempotent by Algebra, Lemma 00EF. By Lemma 09XI this would imply \(A\) has a nontrivial idempotent. This is a contradiction.

Henselization of pairs

We continue the discussion started in Section 09XD.

Lemma

The inclusion functor \[\text{category of henselian pairs} \longrightarrow \text{category of pairs}\] has a left adjoint \((A, I) \mapsto (A^h, I^h)\).

Proof

Let \((A, I)\) be a pair. Consider the category \(\mathcal{C}\) consisting of étale ring maps \(A \to B\) such that \(A/I \to B/IB\) is an isomorphism. We will show that the category \(\mathcal{C}\) is directed and that \(A^h = \colim_{B \in \mathcal{C}} B\) with ideal \(I^h = IA^h\) gives the desired adjoint.

We first prove that \(\mathcal{C}\) is directed (Categories, Definition 002V). It is nonempty because \(\text{id} : A \to A\) is an object. If \(B\) and \(B'\) are two objects of \(\mathcal{C}\), then \(B'' = B \otimes_A B'\) is an object of \(\mathcal{C}\) (use Algebra, Lemma 00U2) and there are morphisms \(B \to B''\) and \(B' \to B''\). Suppose that \(f, g : B \to B'\) are two maps between objects of \(\mathcal{C}\). Then a coequalizer is \[(B' \otimes_{f, B, g} B') \otimes_{(B' \otimes_A B')} B'\] which is étale over \(A\) by Algebra, Lemmas 00U2 and 00U7. Thus the category \(\mathcal{C}\) is directed.

Since \(B/IB = A/I\) for all objects \(B\) of \(\mathcal{C}\) we see that \(A^h/I^h = A^h/IA^h = \colim B/IB = \colim A/I = A/I\).

Next, we show that \(A^h = \colim_{B \in \mathcal{C}} B\) with \(I^h = IA^h\) is a henselian pair. To do this we will verify condition (2) of Lemma 09XI. Namely, suppose given an étale ring map \(A^h \to A'\) and \(A^h\)-algebra map \(\sigma : A' \to A^h/I^h\). Then there exists a \(B \in \mathcal{C}\) and an étale ring map \(B \to B'\) such that \(A' = B' \otimes_B A^h\). See Algebra, Lemma 00U2. Since \(A^h/I^h = A/I\), the map \(\sigma\) induces an \(A\)-algebra map \(s : B' \to A/I\). Then \(B'/IB' = A/I \times C\) as \(A/I\)-algebra, where \(C\) is the kernel of the map \(B'/IB' \to A/I\) induced by \(s\). Let \(g \in B'\) map to \((1, 0) \in A/I \times C\). Then \(B \to B'_g\) is étale and \(A/I \to B'_g/IB'_g\) is an isomorphism, i.e., \(B'_g\) is an object of \(\mathcal{C}\). Thus we obtain a canonical map \(B'_g \to A^h\) such that \[\vcenter{ \xymatrix{ B'_g \ar[r] & A^h \\ B \ar[u] \ar[ur] } } \quad\text{and}\quad \vcenter{ \xymatrix{ B' \ar[r] \ar[rrd]_s & B'_g \ar[r] & A^h \ar[d] \\ & & A/I } }\] commute. This induces a map \(A' = B' \otimes_B A^h \to A^h\) compatible with \(\sigma\) as desired.

Let \((A, I) \to (A', I')\) be a morphism of pairs with \((A', I')\) henselian. We will show there is a unique factorization \(A \to A^h \to A'\) which will finish the proof. Namely, for each \(A \to B\) in \(\mathcal{C}\) the ring map \(A' \to B' = A' \otimes_A B\) is étale and induces an isomorphism \(A'/I' \to B'/I'B'\). Hence there is a section \(\sigma_B : B' \to A'\) by Lemma 09XI. Given a morphism \(B_1 \to B_2\) in \(\mathcal{C}\) we claim the diagram \[\xymatrix{ B'_1 \ar[rr] \ar[rd]_{\sigma_{B_1}} & & B'_2 \ar[ld]^{\sigma_{B_2}} \\ & A' }\] commutes. This follows once we prove that for every \(B\) in \(\mathcal{C}\) the section \(\sigma_B\) is the unique \(A'\)-algebra map \(B' \to A'\). We have \(B' \otimes_{A'} B' = B' \times R\) for some ring \(R\), see Algebra, Lemma 02FL. In our case \(R/I'R = 0\) as \(B'/I'B' = A'/I'\). Thus given two \(A'\)-algebra maps \(\sigma_B, \sigma_B' : B' \to A'\) then \(e = (\sigma_B \otimes \sigma_B')(0, 1) \in A'\) is an idempotent contained in \(I'\). We conclude that \(e = 0\) by Lemma 09XF. Hence \(\sigma_B = \sigma_B'\) as desired. Using the commutativity we obtain \[A^h = \colim_{B \in \mathcal{C}} B \to \colim_{B \in \mathcal{C}} A' \otimes_A B \xrightarrow{\colim \sigma_B} A'\] as desired. The uniqueness of the maps \(\sigma_B\) also guarantees that this map is unique. Hence \((A, I) \mapsto (A^h, I^h)\) is the desired adjoint.

Lemma

Let \((A, I)\) be a pair. Let \((A^h, I^h)\) be as in Lemma 0A02. Then \(A \to A^h\) is flat, \(I^h = IA^h\) and \(A/I^n \to A^h/I^nA^h\) is an isomorphism for all \(n\).

Proof

In the proof of Lemma 0A02 we have seen that \(A^h\) is a filtered colimit of étale \(A\)-algebras \(B\) such that \(A/I \to B/IB\) is an isomorphism and we have seen that \(I^h = IA^h\). As an étale ring map is flat (Algebra, Lemma 00U2) we conclude that \(A \to A^h\) is flat by Algebra, Lemma 05UT. Since each \(A \to B\) is flat we find that the maps \(A/I^n \to B/I^nB\) are isomorphisms as well (for example by Algebra, Lemma 051H). Taking the colimit we find that \(A/I^n = A^h/I^nA^h\) as desired.

Lemma

The functor of Lemma 0A02 associates to a local ring \((A, \mathfrak m)\) its henselization.

Proof

Let \((A^h, \mathfrak m^h)\) be the henselization of the pair \((A, \mathfrak m)\) constructed in Lemma 0A02. Then \(\mathfrak m^h = \mathfrak m A^h\) is a maximal ideal by Lemma 0AGU and since it is contained in the Jacobson radical, we conclude \(A^h\) is local with maximal ideal \(\mathfrak m^h\). Having said this there are two ways to finish the proof.

First proof: observe that the construction in the proof of Algebra, Lemma 04GN as a colimit is the same as the colimit used to construct \(A^h\) in Lemma 0A02. Second proof: Both the henselization \(A \to S\) and \(A \to A^h\) of Lemma 0A02 are local ring homomorphisms, both \(S\) and \(A^h\) are filtered colimits of étale \(A\)-algebras, both \(S\) and \(A^h\) are henselian local rings, and both \(S\) and \(A^h\) have residue fields equal to \(\kappa(\mathfrak m)\) (by Lemma 0AGU for the second case). Hence they are canonically isomorphic by Algebra, Lemma 08HT.

Lemma

Let \((A, I)\) be a pair with \(A\) Noetherian. Let \((A^h, I^h)\) be as in Lemma 0A02. Then the map of \(I\)-adic completions \[A^\wedge \to (A^h)^\wedge\] is an isomorphism. Moreover, \(A^h\) is Noetherian, the maps \(A \to A^h \to A^\wedge\) are flat, and \(A^h \to A^\wedge\) is faithfully flat.

Proof

The first statement is an immediate consequence of Lemma 0AGU and in fact holds without assuming \(A\) is Noetherian. In the proof of Lemma 0A02 we have seen that \(A^h\) is a filtered colimit of étale \(A\)-algebras \(B\) such that \(A/I \to B/IB\) is an isomorphism. For each such \(A \to B\) the induced map \(A^\wedge \to B^\wedge\) is an isomorphism (see proof of Lemma 0AGU). By Algebra, Lemma 00MB the ring map \(B \to A^\wedge = B^\wedge = (A^h)^\wedge\) is flat for each \(B\). Thus \(A^h \to A^\wedge = (A^h)^\wedge\) is flat by Algebra, Lemma 05UU. Since \(I^h = IA^h\) is contained in the Jacobson radical of \(A^h\) and since \(A^h \to A^\wedge\) induces an isomorphism \(A^h/I^h \to A/I\) we see that \(A^h \to A^\wedge\) is faithfully flat by Algebra, Lemma 00HP. By Algebra, Lemma 0316 the ring \(A^\wedge\) is Noetherian. Hence we conclude that \(A^h\) is Noetherian by Algebra, Lemma 033E.

Lemma

Let \((A, I) = \colim (A_i, I_i)\) be a filtered colimit of pairs. The functor of Lemma 0A02 gives \(A^h = \colim A_i^h\) and \(I^h = \colim I_i^h\).

This lemma is false for non-filtered colimits, see Example 0FWU.

Proof

By Categories, Lemma 0038 we see that \((A^h, I^h)\) is the colimit of the system \((A_i^h, I_i^h)\) in the category of henselian pairs. Thus for a henselian pair \((B, J)\) we have \[\Mor((A^h, I^h), (B, J)) = \lim \Mor((A_i^h, I_i^h), (B, J)) = \Mor(\colim (A_i^h, I_i^h), (B, J))\] Here the colimit is in the category of pairs. Since the colimit is filtered we obtain \(\colim (A_i^h, I_i^h) = (\colim A_i^h, \colim I_i^h)\) in the category of pairs; details omitted. Again using the colimit is filtered, this is a henselian pair (Lemma 0FWT). Hence by the Yoneda lemma we find \((A^h, I^h) = (\colim A_i^h, \colim I_i^h)\).

Lemma

Let \(A\) be a ring with ideals \(I\) and \(J\). If \(V(I) = V(J)\) then the functor of Lemma 0A02 produces the same ring for the pair \((A, I)\) as for the pair \((A, J)\).

Proof

Let \((A', IA')\) be the pair produced by Lemma 0A02 starting with the pair \((A, I)\), see Lemma 0AGU. Let \((A'', JA'')\) be the pair produced by Lemma 0A02 starting with the pair \((A, J)\). By Lemma 09XJ we see that \((A', JA')\) is a henselian pair and \((A'', IA'')\) is a henselian pair. By the universal property of the construction we obtain unique \(A\)-algebra maps \(A'' \to A'\) and \(A' \to A''\). The uniqueness shows that these are mutually inverse.

Lemma

Let \((A, I) \to (B, J)\) be a map of pairs such that \(V(J) = V(IB)\). Let \((A^h , I^h) \to (B^h, J^h)\) be the induced map on henselizations (Lemma 0A02). If \(A \to B\) is integral, then the induced map \(A^h \otimes_A B \to B^h\) is an isomorphism.

Proof

By Lemma 0F0L we may assume \(J = IB\). By Lemma 09XK the pair \((A^h \otimes_A B, I^h(A^h \otimes_A B))\) is henselian. By the universal property of \((B^h, IB^h)\) we obtain a map \(B^h \to A^h \otimes_A B\). We omit the proof that this map is the inverse of the map in the lemma.

Lemma

Let \(I_1, I_2, \dots, I_n \subset A\) be ideals in a ring \(A\). Suppose \(I_i + I_j = A\) for \(i \not = j\), \(1 \leq i, j \leq n\). Denote \(A^h\) the henselization of the pair \((A, I_1 \cap \ldots \cap I_n)\) and \(A_i^h\) the henselization of the pair \((A, I_i)\). Then the natural morphism \[A^h \to \prod\nolimits_{i = 1, \ldots, n} A^h_i\] is an isomorphism.

Proof

The map comes from the functoriality of the construction. If \(n = 1\) the result is immediate. If \(n > 1\), observe that \((I_1 \cap \ldots \cap I_{n - 1}) + I_n = A\) by our condition on the ideals. Thus if we prove the lemma for \(n = 2\), then the lemma follows for all \(n\) by induction.

The case \(n = 2\). Choose \(f_i \in I_i\) such that \(f_1 + f_2 = 1\). Recall that \(A^h\) is constructed as the colimit of \(A\)-algebras \(B\) such that \(A \to B\) is étale and \(A/I_1 \cap I_2 \to B/(I_1 \cap I_2)B\) is an isomorphism. See proof of Lemma 0A02. Denote \(Hens\) the category of such \(A\)-algebras \(B\). Similarly, denote \(Hens_i\), \(i = 1, 2\) the category of étale \(A\)-algebras \(B\) such that \(A/I_i \to B/I_iB\) is an isomorphism.

Consider the subcategory \(Hens_1(f_2) \subset Hens_1\) consisting of those \(B\) such that \(f_2\) is invertible. This is a cofinal subcategory since if \(B\) is in \(Hens_1\), then \(B_{f_2}\) is in \(Hens_1\) as well (hint: \(B_{f_2}/I_1B_{f_2} = (B/I_1B)_{f_2} = B/I_1B\) because \(f_2\) is a unit modulo \(I_1\)). Similarly, the subcategory \(Hens_2(f_1) \subset Hens_2\) is cofinal. Hence to compute \(A_1^h\) and \(A_2^h\) we may work with this smaller categories, see Categories, Lemma 04E7.

Consider the functor \[Hens_1(f_2) \times Hens_2(f_1) \to Hens,\quad (B_1, B_2) \mapsto B_1 \times B_2\] This functor makes sense as \[\begin{align*} (B_1 \times B_2)/(I_1 \cap I_2)(B_1 \times B_2) & = B_1/(I_1 \cap I_2)B_1 \times B_2/(I_1 \cap I_2)B_2 \\ & = B_1/I_1B_1 \times B_2/I_2B_2 \\ & = A/I_1 \times A/I_2 \\ & = A/(I_1 \cap I_2) \end{align*}\] The second equality holds because \((I_1 \cap I_2)B_1 = I_1B_1 \cap I_2B_1 = I_1B_1\) as \(f_2 \in I_2\) is invertible in \(B_1\) (we have also used Algebra, Lemma 0BBY). The last equality holds by Algebra, Lemma 00DT. To finish the proof, it suffices to prove that the displayed functor is cofinal (see lemma quoted above). The first condition of Categories, Definition 04E6 holds because an object \(B\) of \(Hens\) maps to \(B_{f_2} \times B_{f_1}\) which is in the image of the functor. The second one always holds because if we have \(B \to B_1 \times B_2\) and \(B \to B'_1 \times B'_2\) with \(B_1, B'_1 \in Hens_1(f_2)\) and \(B_2, B'_2 \in Hens_2(f_1)\), then we can set \(B''_1 = B_1 \otimes_B B'_1\) which is in \(Hens_1(f_2)\) and similarly \(B''_2 = B_2 \otimes_B B''_2\) in \(Hens_2(f_1)\) to find a map \(B \to B''_1 \times B''_2\) fitting into the commutative diagram \[\xymatrix{ & B \ar[ld] \ar[d] \ar[rd] \\ B_1 \times B_2 \ar[r] & B_1'' \times B_2'' & B_1' \times B_2' \ar[l] }\] This finishes the proof.

Lifting and henselian pairs

In this section we mostly combine results from Sections 07LW and 09XD.

Lemma

Let \((R, I)\) be a henselian pair. The map \[P \longrightarrow P/IP\] induces a bijection between the sets of isomorphism classes of finite projective \(R\)-modules and finite projective \(R/I\)-modules. In particular, any finite projective \(R/I\)-module is isomorphic to \(P/IP\) for some finite projective \(R\)-module \(P\).

Proof

We first prove the final statement. Let \(\overline{P}\) be a finite projective \(R/I\)-module. We can find a finite projective module \(P'\) over some \(R'\) étale over \(R\) with \(R/I = R'/IR'\) such that \(P'/IP'\) is isomorphic to \(\overline{P}\), see Lemma 07M5. Then, since \((R, I)\) is a henselian pair, the étale ring map \(R \to R'\) has a section \(\tau : R' \to R\) (Lemma 09XI). Setting \(P = P' \otimes_{R', \tau} R\) we conclude that \(P/IP\) is isomorphic to \(\overline{P}\). Of course, this tells us that the map in the statement of the lemma is surjective.

Injectivity. Suppose that \(P_1\) and \(P_2\) are finite projective \(R\)-modules such that \(P_1/IP_1 \cong P_2/IP_2\) as \(R/I\)-modules. Since \(P_1\) is projective, we can find an \(R\)-module map \(u : P_1 \to P_2\) lifting the given isomorphism. Then \(u\) is surjective by Nakayama’s lemma (Algebra, Lemma 00DV). We similarly find a surjection \(v : P_2 \to P_1\). By Algebra, Lemma 05G8 the map \(v \circ u\) is an isomorphism and we conclude \(u\) is an isomorphism.

Lemma

Let \((A, I)\) be a henselian pair. The functor \(B \to B/IB\) determines an equivalence between finite étale \(A\)-algebras and finite étale \(A/I\)-algebras.

Proof

Let \(B, B'\) be two \(A\)-algebras finite étale over \(A\). Then \(B' \to B'' = B \otimes_A B'\) is finite étale as well (Algebra, Lemmas 00U2 and 02JK). Now we have \(1\)-to-\(1\) correspondences between

1. \(A\)-algebra maps \(B \to B'\),

2. sections of \(B' \to B''\), and

3. idempotents \(e\) of \(B''\) such that \(B' \to B'' \to eB''\) is an isomorphism.

The bijection between (2) and (3) sends \(\sigma : B'' \to B'\) to \(e\) such that \((1 - e)\) is the idempotent that generates the kernel of \(\sigma\) which exists by Algebra, Lemmas 00U7 and 00U8. There is a similar correspondence between \(A/I\)-algebra maps \(B/IB \to B'/IB'\) and idempotents \(\overline{e}\) of \(B''/IB''\) such that \(B'/IB' \to B''/IB'' \to \overline{e}(B''/IB'')\) is an isomorphism. However every idempotent \(\overline{e}\) of \(B''/IB''\) lifts uniquely to an idempotent \(e\) of \(B''\) (Lemma 09XI). Moreover, if \(B'/IB' \to \overline{e}(B''/IB'')\) is an isomorphism, then \(B' \to eB''\) is an isomorphism too by Nakayama’s lemma (Algebra, Lemma 00DV). In this way we see that the functor is fully faithful.

Essential surjectivity. Let \(A/I \to C\) be a finite étale map. By Algebra, Lemma 04D1 there exists an étale map \(A \to B\) such that \(B/IB \cong C\). Let \(B'\) be the integral closure of \(A\) in \(B\). By Lemma 09XH we have \(B'/IB' = C \times C'\) for some ring \(C'\) and \(B'_g \cong B_g\) for some \(g \in B'\) mapping to \((1, 0) \in C \times C'\). Since idempotents lift (Lemma 09XI) we get \(B' = B'_1 \times B'_2\) with \(C = B'_1/IB'_1\) and \(C' = B'_2/IB'_2\). The image of \(g\) in \(B'_1\) is invertible. Then \(B_g = B'_g = B'_1 \times (B_2)_g\) and this implies that \(A \to B'_1\) is étale. We conclude that \(B'_1\) is finite étale over \(A\) (integral étale implies finite étale by Algebra, Lemma 02JJ for example) and the proof is done.

Lemma

Let \(R\) be a ring and \(S\) a smooth \(R\)-algebra. Assume that \(A\) is an \(R\)-algebra and \((A,I)\) is a henselian pair. Then any \(R\)-algebra map \(S \to A/I\) can be lifted to an \(R\)-algebra map \(S \to A\).

Proof

Let \(\tau : S \to A/I\) be an \(R\)-algebra map. Observe that \(S \otimes_R A\) is a smooth \(A\)-algebra by Algebra, Lemma 00T4. Thus by Lemma 07M7 we can lift the induced map \(S \otimes_R A \to A/I\) to an \(A\)-algebra homorphism \(S \otimes_R A \to A'\) where \(A \to A'\) is étale and induces an isomorphism \(A/I \to A'/IA'\). Since \((A, I)\) is henselian there is an \(A\)-algebra map \(A' \to A\), see Lemma 09XI. The composition \(S \to S \otimes_R A \to A' \to A\) is the desired lift.

Lemma

Let \(A = \lim A_n\) be a limit of an inverse system \((A_n)\) of rings. Suppose given \(A_n\)-modules \(M_n\) and \(A_{n + 1}\)-module maps \(M_{n + 1} \to M_n\). Assume

1. the transition maps \(A_{n + 1} \to A_n\) are surjective with locally nilpotent kernels,

2. \(M_1\) is a finite projective \(A_1\)-module,

3. \(M_n\) is a finite flat \(A_n\)-module, and

4. the maps induce isomorphisms \(M_{n + 1} \otimes_{A_{n + 1}} A_n \to M_n\).

Then \(M = \lim M_n\) is a finite projective \(A\)-module and \(M \otimes_A A_n \to M_n\) is an isomorphism for all \(n\).

Proof

By Lemma 0CT7 the pair \((A, \Ker(A \to A_1))\) is henselian. By Lemma 0D4A we can choose a finite projective \(A\)-module \(P\) and an isomorphism \(P \otimes_A A_1 \to M_1\). Since \(P\) is projective, we can successively lift the \(A\)-module map \(P \to M_1\) to \(A\)-module maps \(P \to M_2\), \(P \to M_3\), and so on. Thus we obtain a map \[P \longrightarrow M\] Since \(P\) is finite projective, we can write \(A^{\oplus m} = P \oplus Q\) for some \(m \geq 0\) and \(A\)-module \(Q\). Since \(A = \lim A_n\) we conclude that \(P = \lim P \otimes_A A_n\). Hence, in order to show that the displayed \(A\)-module map is an isomorphism, it suffices to show that the maps \(P \otimes_A A_n \to M_n\) are isomorphisms. From Lemma 0D48 we see that \(M_n\) is a finite projective module. By Lemma 0BC6 the maps \(P \otimes_A A_n \to M_n\) are isomorphisms.

Absolute integral closure

Here is our definition.

Definition

A ring \(A\) is absolutely integrally closed if every monic \(f \in A[T]\) is a product of linear factors.

Be careful: it may be possible to write \(f\) as a product of linear factors in many different ways.

Lemma

Let \(A\) be a ring. The following are equivalent

1. \(A\) is absolutely integrally closed, and

2. any monic \(f \in A[T]\) has a root in \(A\).

Proof

Omitted.

Lemma

Let \(A\) be absolutely integrally closed.

1. Any quotient ring \(A/I\) of \(A\) is absolutely integrally closed.

2. Any localization \(S^{-1}A\) is absolutely integrally closed.

Proof

Omitted.

Lemma

Let \(A\) be a ring. Let \(S \subset A\) be a multiplicative subset consisting of nonzerodivisors. If \(S^{-1}A\) is absolutely integrally closed and \(A \subset S^{-1}A\) is integrally closed in \(S^{-1}A\), then \(A\) is absolutely integrally closed.

Proof

Omitted.

Lemma

Let \(A\) be a normal domain. Then \(A\) is absolutely integrally closed if and only if its fraction field is algebraically closed.

Proof

Observe that a field is algebraically closed if and only if it is absolutely integrally closed as a ring. Hence the lemma follows from Lemmas 0DCN and 0DCP.

Lemma

For any ring \(A\) there exists an extension \(A \subset B\) such that

1. \(B\) is a filtered colimit of finite free \(A\)-algebras,

2. \(B\) is free as an \(A\)-module, and

3. \(B\) is absolutely integrally closed.

Proof

Let \(I\) be the set of monic polynomials over \(A\). For \(i \in I\) denote \(x_i\) a variable and \(P_i\) the corresponding monic polynomial in the variable \(x_i\). Then we set \[F(A) = A[x_i; i \in I]/(P_i; i \in I)\] As the notation suggests \(F\) is a functor from the category of rings to itself. Note that \(A \subset F(A)\), that \(F(A)\) is free as an \(A\)-module, and that \(F(A)\) is a filtered colimit of finite free \(A\)-algebras. Then we take \[B = \colim F^n(A)\] where the transition maps are the inclusions \(F^n(A) \subset F(F^n(A)) = F^{n + 1}(A)\). Any monic polynomial with coefficients in \(B\) actually has coefficients in \(F^n(A)\) for some \(n\) and hence has a solution in \(F^{n + 1}(A)\) by construction. This implies that \(B\) is absolutely integrally closed by Lemma 0DCM. We omit the proof of the other properties.

Lemma

Let \(A\) be absolutely integrally closed. Let \(\mathfrak p \subset A\) be a prime. Then the local ring \(A_\mathfrak p\) is strictly henselian.

Proof

By Lemma 0DCN we may assume \(A\) is a local ring and \(\mathfrak p\) is its maximal ideal. The residue field is algebraically closed by Lemma 0DCN. Every monic polynomial decomposes completely into linear factors hence Algebra, Definition 04GF applies directly.

Lemma

Let \(A\) be absolutely integrally closed. Let \(I \subset A\) be an ideal. Then \((A, I)\) is a henselian pair if (and only if) the following conditions hold

1. \(I\) is contained in the Jacobson radical of \(A\),

2. \(A \to A/I\) induces a bijection on idempotents.

Proof

Let \(f \in A[T]\) be a monic polynomial and let \(f \bmod I = g_0 h_0\) be a factorization over \(A/I\) with \(g_0\), \(h_0\) monic such that \(g_0\) and \(h_0\) generate the unit ideal of \(A/I[T]\). This means that \[A/I[T]/(f) = A/I[T]/(g_0) \times A/I[T]/(h_0)\] Denote \(e \in A/I[T]/(f)\) the element correspoing to the idempotent \((1, 0)\) in the ring on the right. Write \(f = (T - a_1) \ldots (T - a_d)\) with \(a_i \in A\). For each \(i \in \{1, \ldots, d\}\) we obtain an \(A\)-algebra map \(\varphi_i : A[T]/(f) \to A\), \(T \mapsto a_i\) which induces a similar \(A/I\)-algebra map \(\overline{\varphi}_i : A/I[T]/(f) \to A/I\). Denote \(e_i = \overline{\varphi}_i(e) \in A/I\). These are idempotents. By our assumption (2) we can lift \(e_i\) to an idempotent in \(A\). This means we can write \(A = \prod A_j\) as a finite product of rings such that in \(A_j/IA_j\) each \(e_i\) is either \(0\) or \(1\). Some details omitted. Observe that \(A_j\) is absolutely integrally closed as a factor ring of \(A\). It suffices to lift the factorization of \(f\) over \(A_j/IA_j\) to \(A_j\). This reduces us to the situation discussed in the next paragraph.

Assume \(e_i = 1\) for \(i = 1, \ldots, r\) and \(e_i = 0\) for \(i = r + 1, \ldots, d\). From \((g_0, h_0) = A/I[T]\) we have that there are \(k_0, l_0 \in A/I[T]\) such that \(g_0 k_0 + h_0 l_0 = 1\). We see that \(e = h_0 l_0\) and \(e_i = h_0(a_i) l_0(a_i)\). We conclude that \(h_0(a_i)\) is a unit for \(i = 1, \dots ,r\). Since \(f(a_i) = 0\) we find \(0 = h_0(a_i)g_0(a_i)\) and we conclude that \(g_0(a_i) = 0\) for \(i = 1, \ldots, r\). Thus \((T - a_1)\) divides \(g_0\) in \(A/I[T]\), say \(g_0 = (T - a_1) g_0'\). Set \(f' = (T - a_2) \ldots (T - a_d)\) and \(h'_0 = h_0\). By induction on \(d\) we can lift the factorization \(f' \bmod I = g'_0 h'_0\) to a factorization of \(f' = g' h'\) over over \(A\) which gives the factorization \(f = (T - a_1) g' h'\) lifting the factorization \(f \bmod I = g_0 h_0\) as desired.

Auto-associated rings

Some of this material is in [Autour].

Definition

A ring \(R\) is said to be auto-associated if \(R\) is local and its maximal ideal \(\mathfrak m\) is weakly associated to \(R\).

Lemma

An auto-associated ring \(R\) has the following property: (P) Every proper finitely generated ideal \(I \subset R\) has a nonzero annihilator.

Proof

By assumption there exists a nonzero element \(x \in R\) such that for every \(f \in \mathfrak m\) we have \(f^n x = 0\). Say \(I = (f_1, \ldots, f_r)\). Then \(x\) is in the kernel of \(R \to \bigoplus R_{f_i}\). Hence we see that there exists a nonzero \(y \in R\) such that \(f_i y = 0\) for all \(i\), see Algebra, Lemma 0565. As \(y \in \text{Ann}_R(I)\) we win.

Lemma

Let \(R\) be a ring having property (P) of Lemma 05GN. Let \(u : N \to M\) be a homomorphism of projective \(R\)-modules. Then \(u\) is universally injective if and only if \(u\) is injective.

Proof

Assume \(u\) is injective. Our goal is to show \(u\) is universally injective. First we choose a module \(Q\) such that \(N \oplus Q\) is free. On considering the map \(N \oplus Q \to M \oplus Q\) we see that it suffices to prove the lemma in case \(N\) is free. In this case \(N\) is a directed colimit of finite free \(R\)-modules. Thus we reduce to the case that \(N\) is a finite free \(R\)-module, say \(N = R^{\oplus n}\). We prove the lemma by induction on \(n\). The case \(n = 0\) is trivial.

Let \(u : R^{\oplus n} \to M\) be an injective module map with \(M\) projective. Choose an \(R\)-module \(Q\) such that \(M \oplus Q\) is free. After replacing \(u\) by the composition \(R^{\oplus n} \to M \to M \oplus Q\) we see that we may assume that \(M\) is free. Then we can find a direct summand \(R^{\oplus m} \subset M\) such that \(u(R^{\oplus n}) \subset R^{\oplus m}\). Hence we may assume that \(M = R^{\oplus m}\). In this case \(u\) is given by a matrix \(A = (a_{ij})\) so that \(u(x_1, \ldots, x_n) = (\sum x_i a_{i1}, \ldots, \sum x_i a_{im})\). As \(u\) is injective, in particular \(u(x, 0, \ldots, 0) = (xa_{11}, xa_{12}, \ldots, xa_{1m}) \not = 0\) if \(x \not = 0\), and as \(R\) has property (P) we see that \(a_{11}R + a_{12}R + \ldots + a_{1m}R = R\). Hence see that \(R(a_{11}, \ldots, a_{1m}) \subset R^{\oplus m}\) is a direct summand of \(R^{\oplus m}\), in particular \(R^{\oplus m}/R(a_{11}, \ldots, a_{1m})\) is a projective \(R\)-module. We get a commutative diagram \[\xymatrix{ 0 \ar[r] & R \ar[rr] \ar[d]^1 & & R^{\oplus n} \ar[r] \ar[d]^u & R^{\oplus n - 1} \ar[r] \ar[d] & 0 \\ 0 \ar[r] & R \ar[rr]^{(a_{11}, \ldots, a_{1m})} & & R^{\oplus m} \ar[r] & R^{\oplus m}/R(a_{11}, \ldots, a_{1m}) \ar[r] & 0 }\] with split exact rows. Thus the right vertical arrow is injective and we may apply the induction hypothesis to conclude that the right vertical arrow is universally injective. It follows that the middle vertical arrow is universally injective.

Lemma

Let \(R\) be a ring. The following are equivalent

1. \(R\) has property (P) of Lemma 05GN,

2. any injective map of projective \(R\)-modules is universally injective,

3. if \(u : N \to M\) is injective and \(N\), \(M\) are finite projective \(R\)-modules then \(\Coker(u)\) is a finite projective \(R\)-module,

4. if \(N \subset M\) and \(N\), \(M\) are finite projective as \(R\)-modules, then \(N\) is a direct summand of \(M\), and

5. any injective map \(R \to R^{\oplus n}\) is a split injection.

Proof

The implication (1) \(\Rightarrow\) (2) is Lemma 05GP. It is clear that (3) and (4) are equivalent. We have (2) \(\Rightarrow\) (3), (4) by Algebra, Lemma 058L. Part (5) is a special case of (4). Assume (5). Let \(I = (a_1, \ldots, a_n)\) be a proper finitely generated ideal of \(R\). As \(I \not = R\) we see that \(R \to R^{\oplus n}\), \(x \mapsto (xa_1, \ldots, xa_n)\) is not a split injection. Hence it has a nonzero kernel and we conclude that \(\text{Ann}_R(I) \not = 0\). Thus (1) holds.

Example

If the equivalent conditions of Lemma 05GQ hold, then it is not always the case that every injective map of free \(R\)-modules is a split injection. For example suppose that \(R = k[x_1, x_2, x_3, \ldots]/(x_i^2)\). This is an auto-associated ring. Consider the map of free \(R\)-modules \[u : \bigoplus\nolimits_{i \geq 1} Re_i \longrightarrow \bigoplus\nolimits_{i \geq 1} Rf_i, \quad e_i \longmapsto f_i - x_if_{i + 1}.\] For any integer \(n\) the restriction of \(u\) to \(\bigoplus_{i = 1, \ldots, n} Re_i\) is injective as the images \(u(e_1), \ldots, u(e_n)\) are \(R\)-linearly independent. Hence \(u\) is injective and hence universally injective by the lemma. Since \(u \otimes \text{id}_k\) is bijective we see that if \(u\) were a split injection then \(u\) would be surjective. But \(u\) is not surjective because the inverse image of \(f_1\) would be the element \[\sum\nolimits_{i \geq 0} x_1 \ldots x_ie_{i + 1} = e_1 + x_1e_2 + x_1x_2e_3 + \ldots\] which is not an element of the direct sum. A side remark is that \(\Coker(u)\) is a flat (because \(u\) is universally injective), countably generated \(R\)-module which is not projective (as \(u\) is not split), hence not Mittag-Leffler (see Algebra, Lemma 059X).

The following lemma is a special case of Algebra, Proposition 00N1 in case the local ring is Noetherian.

Lemma

Let \((R, \mathfrak m)\) be a local ring. Suppose that \(\varphi : R^m \to R^n\) is a map of finite free modules. The following are equivalent

1. \(\varphi\) is injective,

2. the rank of \(\varphi\) is \(m\) and the annihilator of \(I(\varphi)\) in \(R\) is zero.

If \(R\) is Noetherian these are also equivalent to

1. the rank of \(\varphi\) is \(m\) and either \(I(\varphi) = R\) or it contains a nonzerodivisor.

Here the rank of \(\varphi\) and \(I(\varphi)\) are defined as in Algebra, Definition 00MV.

Proof

If any matrix coefficient of \(\varphi\) is not in \(\mathfrak m\), then we apply Algebra, Lemma 00MT to write \(\varphi\) as the sum of \(1 : R \to R\) and a map \(\varphi' : R^{m-1} \to R^{n-1}\). It is easy to see that the lemma for \(\varphi'\) implies the lemma for \(\varphi\). Thus we may assume from the outset that all the matrix coefficients of \(\varphi\) are in \(\mathfrak m\).

Suppose \(\varphi\) is injective. We may assume \(m > 0\). Let \(\mathfrak q \in \text{WeakAss}(R)\) so that \(R_\mathfrak q\) is an auto-associated ring. Then \(\varphi\) induces an injective map \(R_\mathfrak q^m \to R_\mathfrak q^n\) which is universally injective by Lemmas 05GN and 05GP. Thus \(\varphi : \kappa(\mathfrak q)^m \to \kappa(\mathfrak q)^n\) is injective. Hence the rank of \(\varphi \bmod \mathfrak q\) is \(m\) and \(I(\varphi \otimes \kappa(\mathfrak q))\) is not the zero ideal. Since \(m\) is the maximum rank \(\varphi\) can have, we conclude that \(\varphi\) has rank \(m\) as well (ranks of matrices can only drop after base change). Hence \(I(\varphi) \cdot \kappa(\mathfrak q) = I(\varphi \otimes \kappa(\mathfrak q))\) is not zero. Thus \(I(\varphi)\) is not contained in \(\mathfrak q\). Thus none of the weakly associated primes of \(R\) are weakly associated primes of the \(R\)-module \(\text{Ann}_R I(\varphi)\). Thus \(\text{Ann}_R I(\varphi)\) has no weakly associated primes, see Algebra, Lemma 0548. It follows from Algebra, Lemma 0588 that \(\text{Ann}_R I(\varphi)\) is zero.

Conversely, assume (2). The rank being \(m\) implies \(n \geq m\). Write \(I(\varphi) = (f_1, \ldots, f_r)\) which is possible as \(I(\varphi)\) is finitely generated. By Algebra, Lemma 07DQ we can find maps \(\psi_i : R^n \to R^m\) such that \(\psi \circ \varphi = f_i \text{id}_{R^m}\). Thus \(\varphi(x) = 0\) implies \(f_i x = 0\) for \(i = 1, \ldots, r\). This implies \(x = 0\) and hence \(\varphi\) is injective.

For the equivalence of (1) and (3) in the Noetherian local case we refer to Algebra, Proposition 00N1. If the ring \(R\) is Noetherian but not local, then the reader can deduce it from the local case; details omitted. Another option is to redo the argument above using associated primes, using that there are finitely many of these, using prime avoidance, and using the characterization of nonzerodivisors as elements of a Noetherian ring not contained in any associated prime.

Lemma

Let \(R\) be a ring. Suppose that \(\varphi : R^n \to R^n\) be an injective map of finite free modules of the same rank. Then \(\Hom_R(\Coker(\varphi), R) = 0\).

Proof

Let \(\varphi^t : R^n \to R^n\) be the transpose of \(\varphi\). The lemma claims that \(\varphi^t\) is injective. With notation as in Lemma 00MX we see that the rank of \(\varphi^t\) is \(n\) and that \(I(\varphi) = I(\varphi^t)\). Thus we conclude by the equivalence of (1) and (2) of the lemma.

Flattening stratification

Let \(R \to S\) be a ring map and let \(M\) be an \(S\)-module. For any \(R\)-algebra \(R'\) we can consider the base changes \(S' = S \otimes_R R'\) and \(M' = M \otimes_R R'\). We say \(R \to R'\) flattens \(M\) if the module \(M'\) is flat over \(R'\). We would like to understand the structure of the collection of ring maps \(R \to R'\) which flatten \(M\). In particular we would like to know if there exists a universal flattening \(R \to R_{univ}\) of \(M\), i.e., a ring map \(R \to R_{univ}\) which flattens \(M\) and has the property that any ring map \(R \to R'\) which flattens \(M\) factors through \(R \to R_{univ}\). It turns out that such a universal solution usually does not exist.

We will discuss universal flattenings and flattening stratifications in a scheme theoretic setting \(\mathcal{F}/X/S\) in More on Flatness, Section 052F. If the universal flattening \(R \to R_{univ}\) exists then the morphism of schemes \(\Spec(R_{univ}) \to \Spec(R)\) is the universal flattening of the quasi-coherent module \(\widetilde{M}\) on \(\Spec(S)\).

In this and the next few sections we prove some basic algebra facts related to this. The most basic result is perhaps the following.

Lemma

Let \(R\) be a ring. Let \(M\) be an \(R\)-module. Let \(I_1\), \(I_2\) be ideals of \(R\). If \(M/I_1M\) is flat over \(R/I_1\) and \(M/I_2M\) is flat over \(R/I_2\), then \(M/(I_1 \cap I_2)M\) is flat over \(R/(I_1 \cap I_2)\).

Proof

By replacing \(R\) with \(R/(I_1 \cap I_2)\) and \(M\) by \(M/(I_1 \cap I_2)M\) we may assume that \(I_1 \cap I_2 = 0\). Let \(J \subset R\) be an ideal. To prove that \(M\) is flat over \(R\) we have to show that \(J \otimes_R M \to M\) is injective, see Algebra, Lemma 00HD. By flatness of \(M/I_1M\) over \(R/I_1\) the map \[J/(J \cap I_1) \otimes_R M = (J + I_1)/I_1 \otimes_{R/I_1} M/I_1M \longrightarrow M/I_1M\] is injective. As \(0 \to (J \cap I_1) \to J \to J/(J \cap I_1) \to 0\) is exact we obtain a diagram \[\xymatrix{ (J \cap I_1) \otimes_R M \ar[r] \ar[d] & J \otimes_R M \ar[r] \ar[d] & J/(J \cap I_1) \otimes_R M \ar[r] \ar[d] & 0 \\ M \ar@{=}[r] & M \ar[r] & M/I_1M }\] hence it suffices to show that \((J \cap I_1) \otimes_R M \to M\) is injective. Since \(I_1 \cap I_2 = 0\) the ideal \(J \cap I_1\) maps isomorphically to an ideal \(J' \subset R/I_2\) and we see that \((J \cap I_1) \otimes_R M = J' \otimes_{R/I_2} M/I_2M\). By flatness of \(M/I_2M\) over \(R/I_2\) the map \(J' \otimes_{R/I_2} M/I_2M \to M/I_2M\) is injective, which clearly implies that \((J \cap I_1) \otimes_R M \to M\) is injective.

Flattening over an Artinian ring

A universal flattening exists when the base ring is an Artinian local ring. It exists for an arbitrary module. Hence, as we will see later, a flatting stratification exists when the base scheme is the spectrum of an Artinian local ring.

Lemma

Let \(R\) be an Artinian ring. Let \(M\) be an \(R\)-module. Then there exists a smallest ideal \(I \subset R\) such that \(M/IM\) is flat over \(R/I\).

Proof

This follows directly from Lemma 0522 and the Artinian property.

This ideal has the following universal property.

Lemma

Let \(R\) be an Artinian ring. Let \(M\) be an \(R\)-module. Let \(I \subset R\) be the smallest ideal \(I \subset R\) such that \(M/IM\) is flat over \(R/I\). Then \(I\) has the following universal property: For every ring map \(\varphi : R \to R'\) we have \[R' \otimes_R M\text{ is flat over }R' \Leftrightarrow \text{we have }\varphi(I) = 0.\]

Proof

Note that \(I\) exists by Lemma 0524. The implication \(\Rightarrow\) follows from Algebra, Lemma 00HI. Let \(\varphi : R \to R'\) be such that \(M \otimes_R R'\) is flat over \(R'\). Let \(J = \Ker(\varphi)\). By Algebra, Lemma 051L and as \(R' \otimes_R M = R' \otimes_{R/J} M/JM\) is flat over \(R'\) we conclude that \(M/JM\) is flat over \(R/J\). Hence \(I \subset J\) as desired.

Flattening over a closed subset of the base

Let \(R \to S\) be a ring map. Let \(I \subset R\) be an ideal. Let \(M\) be an \(S\)-module. In the following we will consider the following condition [052W]\[\begin{equation} \forall \mathfrak q \in V(IS) \subset \Spec(S) : M_{\mathfrak q}\text{ is flat over }R. \end{equation}\] Geometrically, this means that \(M\) is flat over \(R\) along the inverse image of \(V(I)\) in \(\Spec(S)\). If \(R\) and \(S\) are Noetherian rings and \(M\) is a finite \(S\)-module, then (052W) is equivalent to the condition that \(M/I^nM\) is flat over \(R/I^n\) for all \(n \geq 1\), see Algebra, Lemma 0523.

Lemma

Let \(R \to S\) be a ring map. Let \(I \subset R\) be an ideal. Let \(M\) be an \(S\)-module. Let \(R \to R'\) be a ring map and \(IR' \subset I' \subset R'\) an ideal. If (052W) holds for \((R \to S, I, M)\), then (052W) holds for \((R' \to S \otimes_R R', I', M \otimes_R R')\).

Proof

Assume (052W) holds for \((R \to S, I \subset R, M)\). Let \(I'(S \otimes_R R') \subset \mathfrak q'\) be a prime of \(S \otimes_R R'\). Let \(\mathfrak q \subset S\) be the corresponding prime of \(S\). Then \(IS \subset \mathfrak q\). Note that \((M \otimes_R R')_{\mathfrak q'}\) is a localization of the base change \(M_{\mathfrak q} \otimes_R R'\). Hence \((M \otimes_R R')_{\mathfrak q'}\) is flat over \(R'\) as a localization of a flat module, see Algebra, Lemmas 00HI and 00HT.

Lemma

Let \(R \to S\) be a ring map. Let \(I \subset R\) be an ideal. Let \(M\) be an \(S\)-module. Let \(R \to R'\) be a ring map and \(IR' \subset I' \subset R'\) an ideal such that

1. the map \(V(I') \to V(I)\) induced by \(\Spec(R') \to \Spec(R)\) is surjective, and

2. \(R'_{\mathfrak p'}\) is flat over \(R\) for all primes \(\mathfrak p' \in V(I')\).

If (052W) holds for \((R' \to S \otimes_R R', I', M \otimes_R R')\), then (052W) holds for \((R \to S, I, M)\).

Proof

Assume (052W) holds for \((R' \to S \otimes_R R', IR', M \otimes_R R')\). Pick a prime \(IS \subset \mathfrak q \subset S\). Let \(I \subset \mathfrak p \subset R\) be the corresponding prime of \(R\). By assumption there exists a prime \(\mathfrak p' \in V(I')\) of \(R'\) lying over \(\mathfrak p\) and \(R_{\mathfrak p} \to R'_{\mathfrak p'}\) is flat. Choose a prime \(\overline{\mathfrak q}' \subset \kappa(\mathfrak q) \otimes_{\kappa(\mathfrak p)} \kappa(\mathfrak p')\) which corresponds to a prime \(\mathfrak q' \subset S \otimes_R R'\) which lies over \(\mathfrak q\) and over \(\mathfrak p'\). Note that \((S \otimes_R R')_{\mathfrak q'}\) is a localization of \(S_{\mathfrak q} \otimes_{R_{\mathfrak p}} R'_{\mathfrak p'}\). By assumption the module \((M \otimes_R R')_{\mathfrak q'}\) is flat over \(R'_{\mathfrak p'}\). Hence Algebra, Lemma 00MQ implies that \(M_{\mathfrak q}\) is flat over \(R_{\mathfrak p}\) which is what we wanted to prove.

Lemma

Let \(R \to S\) be a ring map of finite presentation. Let \(M\) be an \(S\)-module of finite presentation. Let \(R' = \colim_{\lambda \in \Lambda} R_\lambda\) be a directed colimit of \(R\)-algebras. Let \(I_\lambda \subset R_\lambda\) be ideals such that \(I_\lambda R_\mu \subset I_\mu\) for all \(\mu \geq \lambda\) and set \(I' = \colim_\lambda I_\lambda\). If (052W) holds for \((R' \to S \otimes_R R', I', M \otimes_R R')\), then there exists a \(\lambda \in \Lambda\) such that (052W) holds for \((R_\lambda \to S \otimes_R R_\lambda, I_\lambda, M \otimes_R R_\lambda)\).

Proof

We are going to write \(S_\lambda = S \otimes_R R_\lambda\), \(S' = S \otimes_R R'\), \(M_\lambda = M \otimes_R R_\lambda\), and \(M' = M \otimes_R R'\). The base change \(S'\) is of finite presentation over \(R'\) and \(M'\) is of finite presentation over \(S'\) and similarly for the versions with subscript \(\lambda\), see Algebra, Lemma 05G5. By Algebra, Theorem 00RC the set \[U' = \{\mathfrak q' \in \Spec(S') \mid M'_{\mathfrak q'}\text{ is flat over }R'\}\] is open in \(\Spec(S')\). Note that \(V(I'S')\) is a quasi-compact space which is contained in \(U'\) by assumption. Hence there exist finitely many \(g'_j \in S'\), \(j = 1, \ldots, m\) such that \(D(g'_j) \subset U'\) and such that \(V(I'S') \subset \bigcup D(g'_j)\). Note that in particular \((M')_{g'_j}\) is a flat module over \(R'\).

We are going to pick increasingly large elements \(\lambda \in \Lambda\). First we pick it large enough so that we can find \(g_{j, \lambda} \in S_{\lambda}\) mapping to \(g'_j\). The inclusion \(V(I'S') \subset \bigcup D(g'_j)\) means that \(I'S' + (g'_1, \ldots, g'_m) = S'\) which can be expressed as \(1 = \sum z_sh_s + \sum f_jg'_j\) for some \(z_s \in I'\), \(h_s, f_j \in S'\). After increasing \(\lambda\) we may assume such an equation holds in \(S_\lambda\). Hence we may assume that \(V(I_\lambda S_\lambda) \subset \bigcup D(g_{j, \lambda})\). By Algebra, Lemma 02JO we see that for some sufficiently large \(\lambda\) the modules \((M_\lambda)_{g_{j, \lambda}}\) are flat over \(R_\lambda\). In particular the module \(M_\lambda\) is flat over \(R_\lambda\) at all the primes lying over the ideal \(I_\lambda\).

Flattening over a closed subsets of source and base

In this section we slightly generalize the discussion in Section 05LK. We strongly suggest the reader first read and understand that section.

Situation

Let \(R \to S\) be a ring map. Let \(J \subset S\) be an ideal. Let \(M\) be an \(S\)-module.

In this situation, given an \(R\)-algebra \(R'\) and an ideal \(I' \subset R'\) we set \(S' = S \otimes_R R'\) and \(M' = M \otimes_R R'\). We will consider the condition [05LQ]\[\begin{equation} \forall \mathfrak q' \in V(I'S' + JS') \subset \Spec(S') : M'_{\mathfrak q'}\text{ is flat over }R'. \end{equation}\] Geometrically, this means that \(M'\) is flat over \(R'\) along the intersection of the inverse image of \(V(I')\) with the inverse image of \(V(J)\). Since \((R \to S, J, M)\) are fixed, condition (05LQ) only depends on the pair \((R', I')\) where \(R'\) is viewed as an \(R\)-algebra.

Lemma

In Situation 05LP let \(R' \to R''\) be an \(R\)-algebra map. Let \(I' \subset R'\) and \(I'R'' \subset I'' \subset R''\) be ideals. If (05LQ) holds for \((R', I')\), then (05LQ) holds for \((R'', I'')\).

Proof

Assume (05LQ) holds for \((R', I')\). Let \(I''S'' + JS'' \subset \mathfrak q''\) be a prime of \(S''\). Let \(\mathfrak q' \subset S'\) be the corresponding prime of \(S'\). Then both \(I'S' \subset \mathfrak q'\) and \(JS' \subset \mathfrak q'\) because the corresponding conditions hold for \(\mathfrak q''\). Note that \((M'')_{\mathfrak q''}\) is a localization of the base change \(M'_{\mathfrak q'} \otimes_R R''\). Hence \((M'')_{\mathfrak q''}\) is flat over \(R''\) as a localization of a flat module, see Algebra, Lemmas 00HI and 00HT.

Lemma

In Situation 05LP let \(R' \to R''\) be an \(R\)-algebra map. Let \(I' \subset R'\) and \(I'R'' \subset I'' \subset R''\) be ideals. Assume

1. the map \(V(I'') \to V(I')\) induced by \(\Spec(R'') \to \Spec(R')\) is surjective, and

2. \(R''_{\mathfrak p''}\) is flat over \(R'\) for all primes \(\mathfrak p'' \in V(I'')\).

If (05LQ) holds for \((R'', I'')\), then (05LQ) holds for \((R', I')\).

Proof

Assume (05LQ) holds for \((R'', I'')\). Pick a prime \(I'S' + JS' \subset \mathfrak q' \subset S'\). Let \(I' \subset \mathfrak p' \subset R'\) be the corresponding prime of \(R'\). By assumption there exists a prime \(\mathfrak p'' \in V(I'')\) of \(R''\) lying over \(\mathfrak p'\) and \(R'_{\mathfrak p'} \to R''_{\mathfrak p''}\) is flat. Choose a prime \(\overline{\mathfrak q}'' \subset \kappa(\mathfrak q') \otimes_{\kappa(\mathfrak p')} \kappa(\mathfrak p'')\). This corresponds to a prime \(\mathfrak q'' \subset S'' = S' \otimes_{R'} R''\) which lies over \(\mathfrak q'\) and over \(\mathfrak p''\). In particular we see that \(I''S'' \subset \mathfrak q''\) and that \(JS'' \subset \mathfrak q''\). Note that \((S' \otimes_{R'} R'')_{\mathfrak q''}\) is a localization of \(S'_{\mathfrak q'} \otimes_{R'_{\mathfrak p'}} R''_{\mathfrak p''}\). By assumption the module \((M' \otimes_{R'} R'')_{\mathfrak q''}\) is flat over \(R''_{\mathfrak p''}\). Hence Algebra, Lemma 00MQ implies that \(M'_{\mathfrak q'}\) is flat over \(R'_{\mathfrak p'}\) which is what we wanted to prove.

Lemma

In Situation 05LP assume \(R \to S\) is essentially of finite presentation and \(M\) is an \(S\)-module of finite presentation. Let \(R' = \colim_{\lambda \in \Lambda} R_\lambda\) be a directed colimit of \(R\)-algebras. Let \(I_\lambda \subset R_\lambda\) be ideals such that \(I_\lambda R_\mu \subset I_\mu\) for all \(\mu \geq \lambda\) and set \(I' = \colim_\lambda I_\lambda\). If (05LQ) holds for \((R', I')\), then there exists a \(\lambda \in \Lambda\) such that (05LQ) holds for \((R_\lambda, I_\lambda)\).

Proof

We first prove the lemma in case \(R \to S\) is of finite presentation and then we explain what needs to be changed in the general case. We are going to write \(S_\lambda = S \otimes_R R_\lambda\), \(S' = S \otimes_R R'\), \(M_\lambda = M \otimes_R R_\lambda\), and \(M' = M \otimes_R R'\). The base change \(S'\) is of finite presentation over \(R'\) and \(M'\) is of finite presentation over \(S'\) and similarly for the versions with subscript \(\lambda\), see Algebra, Lemma 05G5. By Algebra, Theorem 00RC the set \[U' = \{\mathfrak q' \in \Spec(S') \mid M'_{\mathfrak q'}\text{ is flat over }R'\}\] is open in \(\Spec(S')\). Note that \(V(I'S' + JS')\) is a quasi-compact space which is contained in \(U'\) by assumption. Hence there exist finitely many \(g'_j \in S'\), \(j = 1, \ldots, m\) such that \(D(g'_j) \subset U'\) and such that \(V(I'S' + JS') \subset \bigcup D(g'_j)\). Note that in particular \((M')_{g'_j}\) is a flat module over \(R'\).

We are going to pick increasingly large elements \(\lambda \in \Lambda\). First we pick it large enough so that we can find \(g_{j, \lambda} \in S_{\lambda}\) mapping to \(g'_j\). The inclusion \(V(I'S' + JS') \subset \bigcup D(g'_j)\) means that \(I'S' + JS' + (g'_1, \ldots, g'_m) = S'\) which can be expressed as \[1 = \sum y_tk_t + \sum z_sh_s + \sum f_jg'_j\] for some \(z_s \in I'\), \(y_t \in J\), \(k_t, h_s, f_j \in S'\). After increasing \(\lambda\) we may assume such an equation holds in \(S_\lambda\). Hence we may assume that \(V(I_\lambda S_\lambda + J S_\lambda) \subset \bigcup D(g_{j, \lambda})\). By Algebra, Lemma 02JO we see that for some sufficiently large \(\lambda\) the modules \((M_\lambda)_{g_{j, \lambda}}\) are flat over \(R_\lambda\). In particular the module \(M_\lambda\) is flat over \(R_\lambda\) at all the primes corresponding to points of \(V(I_\lambda S_\lambda + J S_\lambda)\).

In the case that \(S\) is essentially of finite presentation, we can write \(S = \Sigma^{-1}C\) where \(R \to C\) is of finite presentation and \(\Sigma \subset C\) is a multiplicative subset. We can also write \(M = \Sigma^{-1}N\) for some finitely presented \(C\)-module \(N\), see Algebra, Lemma 05N5. At this point we introduce \(C_\lambda\), \(C'\), \(N_\lambda\), \(N'\). Then in the discussion above we obtain an open \(U' \subset \Spec(C')\) over which \(N'\) is flat over \(R'\). The assumption that (05LQ) is true means that \(V(I'S' + JS')\) maps into \(U'\), because for a prime \(\mathfrak q' \subset S'\), corresponding to a prime \(\mathfrak r' \subset C'\) we have \(M'_{\mathfrak q'} = N'_{\mathfrak r'}\). Thus we can find \(g'_j \in C'\) such that \(\bigcup D(g'_j)\) contains the image of \(V(I'S' + JS')\). The rest of the proof is exactly the same as before.

Lemma

In Situation 05LP. Let \(I \subset R\) be an ideal. Assume

1. \(R\) is a Noetherian ring,

2. \(S\) is a Noetherian ring,

3. \(M\) is a finite \(S\)-module, and

4. for each \(n \geq 1\) and any prime \(\mathfrak q \in V(J + IS)\) the module \((M/I^n M)_{\mathfrak q}\) is flat over \(R/I^n\).

Then (05LQ) holds for \((R, I)\), i.e., for every prime \(\mathfrak q \in V(J + IS)\) the localization \(M_{\mathfrak q}\) is flat over \(R\).

Proof

Let \(\mathfrak q \in V(J + IS)\). Then Algebra, Lemma 0523 applied to \(R \to S_{\mathfrak q}\) and \(M_{\mathfrak q}\) implies that \(M_{\mathfrak q}\) is flat over \(R\).

Flattening over a Noetherian complete local ring

The following three lemmas give a completely algebraic proof of the existence of the “local” flattening stratification when the base is a complete local Noetherian ring \(R\) and the given module is finite over a finite type \(R\)-algebra \(S\).

Lemma

Let \(R \to S\) be a ring map. Let \(M\) be an \(S\)-module. Assume

1. \((R, \mathfrak m)\) is a complete local Noetherian ring,

2. \(S\) is a Noetherian ring, and

3. \(M\) is finite over \(S\).

Then there exists an ideal \(I \subset \mathfrak m\) such that

1. \((M/IM)_{\mathfrak q}\) is flat over \(R/I\) for all primes \(\mathfrak q\) of \(S/IS\) lying over \(\mathfrak m\), and

2. if \(J \subset R\) is an ideal such that \((M/JM)_{\mathfrak q}\) is flat over \(R/J\) for all primes \(\mathfrak q\) lying over \(\mathfrak m\), then \(I \subset J\).

In other words, \(I\) is the smallest ideal of \(R\) such that (052W) holds for \((\overline{R} \to \overline{S}, \overline{\mathfrak m}, \overline{M})\) where \(\overline{R} = R/I\), \(\overline{S} = S/IS\), \(\overline{\mathfrak m} = \mathfrak m/I\) and \(\overline{M} = M/IM\).

Proof

Let \(J \subset R\) be an ideal. Apply Algebra, Lemma 0523 to the module \(M/JM\) over the ring \(R/J\). Then we see that \((M/JM)_{\mathfrak q}\) is flat over \(R/J\) for all primes \(\mathfrak q\) of \(S/JS\) if and only if \(M/(J + \mathfrak m^n)M\) is flat over \(R/(J + \mathfrak m^n)\) for all \(n \geq 1\). We will use this remark below.

For every \(n \geq 1\) the local ring \(R/\mathfrak m^n\) is Artinian. Hence, by Lemma 0524 there exists a smallest ideal \(I_n \supset \mathfrak m^n\) such that \(M/I_nM\) is flat over \(R/I_n\). It is clear that \(I_{n + 1} + \mathfrak m^n\) is contains \(I_n\) and applying Lemma 0522 we see that \(I_n = I_{n + 1} + \mathfrak m^n\). Since \(R = \lim_n\ R/\mathfrak m^n\) we see that \(I = \lim_n\ I_n/\mathfrak m^n\) is an ideal in \(R\) such that \(I_n = I + \mathfrak m^n\) for all \(n \geq 1\). By the initial remarks of the proof we see that \(I\) verifies (1) and (2). Some details omitted.

Lemma

With notation \(R \to S\), \(M\), and \(I\) and assumptions as in Lemma 0526. Consider a local homomorphism of local rings \(\varphi : (R, \mathfrak m) \to (R', \mathfrak m')\) such that \(R'\) is Noetherian. Then the following are equivalent

1. condition (052W) holds for \((R' \to S \otimes_R R', \mathfrak m', M \otimes_R R')\), and

2. \(\varphi(I) = 0\).

Proof

The implication (2) \(\Rightarrow\) (1) follows from Lemma 052X. Let \(\varphi : R \to R'\) be as in the lemma satisfying (1). We have to show that \(\varphi(I) = 0\). This is equivalent to the condition that \(\varphi(I)R' = 0\). By Artin-Rees in the Noetherian local ring \(R'\) (see Algebra, Lemma 00IP) this is equivalent to the condition that \(\varphi(I)R' + (\mathfrak m')^n = (\mathfrak m')^n\) for all \(n > 0\). Hence this is equivalent to the condition that the composition \(\varphi_n : R \to R' \to R'/(\mathfrak m')^n\) annihilates \(I\) for each \(n\). Now assumption (1) for \(\varphi\) implies assumption (1) for \(\varphi_n\) by Lemma 052X. This reduces us to the case where \(R'\) is Artinian local.

Assume \(R'\) Artinian. Let \(J = \Ker(\varphi)\). We have to show that \(I \subset J\). By the construction of \(I\) in Lemma 0526 it suffices to show that \((M/JM)_{\mathfrak q}\) is flat over \(R/J\) for every prime \(\mathfrak q\) of \(S/JS\) lying over \(\mathfrak m\). As \(R'\) is Artinian, condition (1) signifies that \(M \otimes_R R'\) is flat over \(R'\). As \(R'\) is Artinian and \(R/J \to R'\) is a local injective ring map, it follows that \(R/J\) is Artinian too. Hence the flatness of \(M \otimes_R R' = M/JM \otimes_{R/J} R'\) over \(R'\) implies that \(M/JM\) is flat over \(R/J\) by Algebra, Lemma 051L. This concludes the proof.

Lemma

With notation \(R \to S\), \(M\), and \(I\) and assumptions as in Lemma 0526. In addition assume that \(R \to S\) is of finite type. Then for any local homomorphism of local rings \(\varphi : (R, \mathfrak m) \to (R', \mathfrak m')\) the following are equivalent

1. condition (052W) holds for \((R' \to S \otimes_R R', \mathfrak m', M \otimes_R R')\), and

2. \(\varphi(I) = 0\).

Proof

The implication (2) \(\Rightarrow\) (1) follows from Lemma 052X. Let \(\varphi : R \to R'\) be as in the lemma satisfying (1). As \(R\) is Noetherian we see that \(R \to S\) is of finite presentation and \(M\) is an \(S\)-module of finite presentation. Write \(R' = \colim_\lambda R_\lambda\) as a directed colimit of local \(R\)-subalgebras \(R_\lambda \subset R'\), with maximal ideals \(\mathfrak m_\lambda = R_\lambda \cap \mathfrak m'\) such that each \(R_\lambda\) is essentially of finite type over \(R\). By Lemma 05LM we see that condition (052W) holds for \((R_\lambda \to S \otimes_R R_\lambda, \mathfrak m_\lambda, M \otimes_R R_\lambda)\) for some \(\lambda\). Hence Lemma 0527 applies to the ring map \(R \to R_\lambda\) and we see that \(I\) maps to zero in \(R_\lambda\), a fortiori it maps to zero in \(R'\).

Descent of flatness along integral maps

First a few simple lemmas.

Lemma

Let \(R\) be a ring. Let \(P(T)\) be a monic polynomial with coefficients in \(R\). Let \(\alpha \in R\) be such that \(P(\alpha) = 0\). Then \(P(T) = (T - \alpha)Q(T)\) for some monic polynomial \(Q(T) \in R[T]\).

Proof

By induction on the degree of \(P\). If \(\deg(P) = 1\), then \(P(T) = T - \alpha\) and the result is true. If \(\deg(P) > 1\), then we can write \(P(T) = (T - \alpha)Q(T) + r\) for some polynomial \(Q \in R[T]\) of degree \(< \deg(P)\) and some \(r \in R\) by long division. By assumption \(0 = P(\alpha) = (\alpha - \alpha)Q(\alpha) + r = r\) and we conclude that \(r = 0\) as desired.

Lemma

Let \(R\) be a ring. Let \(P(T)\) be a monic polynomial with coefficients in \(R\). There exists a finite free ring map \(R \to R'\) such that \(P(T) = (T - \alpha)Q(T)\) for some \(\alpha \in R'\) and some monic polynomial \(Q(T) \in R'[T]\).

Proof

Write \(P(T) = T^d + a_1T^{d - 1} + \ldots + a_0\). Set \(R' = R[x]/(x^d + a_1x^{d - 1} + \ldots + a_0)\). Set \(\alpha\) equal to the congruence class of \(x\). Then it is clear that \(P(\alpha) = 0\). Thus we win by Lemma 052Z.

Lemma

Let \(R \to S\) be a finite ring map. There exists a finite free ring extension \(R \subset R'\) such that \(S \otimes_R R'\) is a quotient of a ring of the form \[R'[T_1, \ldots, T_n]/(P_1(T_1), \ldots, P_n(T_n))\] with \(P_i(T) = \prod_{j = 1, \ldots, d_i} (T - \alpha_{ij})\) for some \(\alpha_{ij} \in R'\).

Proof

Let \(x_1, \ldots, x_n \in S\) be generators of \(S\) over \(R\). For each \(i\) we can choose a monic polynomial \(P_i(T) \in R[T]\) such that \(P_i(x_i) = 0\) in \(S\), see Algebra, Lemma 00GK. Say \(\deg(P_i) = d_i\). By Lemma 0530 (applied \(\sum d_i\) times) there exists a finite free ring extension \(R \subset R'\) such that each \(P_i\) splits completely: \[P_i(T) = \prod\nolimits_{j = 1, \ldots, d_i} (T - \alpha_{ij})\] for certain \(\alpha_{ik} \in R'\). Let \(R'[T_1, \ldots, T_n] \to S \otimes_R R'\) be the \(R'\)-algebra map which maps \(T_i\) to \(x_i \otimes 1\). As this maps \(P_i(T_i)\) to zero, this induces the desired surjection.

Lemma

Let \(R\) be a ring. Let \(S = R[T_1, \ldots, T_n]/J\). Assume \(J\) contains elements of the form \(P_i(T_i)\) with \(P_i(T) = \prod_{j = 1, \ldots, d_i} (T - \alpha_{ij})\) for some \(\alpha_{ij} \in R\). For \(\underline{k} = (k_1, \ldots, k_n)\) with \(1 \leq k_i \leq d_i\) consider the ring map \[\Phi_{\underline{k}} : R[T_1, \ldots, T_n] \to R, \quad T_i \longmapsto \alpha_{ik_i}\] Set \(J_{\underline{k}} = \Phi_{\underline{k}}(J)\). Then the image of \(\Spec(S) \to \Spec(R)\) is equal to \(V(\bigcap J_{\underline{k}})\).

Proof

This lemma proves itself. Hint: \(V(\bigcap J_{\underline{k}}) = \bigcup V(J_{\underline{k}})\).

The following result is due to Ferrand, see [Ferrand].

Lemma

Let \(R \to S\) be a finite injective homomorphism of Noetherian rings. Let \(M\) be an \(R\)-module. If \(M \otimes_R S\) is a flat \(S\)-module, then \(M\) is a flat \(R\)-module.

Proof

Let \(M\) be an \(R\)-module such that \(M \otimes_R S\) is flat over \(S\). By Algebra, Lemma 00HJ in order to prove that \(M\) is flat we may replace \(R\) by any faithfully flat ring extension. By Lemma 0531 we can find a finite locally free ring extension \(R \subset R'\) such that \(S' = S \otimes_R R' = R'[T_1, \ldots, T_n]/J\) for some ideal \(J \subset R'[T_1, \ldots, T_n]\) which contains the elements of the form \(P_i(T_i)\) with \(P_i(T) = \prod_{j = 1, \ldots, d_i} (T - \alpha_{ij})\) for some \(\alpha_{ij} \in R'\). Note that \(R'\) is Noetherian and that \(R' \subset S'\) is a finite extension of rings. Hence we may replace \(R\) by \(R'\) and assume that \(S\) has a presentation as in Lemma 0532. Note that \(\Spec(S) \to \Spec(R)\) is surjective, see Algebra, Lemma 00GQ. Thus, using Lemma 0532 we conclude that \(I = \bigcap J_{\underline{k}}\) is an ideal such that \(V(I) = \Spec(R)\). This means that \(I \subset \sqrt{(0)}\), and since \(R\) is Noetherian that \(I\) is nilpotent. The maps \(\Phi_{\underline{k}}\) induce commutative diagrams \[\xymatrix{ S \ar[rr] & & R/J_{\underline{k}} \\ & R \ar[lu] \ar[ru] }\] from which we conclude that \(M/J_{\underline{k}}M\) is flat over \(R/J_{\underline{k}}\). By Lemma 0522 we see that \(M/IM\) is flat over \(R/I\). Finally, applying Algebra, Lemma 051J we conclude that \(M\) is flat over \(R\).

Lemma

Let \(R \to S\) be an injective integral ring map. Let \(M\) be a finitely presented module over \(R[x_1, \ldots, x_n]\). If \(M \otimes_R S\) is flat over \(S\), then \(M\) is flat over \(R\).

Proof

Choose a presentation \[R[x_1, \ldots, x_n]^{\oplus t} \to R[x_1, \ldots, x_n]^{\oplus r} \to M \to 0.\] Let’s say that the first map is given by the \(r \times t\)-matrix \(T = (f_{ij})\) with \(f_{ij} \in R[x_1, \ldots, x_n]\). Write \(f_{ij} = \sum f_{ij, I} x^I\) with \(f_{ij, I} \in R\) (multi-index notation). Consider diagrams \[\xymatrix{ R \ar[r] & S \\ R_\lambda \ar[u] \ar[r] & S_\lambda \ar[u] }\] where \(R_\lambda\) is a finitely generated \(\mathbf{Z}\)-subalgebra of \(R\) containing all \(f_{ij, I}\) and \(S_\lambda\) is a finite \(R_\lambda\)-subalgebra of \(S\). Let \(M_\lambda\) be the finite \(R_\lambda[x_1, \ldots, x_n]\)-module defined by a presentation as above, using the same matrix \(T\) but now viewed as a matrix over \(R_\lambda[x_1, \ldots, x_n]\). Note that \(S\) is the directed colimit of the \(S_\lambda\) (details omitted). By Algebra, Lemma 02JO we see that for some \(\lambda\) the module \(M_\lambda \otimes_{R_\lambda} S_\lambda\) is flat over \(S_\lambda\). By Lemma 0533 we conclude that \(M_\lambda\) is flat over \(R_\lambda\). Since \(M = M_\lambda \otimes_{R_\lambda} R\) we win by Algebra, Lemma 00HI.

Lemma

Let \(R \to S\) be a finite injective homomorphism of Noetherian rings. Let \(P\) be an \(R\)-module. If \(P \otimes_R S\) is a projective \(S\)-module, then \(P\) is a projective \(R\)-module.

Proof

Let \(P\) be an \(R\)-module such that \(P \otimes_R S\) is projective over \(S\). By Algebra, Theorem 05A9 in order to prove that \(P\) is projective we may replace \(R\) by any faithfully flat ring extension. By Lemma 0531 we can find a finite locally free ring extension \(R \subset R'\) such that \(S' = S \otimes_R R' = R'[T_1, \ldots, T_n]/J\) for some ideal \(J \subset R'[T_1, \ldots, T_n]\) which contains the elements of the form \(P_i(T_i)\) with \(P_i(T) = \prod_{j = 1, \ldots, d_i} (T - \alpha_{ij})\) for some \(\alpha_{ij} \in R'\). Note that \(R'\) is Noetherian and that \(R' \subset S'\) is a finite extension of rings. Hence we may replace \(R\) by \(R'\) and assume that \(S\) has a presentation as in Lemma 0532. Note that \(\Spec(S) \to \Spec(R)\) is surjective, see Algebra, Lemma 00GQ. Thus, using Lemma 0532 we conclude that \(I = \bigcap J_{\underline{k}}\) is an ideal such that \(V(I) = \Spec(R)\). This means that \(I \subset \sqrt{(0)}\), and since \(R\) is Noetherian that \(I\) is nilpotent. The maps \(\Phi_{\underline{k}}\) induce commutative diagrams \[\xymatrix{ S \ar[rr] & & R/J_{\underline{k}} \\ & R \ar[lu] \ar[ru] }\] from which we conclude that \(P/J_{\underline{k}}P\) is projective over \(R/J_{\underline{k}}\). By Algebra, Lemma 0H7M we see that \(P/IP\) is projective over \(R/I\). Since \(P\) is flat over \(R\) by Lemma 0533, we may apply Algebra, Lemma 05CG to conclude that \(P\) is projective over \(R\).

Torsion free modules

In this section we discuss torsion free modules and the relationship with flatness (especially over dimension 1 rings).

Definition

Let \(R\) be a domain. Let \(M\) be an \(R\)-module.

1. We say an element \(x \in M\) is torsion if there exists a nonzero \(f \in R\) such that \(fx = 0\).

2. We say \(M\) is torsion free if the only torsion element of \(M\) is \(0\).

3. We say \(M\) is a torsion module if every element of \(M\) is torsion.

Let \(R\) be a domain and let \(S = R \setminus \{0\}\) be the multiplicative set of nonzero elements of \(R\). Then an \(R\)-module \(M\) is torsion free if and only if \(M \to S^{-1}M\) is injective. In other words, if and only if the map \(M \to M \otimes_R K\) is injective where \(K = S^{-1}R\) is the fraction field of \(R\).

Lemma

Let \(R\) be a domain. Let \(M\) be an \(R\)-module. The set of torsion elements \(M_{tors}\) of \(M\) is the kernel of the map \(M \to M \otimes_R K\). Thus \(M_{tors}\) is an \(R\)-submodule of \(M\). The quotient module \(M/M_{tors}\) is torsion free.

Proof

See discussion above.

Lemma

Let \(R\) be a domain. Let \(M\) be a torsion free \(R\)-module. For any multiplicative set \(S \subset R\) the module \(S^{-1}M\) is a torsion free \(S^{-1}R\)-module.

Proof

Omitted.

Lemma

Let \(R \to R'\) be a flat homomorphism of domains. If \(M\) is a torsion free \(R\)-module, then \(M \otimes_R R'\) is a torsion free \(R'\)-module.

Proof

If \(M\) is torsion free, then \(M \subset M \otimes_R K\) is injective where \(K\) is the fraction field of \(R\). Since \(R'\) is flat over \(R\) we see that \(M \otimes_R R' \to (M \otimes_R K) \otimes_R R'\) is injective. Since \(M \otimes_R K\) is isomorphic to a direct sum of copies of \(K\), it suffices to see that \(K \otimes_R R'\) is torsion free. This is true because it is a localization of \(R'\).

Lemma

Let \(R\) be a domain. Let \(0 \to M \to M' \to M'' \to 0\) be a short exact sequence of \(R\)-modules. If \(M\) and \(M''\) are torsion free, then \(M'\) is torsion free.

Proof

Omitted.

Lemma

Let \(R\) be a domain. Let \(M\) be an \(R\)-module. Then \(M\) is torsion free if and only if \(M_\mathfrak m\) is a torsion free \(R_\mathfrak m\)-module for all maximal ideals \(\mathfrak m\) of \(R\).

Proof

Omitted. Hint: Use Lemma 0AUR and Algebra, Lemma 00HN.

Lemma

Let \(R\) be a domain. Let \(M\) be a finite \(R\)-module. Then \(M\) is torsion free if and only if \(M\) is a submodule of a finite free module.

Proof

If \(M\) is a submodule of \(R^{\oplus n}\), then \(M\) is torsion free. For the converse, assume \(M\) is torsion free. Let \(K\) be the fraction field of \(R\). Then \(M \otimes_R K\) is a finite dimensional \(K\)-vector space. Choose a basis \(e_1, \ldots, e_r\) for this vector space. Let \(x_1, \ldots, x_n\) be generators of \(M\). Write \(x_i = \sum (a_{ij}/b_{ij}) e_j\) for some \(a_{ij}, b_{ij} \in R\) with \(b_{ij} \not = 0\). Set \(b = \prod_{i, j} b_{ij}\). Since \(M\) is torsion free the map \(M \to M \otimes_R K\) is injective and the image is contained in \(R^{\oplus r} = R e_1/b \oplus \ldots \oplus Re_r/b\).

Lemma

Let \(R\) be a Noetherian domain. Let \(M\) be a nonzero finite \(R\)-module. The following are equivalent

1. \(M\) is torsion free,

2. \(M\) is a submodule of a finite free module,

3. \((0)\) is the only associated prime of \(M\),

4. \((0)\) is in the support of \(M\) and \(M\) has property \((S_1)\), and

5. \((0)\) is in the support of \(M\) and \(M\) has no embedded associated prime.

Proof

We have seen the equivalence of (1) and (2) in Lemma 0AUU. We have seen the equivalence of (4) and (5) in Algebra, Lemma 031Q. The equivalence between (3) and (5) is immediate from the definition. A localization of a torsion free module is torsion free (Lemma 0AUR), hence it is clear that a \(M\) has no associated primes different from \((0)\). Thus (1) implies (5). Conversely, assume (5). If \(M\) has torsion, then there exists an embedding \(R/I \subset M\) for some nonzero ideal \(I\) of \(R\). Hence \(M\) has an associated prime different from \((0)\) (see Algebra, Lemmas 02M3 and 0587). This is an embedded associated prime which contradicts the assumption.

Lemma

Let \(R\) be a domain. Any flat \(R\)-module is torsion free.

Proof

If \(x \in R\) is nonzero, then \(x : R \to R\) is injective, and hence if \(M\) is flat over \(R\), then \(x : M \to M\) is injective. Thus if \(M\) is flat over \(R\), then \(M\) is torsion free.

Lemma

Let \(A\) be a valuation ring. An \(A\)-module \(M\) is flat over \(A\) if and only if \(M\) is torsion free.

Proof

The implication “flat \(\Rightarrow\) torsion free” is Lemma 0538. For the converse, assume \(M\) is torsion free. By the equational criterion of flatness (see Algebra, Lemma 00HK) we have to show that every relation in \(M\) is trivial. To do this assume that \(\sum_{i = 1, \ldots, n} a_i x_i = 0\) with \(x_i \in M\) and \(a_i \in A\). After renumbering we may assume that \(v(a_1) \leq v(a_i)\) for all \(i\). Hence we can write \(a_i = a'_i a_1\) for some \(a'_i \in A\). Note that \(a'_1 = 1\). As \(M\) is torsion free we see that \(x_1 = - \sum_{i \geq 2} a'_i x_i\). Thus, if we choose \(y_i = x_i\), \(i = 2, \ldots, n\) then \[x_1 = \sum\nolimits_{j \geq 2} -a'_j y_j, \quad x_i = y_i, (i \geq 2)\quad 0 = a_1 \cdot (-a'_j) + a_j \cdot 1 (j \geq 2)\] shows that the relation was trivial (to be explicit the elements \(a_{ij}\) are defined by setting \(a_{11} = 0\), \(a_{1j} = -a'_j\) for \(j > 1\), and \(a_{ij} = \delta_{ij}\) for \(i, j \geq 2\)).

Lemma

Let \(A\) be a Dedekind domain (for example a discrete valuation ring or more generally a PID).

1. An \(A\)-module is flat if and only if it is torsion free.

2. A finite torsion free \(A\)-module is finite locally free.

3. A finite torsion free \(A\)-module is finite free if \(A\) is a PID.

Proof

(For the parenthetical remark in the statement of the lemma, see Algebra, Lemma 0AUQ.) Proof of (1). By Lemma 0AUT and Algebra, Lemma 00HT it suffices to check the statement over \(A_\mathfrak m\) for \(\mathfrak m \subset A\) maximal. Since \(A_\mathfrak m\) is a discrete valuation ring (Algebra, Lemma 034X) we win by Lemma 0539.

Proof of (2). Follows from Algebra, Lemma 00NX and (1).

Proof of (3). Let \(A\) be a PID and let \(M\) be a finite torsion free module. By Lemma 0AUU we see that \(M \subset A^{\oplus n}\) for some \(n\). We argue that \(M\) is free by induction on \(n\). The case \(n = 1\) expresses exactly the fact that \(A\) is a PID. If \(n > 1\) let \(M' \subset A^{\oplus n - 1}\) be the image of the projection onto the last \(n - 1\) summands of \(A^{\oplus n}\). Then we obtain a short exact sequence \(0 \to I \to M \to M' \to 0\) where \(I\) is the intersection of \(M\) with the first summand \(A\) of \(A^{\oplus n}\). By induction we see that \(M\) is an extension of finite free \(A\)-modules, whence finite free.

Lemma

Let \(R\) be a domain. Let \(M\), \(N\) be \(R\)-modules. If \(N\) is torsion free, so is \(\Hom_R(M, N)\).

Proof

Choose a surjection \(\bigoplus_{i \in I} R \to M\). Then \(\Hom_R(M, N) \subset \prod_{i \in I} N\).

Ranks of modules

Here is our definition.

Definition

Let \(R\) be a domain with fraction field \(K\). The rank of an \(R\)-module \(M\) is \(\dim_K(M \otimes_R K)\).

This definition does not conflict with the notion of a locally free module of rank \(r\) in Algebra, Definition 00NW when both definitions apply.

Lemma

Let \(R\) be a domain. If \(M \to M'\) is a map of \(R\)-modules whose kernel and cokernel are torsion, then the rank of \(M\) equals the rank of \(M'\).

Proof

Omitted. Hint: the induced map \(M \otimes_R K \to M' \otimes_R K\) is an isomorphism if \(K\) is the fraction field of \(R\).

Lemma

Let \(R\) be a domain. Let \(0 \to M \to M' \to M'' \to 0\) be a short exact sequence of \(R\)-modules. Then the rank of \(M'\) is the sum of the ranks of \(M\) and \(M''\).

Proof

Omitted.

Lemma

Let \(R\) be a domain. Let \(M\) and \(N\) be \(R\)-modules.

1. The rank of \(M \otimes_R N\) is the product of the ranks of \(M\) and \(N\).

2. If \(M\) is a finitely presented \(R\)-module, then the rank of \(\Hom_R(M, N)\) is the product of the ranks of \(M\) and \(N\).

Proof

Part (1) follows from the same fact for vector spaces. Assume \(M\) is finitely presented as an \(R\)-module. Then \(\Hom_R(M, N) \otimes_R K\) is isomorphic to \(\Hom_K(M \otimes_R K, N \otimes_R K)\) as a \(K\)-vector space by Algebra, Lemma 0583. Since also \(M \otimes_R K\) is finite dimensional, we conclude from the corresponding fact for vector spaces.

Lemma

Let \(R \subset R'\) be an extension of domains. If \(M\) is an \(R\)-module, then the rank of \(M\) over \(R\) is equal to the rank of \(M \otimes_R R'\) over \(R'\).

Proof

This is true because the dimension of a vector space is invariant under extension of ground field.

Lemma

Let \(R\) be a domain. Let \(M\) be a finite \(R\)-module. Then

1. \(M\) has finite rank \(r \geq 0\),

2. there is a map \(M \to R^{\oplus r}\) whose kernel and cokernel are torsion modules,

3. there is an \(f \in R\), \(f \not = 0\) such that \(M_f\) is free of rank \(r\), and

4. there is an injective map \(R^{\oplus r} \to M\) whose cokernel is a torsion module.

Proof

Since \(M\) is finite, \(M \otimes_R K\) is a finite dimensional vector space over the fraction field \(K\) of \(R\). Thus the rank \(r\) is finite.

Choose generators \(x_1, \ldots, x_n\) of \(M\). Choose a basis \(e_1, \ldots, e_r\) for \(M \otimes_R K\). We may write \(x_i \otimes 1 = \sum (a_{ij}/b_{ij}) e_j\) for some \(a_{ij} , b_{ij} \in R\) with \(b_{ij} \not = 0\). Setting \(b = \prod b_{ij}\) and replacing \(a_{ij}\) by \(a_{ij}b/b_{ij}\) we may assume that \(x_i \otimes 1 = \sum (a_{ij}/b) e_j\). Note that if we have a relation \(\sum r_i x_i = 0\) in \(M\) then \(\sum r_ia_{ij} = 0\) in \(R\) for all \(j = 1, \ldots, r\).

Proof of (2). The last comment above tells us the map \(M \to \bigoplus Re_j\) sending \(x_i\) to \(\sum a_{ij} e_j\). is well defined. The fact that \(e_1, \ldots, e_r\) forms a basis of \(M \otimes_R K\) tells us that this map becomes an isomorphism upon tensoring by \(K\). Hence this is a map as in (2).

Proof of (3). The cokernel of the map \(M \to R^{\oplus r}\) constructed in (2) is finite and torsion. Hence we can find a \(g \in R\), \(g \not = 0\) which annihilates this cokernel. In other words, such that the map \(M_g \to R_g^{\oplus r}\) is surjective. Then we can choose a splitting \(M_g = N \oplus R_g^{\oplus r}\). Then \(N\) is a finite \(R_g\)-module with \(N \otimes_{R_g} K = 0\). In other words, \(N\) is torsion and finite over \(R_g\). Thus we can find a \(g' \in R\) which annihilates \(N\). Setting \(f= gg'\) we see that (3) holds.

To prove (4) choose a basis \(y_1, \ldots, y_r\) of \(M_f\) where \(f\) is as in (3). Then write \(y_j = m_j/f^{t_j}\) for some \(m_j \in M\) and \(t_j \geq 0\). The map \(R^{\oplus r} \to M\) sending the \(j\)th basis vector to \(m_j\) will be as in (4). Details omitted.

Reflexive modules

Here is our definition.

Definition

Let \(R\) be a domain. We say an \(R\)-module \(M\) is reflexive if the natural map \[j : M \longrightarrow \Hom_R(\Hom_R(M, R), R)\] which sends \(m \in M\) to the map sending \(\varphi \in \Hom_R(M, R)\) to \(\varphi(m) \in R\) is an isomorphism.

We can make this definition for more general rings, but already the definition above has drawbacks. It would be wise to restrict to Noetherian domains and finite torsion free modules and (perhaps) impose some regularity conditions on \(R\) (e.g., \(R\) is normal).

Lemma

Let \(R\) be a domain and let \(M\) be an \(R\)-module.

1. If \(M\) is reflexive, then \(M\) is torsion free.

2. If \(M\) is finite, then the kernel and cokernel of \(j : M \to \Hom_R(\Hom_R(M, R), R)\) are torsion modules.

3. If \(M\) is finite, then \(j\) is injective if and only if \(M\) is torsion free.

Proof

To see (1) use that the double dual of \(M\) is torsion free by Lemma 0AUX. Assume that \(M\) is finite. Choose generators \(x_1, \ldots, x_n\) of \(M\) over \(R\). Let \(K\) be the fraction field of \(R\). After renumbering we may assume that for some \(0 \leq r \leq n\) the elements \(x_1, \ldots, x_r\) map to a basis of \(M \otimes_R K\) over \(K\). For \(r < i \leq n\) write \(x_i \otimes 1 = \sum_{1 \leq j \leq r} x_j \otimes k_{ij}\) with \(k_{ij} \in K\). Choose \(c \in R\) such that \(a_{ij} = ck_{ij} \in R\) and such that \(cx_i = \sum_{1 \leq j \leq r} a_{ij} x_j\) in \(M\) (small detail omitted). Then we get maps \[R^{\oplus r} \xrightarrow{\alpha} M \xrightarrow{\beta} R^{\oplus r} \xrightarrow{\alpha} M\] with both \(\alpha \circ \beta\) and \(\beta \circ \alpha\) given by multiplication with \(c\). Here \(\alpha\) uses \(x_1, \ldots, x_r\) and \(\beta\) sends \(x_i\) to \(c\) times the \(i\)th basis vector for \(1 \leq i \leq r\) and to the vector \((a_{i1}, \ldots, a_{ir})\) for \(r < i \leq n\). Consider the commutative diagram \[\xymatrix{ R^{\oplus r} \ar[r] \ar[d]^{\text{id}} & M \ar[r] \ar[d]^j & R^{\oplus r} \ar[d]^{\text{id}} \ar[r] & M \ar[d]^j \\ R^{\oplus r} \ar[r] & \Hom_R(\Hom_R(M, R), R) \ar[r] & R^{\oplus r} \ar[r] & \Hom_R(\Hom_R(M, R), R) }\] Of course the compositions of consecutive arrows on the bottom row are equal to multiplication by \(c\). The diagram shows that the kernel and cokernel of \(j\) are annihilated by \(c\) which proves (2). For (3) note that if \(M\) is torsion free, then \(\beta\) is injective which implies that \(j\) is injective. The converse is clear because we’ve seen already that the double dual of \(M\) is torsion free.

Lemma

Let \(R\) be a discrete valuation ring and let \(M\) be a finite \(R\)-module. Then the map \(j : M \to \Hom_R(\Hom_R(M, R), R)\) is surjective.

Proof

Let \(M_{tors} \subset M\) be the torsion submodule. Then we have \(\Hom_R(M, R) = \Hom_R(M/M_{tors}, R)\) (holds over any domain). Hence we may assume that \(M\) is torsion free. Then \(M\) is free by Lemma 0AUW and the lemma is clear.

Lemma

Let \(R\) be a Noetherian domain. Let \(M\) be a finite \(R\)-module. The following are equivalent:

1. \(M\) is reflexive,

2. \(M_\mathfrak p\) is a reflexive \(R_\mathfrak p\)-module for all primes \(\mathfrak p \subset R\), and

3. \(M_\mathfrak m\) is a reflexive \(R_\mathfrak m\)-module for all maximal ideals \(\mathfrak m\) of \(R\).

Proof

The localization of \(j : M \to \Hom_R(\Hom_R(M, R), R)\) at a prime \(\mathfrak p\) is the corresponding map for the module \(M_\mathfrak p\) over the Noetherian local domain \(R_\mathfrak p\). See Algebra, Lemma 0583. Thus the lemma holds by Algebra, Lemma 00HN.

Lemma

Let \(R\) be a Noetherian domain. Let \(0 \to M \to M' \to M''\) an exact sequence of finite \(R\)-modules. If \(M'\) is reflexive and \(M''\) is torsion free, then \(M\) is reflexive.

Proof

We will use without further mention that \(\Hom_R(N, N')\) is a finite \(R\)-module for any finite \(R\)-modules \(N\) and \(N'\), see Algebra, Lemma 08YR. We take duals to get a sequence \[\Hom_R(M, R) \leftarrow \Hom_R(M', R) \leftarrow \Hom_R(M'', R)\] Dualizing again we obtain a commutative diagram \[\xymatrix{ \Hom_R(\Hom_R(M, R), R) \ar[r]_j & \Hom_R(\Hom_R(M', R), R) \ar[r] & \Hom_R(\Hom_R(M'', R), R) \\ M \ar[u] \ar[r] & M' \ar[u] \ar[r] & M'' \ar[u] }\] We do not know the top row is exact. But, by assumption the middle vertical arrow is an isomorphism and the right vertical arrow is injective (Lemma 0AV0). We claim \(j\) is injective. Assuming the claim a diagram chase shows that the left vertical arrow is an isomorphism, i.e., \(M\) is reflexive.

Proof of the claim. Consider the exact sequence \(\Hom_R(M',R)\to \Hom_R(M,R)\to Q \to 0\) defining \(Q\). One applies Algebra, Lemma 0583 to obtain \[\Hom_K(M'\otimes_R K,K) \to \Hom_K(M\otimes_R K,K) \to Q\otimes_R K\to 0\] But \(M \otimes_R K \to M' \otimes_R K\) is an injective map of vector spaces, hence split injective, so \(Q \otimes_R K = 0\), that is, \(Q\) is torsion. Then one gets the exact sequence \[0 \to \Hom_R(Q,R) \to \Hom_R(\Hom_R(M,R),R) \to \Hom_R(\Hom_R(M',R),R)\] and \(\Hom_R(Q,R)=0\) because \(Q\) is torsion.

Lemma

Let \(R\) be a Noetherian domain. Let \(M\) be a finite \(R\)-module. The following are equivalent

1. \(M\) is reflexive,

2. there exists a short exact sequence \(0 \to M \to F \to N \to 0\) with \(F\) finite free and \(N\) torsion free.

Proof

Observe that a finite free module is reflexive. By Lemma 0EB8 we see that (2) implies (1). Assume \(M\) is reflexive. Choose a presentation \(R^{\oplus m} \to R^{\oplus n} \to \Hom_R(M, R) \to 0\). Dualizing we get an exact sequence \[0 \to \Hom_R(\Hom_R(M, R), R) \to R^{\oplus n} \to N \to 0\] with \(N = \Im(R^{\oplus n} \to R^{\oplus m})\) a torsion free module. As \(M = \Hom_R(\Hom_R(M, R), R)\) we get an exact sequence as in (2).

Lemma

Let \(R \to R'\) be a flat homomorphism of Noetherian domains. If \(M\) is a finite reflexive \(R\)-module, then \(M \otimes_R R'\) is a finite reflexive \(R'\)-module.

Proof

Choose a short exact sequence \(0 \to M \to F \to N \to 0\) with \(F\) finite free and \(N\) torsion free, see Lemma 0AV2. Since \(R \to R'\) is flat we obtain a short exact sequence \(0 \to M \otimes_R R' \to F \otimes_R R' \to N \otimes_R R' \to 0\) with \(F \otimes_R R'\) finite free and \(N \otimes_R R'\) torsion free (Lemma 0AXM). Thus \(M \otimes_R R'\) is reflexive by Lemma 0AV2.

Lemma

Let \(R\) be a Noetherian domain. Let \(M\) be a finite \(R\)-module. Let \(N\) be a finite reflexive \(R\)-module. Then \(\Hom_R(M, N)\) is reflexive.

Proof

Choose a presentation \(R^{\oplus m} \to R^{\oplus n} \to M \to 0\). Then we obtain \[0 \to \Hom_R(M, N) \to N^{\oplus n} \to N' \to 0\] with \(N' = \Im(N^{\oplus n} \to N^{\oplus m})\) torsion free. We conclude by Lemma 0EB8.

Definition

Let \(R\) be a Noetherian domain. Let \(M\) be a finite \(R\)-module. The module \(M^{**} = \Hom_R(\Hom_R(M, R), R)\) is called the reflexive hull of \(M\).

This makes sense because the reflexive hull is reflexive by Lemma 0AV3. The assignment \(M \mapsto M^{**}\) is a functor. If \(\varphi : M \to N\) is an \(R\)-module map into a reflexive \(R\)-module \(N\), then \(\varphi\) factors \(M \to M^{**} \to N\) through the reflexive hull of \(M\). Another way to say this is that taking the reflexive hull is the left adjoint to the inclusion functor \[\text{finite reflexive modules} \subset \text{finite modules}\] over a Noetherian domain \(R\).

Lemma

Let \(R\) be a Noetherian local ring. Let \(M\), \(N\) be finite \(R\)-modules.

1. If \(N\) has depth \(\geq 1\), then \(\Hom_R(M, N)\) has depth \(\geq 1\).

2. If \(N\) has depth \(\geq 2\), then \(\Hom_R(M, N)\) has depth \(\geq 2\).

Proof

Choose a presentation \(R^{\oplus m} \to R^{\oplus n} \to M \to 0\). Dualizing we get an exact sequence \[0 \to \Hom_R(M, N) \to N^{\oplus n} \to N' \to 0\] with \(N' = \Im(N^{\oplus n} \to N^{\oplus m})\). A submodule of a module with depth \(\geq 1\) has depth \(\geq 1\); this follows immediately from the definition. Thus part (1) is clear. For (2) note that here the assumption and the previous remark implies \(N'\) has depth \(\geq 1\). The module \(N^{\oplus n}\) has depth \(\geq 2\). From Algebra, Lemma 00LX we conclude \(\Hom_R(M, N)\) has depth \(\geq 2\).

Lemma

Let \(R\) be a Noetherian ring. Let \(M\), \(N\) be finite \(R\)-modules.

1. If \(N\) has property \((S_1)\), then \(\Hom_R(M, N)\) has property \((S_1)\).

2. If \(N\) has property \((S_2)\), then \(\Hom_R(M, N)\) has property \((S_2)\).

3. If \(R\) is a domain, \(N\) is torsion free and \((S_2)\), then \(\Hom_R(M, N)\) is torsion free and has property \((S_2)\).

Proof

Since localizing at primes commutes with taking \(\Hom_R\) for finite \(R\)-modules (Algebra, Lemma 08YR) parts (1) and (2) follow immediately from Lemma 0AV5. Part (3) follows from (2) and Lemma 0AUX.

Lemma

Let \(R\) be a Noetherian ring. Let \(\varphi : M \to N\) be a map of \(R\)-modules. Assume that for every prime \(\mathfrak p\) of \(R\) at least one of the following happens

1. \(M_\mathfrak p \to N_\mathfrak p\) is injective, or

2. \(\mathfrak p \not \in \text{Ass}(M)\).

Then \(\varphi\) is injective.

Proof

Let \(\mathfrak p\) be an associated prime of \(\Ker(\varphi)\). Then there exists an element \(x \in M_\mathfrak p\) which is in the kernel of \(M_\mathfrak p \to N_\mathfrak p\) and whose annihilator is \(\mathfrak pR_\mathfrak p\) (Algebra, Lemma 0310). This is impossible in both cases. Hence \(\text{Ass}(\Ker(\varphi)) = \emptyset\) and we conclude \(\Ker(\varphi) = 0\) by Algebra, Lemma 0587.

Lemma

Let \(R\) be a Noetherian ring. Let \(\varphi : M \to N\) be a map of \(R\)-modules. Assume \(M\) is finite and that for every prime \(\mathfrak p\) of \(R\) one of the following happens

1. \(M_\mathfrak p \to N_\mathfrak p\) is an isomorphism, or

2. \(\text{depth}(M_\mathfrak p) \geq 2\) and \(\mathfrak p \not \in \text{Ass}(N)\).

Then \(\varphi\) is an isomorphism.

Proof

By Lemma 0AV7 we see that \(\varphi\) is injective. Let \(N' \subset N\) be an finitely generated \(R\)-module containing the image of \(M\). Then \(\text{Ass}(N_\mathfrak p) = \emptyset\) implies \(\text{Ass}(N'_\mathfrak p) = \emptyset\). Hence the assumptions of the lemma hold for \(M \to N'\). In order to prove that \(\varphi\) is an isomorphism, it suffices to prove the same thing for every such \(N' \subset N\). Thus we may assume \(N\) is a finite \(R\)-module. In this case, \(\mathfrak p \not \in \text{Ass}(N) \Rightarrow \text{depth}(N_\mathfrak p) \geq 1\), see Algebra, Lemma 00LL. Consider the short exact sequence \[0 \to M \to N \to Q \to 0\] defining \(Q\). Looking at the conditions we see that either \(Q_\mathfrak p = 0\) in case (1) or \(\text{depth}(Q_\mathfrak p) \geq 1\) in case (2) by Algebra, Lemma 00LX. This implies that \(Q\) does not have any associated primes, hence \(Q = 0\) by Algebra, Lemma 0587.

Lemma

Let \(R\) be a Noetherian domain. Let \(\varphi : M \to N\) be a map of \(R\)-modules. Assume \(M\) is finite, \(N\) is torsion free, and that for every prime \(\mathfrak p\) of \(R\) one of the following happens

1. \(M_\mathfrak p \to N_\mathfrak p\) is an isomorphism, or

2. \(\text{depth}(M_\mathfrak p) \geq 2\).

Then \(\varphi\) is an isomorphism.

Proof

This is a special case of Lemma 0AV8.

Lemma

Let \(R\) be a Noetherian domain. Let \(M\) be a finite \(R\)-module. The following are equivalent

1. \(M\) is reflexive,

2. for every prime \(\mathfrak p\) of \(R\) one of the following happens

   1. \(M_\mathfrak p\) is a reflexive \(R_\mathfrak p\)-module, or

   2. \(\text{depth}(M_\mathfrak p) \geq 2\).

Proof

If (1) is true, then \(M_\mathfrak p\) is a reflexive module for all primes of \(\mathfrak p\) by Lemma 0AV1. Thus (1) \(\Rightarrow\) (2). Assume (2). Set \(N = \Hom_R(\Hom_R(M, R), R)\) so that \[N_\mathfrak p = \Hom_{R_\mathfrak p}( \Hom_{R_\mathfrak p}(M_\mathfrak p, R_\mathfrak p), R_\mathfrak p)\] for every prime \(\mathfrak p\) of \(R\). See Algebra, Lemma 0583. We apply Lemma 0AV9 to the map \(j : M \to N\). This is allowed because \(M\) is finite and \(N\) is torsion free by Lemma 0AUX. In case (2)(a) the map \(M_\mathfrak p \to N_\mathfrak p\) is an isomorphism and in case (2)(b) we have \(\text{depth}(M_\mathfrak p) \geq 2\).

Lemma

Let \(R\) be a Noetherian domain. Let \(M\) be a finite reflexive \(R\)-module. Let \(\mathfrak p \subset R\) be a prime ideal.

1. If \(\text{depth}(R_\mathfrak p) \geq 2\), then \(\text{depth}(M_\mathfrak p) \geq 2\).

2. If \(R\) is \((S_2)\), then \(M\) is \((S_2)\).

Proof

Since formation of reflexive hull \(\Hom_R(\Hom_R(M, R), R)\) commutes with localization (Algebra, Lemma 0583) part (1) follows from Lemma 0AV5. Part (2) is immediate from Lemma 0AV6.

Example

The results above and below suggest reflexivity is related to the \((S_2)\) condition; here is an example to prevent too optimistic conjectures. Let \(k\) be a field. Let \(R\) be the \(k\)-subalgebra of \(k[x, y]\) generated by \(1, y, x^2, xy, x^3\). Then \(R\) is not \((S_2)\). So \(R\) as an \(R\)-module is an example of a reflexive \(R\)-module which is not \((S_2)\). Let \(M = k[x, y]\) viewed as an \(R\)-module. Then \(M\) is a reflexive \(R\)-module because \[\Hom_R(M, R) = \mathfrak m = (y, x^2, xy, x^3) \quad\text{and}\quad \Hom_R(\mathfrak m, R) = M\] and \(M\) is \((S_2)\) as an \(R\)-module (computations omitted). Thus \(R\) is a Noetherian domain possessing a reflexive \((S_2)\) module but \(R\) is not \((S_2)\) itself.

Lemma

Let \(R\) be a Noetherian normal domain with fraction field \(K\). Let \(M\) be a finite \(R\)-module. The following are equivalent

1. \(M\) is reflexive,

2. \(M\) is torsion free and has property \((S_2)\),

3. \(M\) is torsion free and \(M = \bigcap_{\text{height}(\mathfrak p) = 1} M_{\mathfrak p}\) where the intersection happens in \(M_K = M \otimes_R K\).

Proof

By Algebra, Lemma 031S we see that \(R\) satisfies \((R_1)\) and \((S_2)\).

Assume (1). Then \(M\) is torsion free by Lemma 0AV0 and satisfies \((S_2)\) by Lemma 0EBA. Thus (2) holds.

Assume (2). By definition \(M' = \bigcap_{\text{height}(\mathfrak p) = 1} M_{\mathfrak p}\) is the kernel of the map \[M_K \longrightarrow \bigoplus\nolimits_{\text{height}(\mathfrak p) = 1} M_K/M_\mathfrak p \subset \prod\nolimits_{\text{height}(\mathfrak p) = 1} M_K/M_\mathfrak p\] Observe that our map indeed factors through the direct sum as indicated since given \(a/b \in K\) there are at most finitely many height \(1\) primes \(\mathfrak p\) with \(b \in \mathfrak p\). Let \(\mathfrak p_0\) be a prime of height \(1\). Then \((M_K/M_\mathfrak p)_{\mathfrak p_0} = 0\) unless \(\mathfrak p = \mathfrak p_0\) in which case we get \((M_K/M_\mathfrak p)_{\mathfrak p_0} = M_K/M_{\mathfrak p_0}\). Thus by exactness of localization and the fact that localization commutes with direct sums, we see that \(M'_{\mathfrak p_0} = M_{\mathfrak p_0}\). Since \(M\) has depth \(\geq 2\) at primes of height \(> 1\), we see that \(M \to M'\) is an isomorphism by Lemma 0AV9. Hence (3) holds.

Assume (3). Let \(\mathfrak p\) be a prime of height \(1\). Then \(R_\mathfrak p\) is a discrete valuation ring by \((R_1)\). By Lemma 0AUW we see that \(M_\mathfrak p\) is finite free, in particular reflexive. Hence the map \(M \to M^{**}\) induces an isomorphism at all the primes \(\mathfrak p\) of height \(1\). Thus the condition \(M = \bigcap_{\text{height}(\mathfrak p) = 1} M_{\mathfrak p}\) implies that \(M = M^{**}\) and (1) holds.

Lemma

Let \(R\) be a Noetherian normal domain. Let \(M\) be a finite \(R\)-module. Then the reflexive hull of \(M\) is the intersection \[M^{**} = \bigcap\nolimits_{\text{height}(\mathfrak p) = 1} M_{\mathfrak p}/(M_\mathfrak p)_{tors} = \bigcap\nolimits_{\text{height}(\mathfrak p) = 1} (M/M_{tors})_\mathfrak p\] taken in \(M \otimes_R K\).

Proof

Let \(\mathfrak p\) be a prime of height \(1\). The kernel of \(M_\mathfrak p \to M \otimes_R K\) is the torsion submodule \((M_\mathfrak p)_{tors}\) of \(M_\mathfrak p\). Moreover, we have \((M/M_{tors})_\mathfrak p = M_\mathfrak p/(M_\mathfrak p)_{tors}\) and this is a finite free module over the discrete valuation ring \(R_\mathfrak p\) (Lemma 0AUW). Then \(M_\mathfrak p/(M_\mathfrak p)_{tors} \to (M_\mathfrak p)^{**} = (M^{**})_\mathfrak p\) is an isomorphism, hence the lemma is a consequence of Lemma 0AVB.

Lemma

Let \(A\) be a Noetherian normal domain with fraction field \(K\). Let \(L\) be a finite extension of \(K\). If the integral closure \(B\) of \(A\) in \(L\) is finite over \(A\), then \(B\) is reflexive as an \(A\)-module.

Proof

It suffices to show that \(B = \bigcap B_\mathfrak p\) where the intersection is over height \(1\) primes \(\mathfrak p \subset A\), see Lemma 0AVB. Let \(b \in \bigcap B_\mathfrak p\). Let \(x^d + a_1x^{d - 1} + \ldots + a_d\) be the minimal polynomial of \(b\) over \(K\). We want to show \(a_i \in A\). By Algebra, Lemma 00H7 we see that \(a_i \in A_\mathfrak p\) for all \(i\) and all height one primes \(\mathfrak p\). Hence we get what we want from Algebra, Lemma 031T (or the lemma already cited as \(A\) is a reflexive module over itself).

Content ideals

The definition may not be what you expect.

Definition

Let \(A\) be a ring. Let \(M\) be a flat \(A\)-module. Let \(x \in M\). If the set of ideals \(I\) in \(A\) such that \(x \in IM\) has a smallest element, we call it the content ideal of \(x\).

Note that since \(M\) is flat over \(A\), for a pair of ideals \(I, I'\) of \(A\) we have \(IM \cap I'M = (I \cap I')M\) as can be seen by tensoring the exact sequence \(0 \to I \cap I' \to I \oplus I' \to I + I' \to 0\) by \(M\).

Lemma

Let \(A\) be a ring. Let \(M\) be a flat \(A\)-module. Let \(x \in M\). The content ideal of \(x\), if it exists, is finitely generated.

Proof

Say \(x \in IM\). Then we can write \(x = \sum_{i = 1, \ldots, n} f_i x_i\) with \(f_i \in I\) and \(x_i \in M\). Hence \(x \in I'M\) with \(I' = (f_1, \ldots, f_n)\).

Lemma

Let \((A, \mathfrak m)\) be a local ring. Let \(u : M \to N\) be a map of flat \(A\)-modules such that \(\overline{u} : M/\mathfrak m M \to N/\mathfrak m N\) is injective. If \(x \in M\) has content ideal \(I\), then \(u(x)\) has content ideal \(I\) as well.

Proof

It is clear that \(u(x) \in IN\). If \(u(x) \in I'N\), then \(u(x) \in (I' \cap I)N\), see discussion following Definition 0ASA. Hence it suffices to show: if \(x \in I'N\) and \(I' \subset I\), \(I' \not = I\), then \(u(x) \not \in I'N\). Since \(I/I'\) is a nonzero finite \(A\)-module (Lemma 0ASB) there is a nonzero map \(\chi : I/I' \to A/\mathfrak m\) of \(A\)-modules by Nakayama’s lemma (Algebra, Lemma 00DV). Since \(I\) is the content ideal of \(x\) we see that \(x \not \in I''M\) where \(I'' = \Ker(\chi)\). Hence \(x\) is not in the kernel of the map \[IM = I \otimes_A M \xrightarrow{\chi \otimes 1} A/\mathfrak m \otimes M \cong M/\mathfrak m M\] Applying our hypothesis on \(\overline{u}\) we conclude that \(u(x)\) does not map to zero under the map \[IN = I \otimes_A N \xrightarrow{\chi \otimes 1} A/\mathfrak m \otimes N \cong N/\mathfrak m N\] and we conclude.

Lemma

Let \(A\) be a ring. Let \(M\) be a flat Mittag-Leffler module. Then every element of \(M\) has a content ideal.

Proof

This is a special case of Algebra, Lemma 059S.

Flatness and finiteness conditions

In this section we discuss some implications of the type “flat \(+\) finite type \(\Rightarrow\) finite presentation”. We will revisit this result in the chapter on flatness, see More on Flatness, Section 057N. A first result of this type was proved in Algebra, Lemma 052U.

Lemma

Let \(R\) be a ring. Let \(S = R[x_1, \ldots, x_n]\) be a polynomial ring over \(R\). Let \(M\) be an \(S\)-module. Assume

1. there exist finitely many primes \(\mathfrak p_1, \ldots, \mathfrak p_m\) of \(R\) such that the map \(R \to \prod R_{\mathfrak p_j}\) is injective,

2. \(M\) is a finite \(S\)-module,

3. \(M\) flat over \(R\), and

4. for every prime \(\mathfrak p\) of \(R\) the module \(M_{\mathfrak p}\) is of finite presentation over \(S_{\mathfrak p}\).

Then \(M\) is of finite presentation over \(S\).

Proof

Choose a presentation \[0 \to K \to S^{\oplus r} \to M \to 0\] of \(M\) as an \(S\)-module. Let \(\mathfrak q\) be a prime ideal of \(S\) lying over a prime \(\mathfrak p\) of \(R\). By assumption there exist finitely many elements \(k_1, \ldots, k_t \in K\) such that if we set \(K' = \sum Sk_j \subset K\) then \(K'_{\mathfrak p} = K_{\mathfrak p}\) and \(K'_{\mathfrak p_j} = K_{\mathfrak p_j}\) for \(j = 1, \ldots, m\). Setting \(M' = S^{\oplus r}/K'\) we deduce that in particular \(M'_{\mathfrak q} = M_{\mathfrak q}\). By openness of flatness, see Algebra, Theorem 00RC we conclude that there exists a \(g \in S\), \(g \not \in \mathfrak q\) such that \(M'_g\) is flat over \(R\). Thus \(M'_g \to M_g\) is a surjective map of flat \(R\)-modules. Consider the commutative diagram \[\xymatrix{ M'_g \ar[r] \ar[d] & M_g \ar[d] \\ \prod (M'_g)_{\mathfrak p_j} \ar[r] & \prod (M_g)_{\mathfrak p_j} }\] The bottom arrow is an isomorphism by choice of \(k_1, \ldots, k_t\). The left vertical arrow is an injective map as \(R \to \prod R_{\mathfrak p_j}\) is injective and \(M'_g\) is flat over \(R\). Hence the top horizontal arrow is injective, hence an isomorphism. This proves that \(M_g\) is of finite presentation over \(S_g\). We conclude by applying Algebra, Lemma 00EO.

Lemma

Let \(R \to S\) be a ring homomorphism. Assume

1. there exist finitely many primes \(\mathfrak p_1, \ldots, \mathfrak p_m\) of \(R\) such that the map \(R \to \prod R_{\mathfrak p_j}\) is injective,

2. \(R \to S\) is of finite type,

3. \(S\) flat over \(R\), and

4. for every prime \(\mathfrak p\) of \(R\) the ring \(S_{\mathfrak p}\) is of finite presentation over \(R_{\mathfrak p}\).

Then \(S\) is of finite presentation over \(R\).

Proof

By assumption \(S\) is a quotient of a polynomial ring over \(R\). Thus the result follows directly from Lemma 053A.

Lemma

Let \(R\) be a ring. Let \(S = R[x_1, \ldots, x_n]\) be a graded polynomial algebra over \(R\), i.e., \(\deg(x_i) > 0\) but not necessarily equal to \(1\). Let \(M\) be a graded \(S\)-module. Assume

1. \(R\) is a local ring,

2. \(M\) is a finite \(S\)-module, and

3. \(M\) is flat over \(R\).

Then \(M\) is finitely presented as an \(S\)-module.

Proof

Let \(M = \bigoplus M_d\) be the grading on \(M\). Pick homogeneous generators \(m_1, \ldots, m_r \in M\) of \(M\). Say \(\deg(m_i) = d_i \in \mathbf{Z}\). This gives us a presentation \[0 \to K \to \bigoplus\nolimits_{i = 1, \ldots, r} S(-d_i) \to M \to 0\] which in each degree \(d\) leads to the short exact sequence \[0 \to K_d \to \bigoplus\nolimits_{i = 1, \ldots, r} S_{d - d_i} \to M_d \to 0.\] By assumption each \(M_d\) is a finite flat \(R\)-module. By Algebra, Lemma 00NZ this implies each \(M_d\) is a finite free \(R\)-module. Hence we see each \(K_d\) is a finite \(R\)-module. Also each \(K_d\) is flat over \(R\) by Algebra, Lemma 00HM. Hence we conclude that each \(K_d\) is finite free by Algebra, Lemma 00NZ again.

Let \(\mathfrak m\) be the maximal ideal of \(R\). By the flatness of \(M\) over \(R\) the short exact sequences above remain short exact after tensoring with \(\kappa = \kappa(\mathfrak m)\). As the ring \(S \otimes_R \kappa\) is Noetherian we see that there exist homogeneous elements \(k_1, \ldots, k_t \in K\) such that the images \(\overline{k}_j\) generate \(K \otimes_R \kappa\) over \(S \otimes_R \kappa\). Say \(\deg(k_j) = e_j\). Thus for any \(d\) the map \[\bigoplus\nolimits_{j = 1, \ldots, t} S_{d - e_j} \longrightarrow K_d\] becomes surjective after tensoring with \(\kappa\). By Nakayama’s lemma (Algebra, Lemma 00DV) this implies the map is surjective over \(R\). Hence \(K\) is generated by \(k_1, \ldots, k_t\) over \(S\) and we win.

Lemma

Let \(R\) be a ring. Let \(S = \bigoplus_{n \geq 0} S_n\) be a graded \(R\)-algebra. Let \(M = \bigoplus_{d \in \mathbf{Z}} M_d\) be a graded \(S\)-module. Assume \(S\) is finitely generated as an \(R\)-algebra, assume \(S_0\) is a finite \(R\)-algebra, and assume there exist finitely many primes \(\mathfrak p_j\), \(i = 1, \ldots, m\) such that \(R \to \prod R_{\mathfrak p_j}\) is injective.

1. If \(S\) is flat over \(R\), then \(S\) is a finitely presented \(R\)-algebra.

2. If \(M\) is flat as an \(R\)-module and finite as an \(S\)-module, then \(M\) is finitely presented as an \(S\)-module.

Proof

As \(S\) is finitely generated as an \(R\)-algebra, it is finitely generated as an \(S_0\) algebra, say by homogeneous elements \(t_1, \ldots, t_n \in S\) of degrees \(d_1, \ldots, d_n > 0\). Set \(P = R[x_1, \ldots, x_n]\) with \(\deg(x_i) = d_i\). The ring map \(P \to S\), \(x_i \to t_i\) is finite as \(S_0\) is a finite \(R\)-module. To prove (1) it suffices to prove that \(S\) is a finitely presented \(P\)-module. To prove (2) it suffices to prove that \(M\) is a finitely presented \(P\)-module. Thus it suffices to prove that if \(S = P\) is a graded polynomial ring and \(M\) is a finite \(S\)-module flat over \(R\), then \(M\) is finitely presented as an \(S\)-module. By Lemma 053C we see \(M_{\mathfrak p}\) is a finitely presented \(S_{\mathfrak p}\)-module for every prime \(\mathfrak p\) of \(R\). Thus the result follows from Lemma 053A.

Remark

Let \(R\) be a ring. When does \(R\) satisfy the condition mentioned in Lemmas 053A, 053B, and 053D? This holds if

1. \(R\) is local,

2. \(R\) is Noetherian,

3. \(R\) is a domain,

4. \(R\) is a reduced ring with finitely many minimal primes, or

5. \(R\) has finitely many weakly associated primes, see Algebra, Lemma 05CB.

Thus these lemmas hold in all cases listed above.

The following lemma will be improved on in More on Flatness, Proposition 053G.

Lemma

Let \(A\) be a valuation ring. Let \(A \to B\) be a ring map of finite type. Let \(M\) be a finite \(B\)-module.

1. If \(B\) is flat over \(A\), then \(B\) is a finitely presented \(A\)-algebra.

2. If \(M\) is flat as an \(A\)-module, then \(M\) is finitely presented as a \(B\)-module.

Proof

We are going to use that an \(A\)-module is flat if and only if it is torsion free, see Lemma 0539. By Algebra, Lemma 052N we can find a graded \(A\)-algebra \(S\) with \(S_0 = A\) and generated by finitely many elements in degree \(1\), an element \(f \in S_1\) and a finite graded \(S\)-module \(N\) such that \(B \cong S_{(f)}\) and \(M \cong N_{(f)}\). If \(M\) is torsion free, then we can take \(N\) torsion free by replacing it by \(N/N_{tors}\), see Lemma 0537. Similarly, if \(B\) is torsion free, then we can take \(S\) torsion free by replacing it by \(S/S_{tors}\). Hence in case (1), we may apply Lemma 053D to see that \(S\) is a finitely presented \(A\)-algebra, which implies that \(B = S_{(f)}\) is a finitely presented \(A\)-algebra. To see (2) we may first replace \(S\) by a graded polynomial ring, and then we may apply Lemma 053C to conclude.

Lemma

Let \(A\) be a valuation ring. Let \(A \to B\) be a local homomorphism which is essentially of finite type. Let \(M\) be a finite \(B\)-module.

1. If \(B\) is flat over \(A\), then \(B\) is essentially of finite presentation over \(A\).

2. If \(M\) is flat as an \(A\)-module, then \(M\) is finitely presented as a \(B\)-module.

Proof

By assumption we can write \(B\) as a quotient of the localization of a polynomial algebra \(P = A[x_1, \ldots, x_n]\) at a prime ideal \(\mathfrak q\). In case (1) we consider \(M = B\) as a finite module over \(P_\mathfrak q\) and in case (2) we consider \(M\) as a finite module over \(P_\mathfrak q\). In both cases, we have to show that this is a finitely presented \(P_\mathfrak q\)-module, see Algebra, Lemma 0561 for case (2).

Choose a presentation \(0 \to K \to P_\mathfrak q^{\oplus r} \to M \to 0\) which is possible because \(M\) is finite over \(P_\mathfrak q\). Let \(L = P^{\oplus r} \cap K\). Then \(K = L_\mathfrak q\), see Algebra, Lemma 00CU. Then \(N = P^{\oplus r}/L\) is a submodule of \(M\) and hence flat by Lemma 0539. Since also \(N\) is a finite \(P\)-module, we see that \(N\) is finitely presented as a \(P\)-module by Lemma 053E. Since localization is exact (Algebra, Proposition 00CS) we see that \(N_\mathfrak q = M\) and we conclude.

Blowing up and flatness

In this section we begin our discussion of results of the form: “After a blowup the strict transform becomes flat”. More results of this type may be found in Divisors, Section 0F84 and More on Flatness, Section 080X.

Definition

Let \(R\) be a ring. Let \(I \subset R\) be an ideal and \(a \in I\). Let \(R[\frac{I}{a}]\) be the affine blowup algebra, see Algebra, Definition 052Q. Let \(M\) be an \(R\)-module. The strict transform of \(M\) along \(R \to R[\frac{I}{a}]\) is the \(R[\frac{I}{a}]\)-module \[M' = \left(M \otimes_R R[\textstyle{\frac{I}{a}}]\right)/a\text{-power torsion}\]

The following is a very weak version of flattening by blowing up, but it is already sometimes a useful result.

Lemma

Let \((R, \mathfrak m)\) be a local domain with fraction field \(K\). Let \(S\) be a finite type \(R\)-algebra. Let \(M\) be a finite \(S\)-module. For every valuation ring \(A \subset K\) dominating \(R\) there exists an ideal \(I \subset \mathfrak m\) and a nonzero element \(a \in I\) such that

1. \(I\) is finitely generated,

2. \(A\) has center on \(R[\frac{I}{a}]\),

3. the fibre ring of \(R \to R[\frac{I}{a}]\) at \(\mathfrak m\) is not zero, and

4. the strict transform \(S_{I, a}\) of \(S\) along \(R \to R[\frac{I}{a}]\) is flat and of finite presentation over \(R\), and the strict transform \(M_{I, a}\) of \(M\) along \(R \to R[\frac{I}{a}]\) is flat over \(R\) and finitely presented over \(S_{I, a}\).

Proof

Write \(S = R[x_1, \ldots, x_n]/J\) and denote \(N = S \oplus M\) viewed as a module over \(P = R[x_1, \ldots, x_n]\). If we can prove the lemma in case \(S\) is a polynomial algebra over \(R\), then we can find \(I, a\) satisfying (1), (2), (3) such that the strict transform \(N_{I, a}\) of \(N\) along \(R \to R[\frac{I}{a}]\) is flat over \(R\) and finitely presented as a module over the strict transform \(P_{I, a}\) of \(P\). Since \(P_{I, a} = R[\frac{I}{a}][x_1, \ldots, x_n]\) (small detail omitted) we find that the summand \(S_{I, a} \subset N_{I, a}\) is flat over \(R\) and finitely presented as a module over \(R[\frac{I}{a}][x_1, \ldots, x_n]\). Hence \(S_{I, a}\) is finitely presented as an \(R[\frac{I}{a}]\)-algebra. Moreover, the summand \(M_{I, a} \subset N_{I, a}\) is flat over \(R\) and finitely presented as a module over \(P_{I, a}\) hence also finitely presented as a module over \(S_{I, a}\), see Algebra, Lemma 0561. This reduces us to the case discussed in the next paragraph.

Assume \(S = R[x_1, \ldots, x_n]\). Choose a presentation \[0 \to K \to S^{\oplus r} \to M \to 0.\] Let \(M_A\) be the quotient of \(M \otimes_R A\) by its torsion submodule, see Lemma 0537. Then \(M_A\) is a finite module over \(S_A = A[x_1, \ldots, x_n]\). By Lemma 0539 we see that \(M_A\) is flat over \(A\). By Lemma 053E we see that \(M_A\) is finitely presented. Hence there exist finitely many elements \(k_1, \ldots, k_t \in S_A^{\oplus r}\) which generate the kernel of the presentation \(S_A^{\oplus r} \to M_A\) as an \(S_A\)-module. For any choice of \(a \in I \subset \mathfrak m\) satisfying (1), (2), and (3) we denote \(M_{I, a}\) the strict transform of \(M\) along \(R \to R[\frac{I}{a}]\). It is a finite module over \(S_{I, a} = R[\frac{I}{a}][x_1, \ldots, x_n]\). By Algebra, Lemma 052M we have \(A = \colim_{I, a} R[\frac{I}{a}]\). This implies that \(S_A = \colim S_{I, a}\) and \[\colim M \otimes_R R[\textstyle{\frac{I}{a}}] = M \otimes_R A\] Choose \(I, a\) and lifts \(k_1, \ldots, k_t \in S_{I, a}^{\oplus r}\). Since \(M_A\) is the quotient of \(M \otimes_R A\) by torsion, we see that the images of \(k_1, \ldots, k_t\) in \(M \otimes_R A\) are annihilated by a nonzero element \(\alpha \in A\). After replacing \(I, a\) by a different pair (recall that the colimit is filtered), we may assume \(\alpha = x/a^n\) for some \(x \in I^n\) nonzero. Then we find that \(x k_1, \ldots, x k_t\) map to zero in \(M \otimes_R A\). Hence after replacing \(I, a\) by a different pair we may assume \(x k_1, \ldots, x k_t\) map to zero in \(M \otimes_R R[\frac{I}{a}]\) for some nonzero \(x \in R\). Then finally replacing \(I, a\) by \(xI, xa\) we find that we may assume \(k_1, \ldots, k_t\) map to \(a\)-power torsion elements of \(M \otimes_R R[\frac{I}{a}]\). For any such pair \((I, a)\) we set \[M'_{I, a} = S_{I, a}^{\oplus r}/ \sum S_{I, a}k_j.\] Since \(M_A = S_A^{\oplus r}/ \sum S_Ak_j\) we see that \(M_A = \colim_{I, a} M'_{I, a}\). At this point we finally apply Algebra, Lemma 02JO (3) to conclude that \(M'_{I, a}\) is flat for some pair \((I, a)\) as above. This lemma does not apply a priori to the system of strict transforms \[M_{I, a} = (M \otimes_R R[\textstyle{\frac{I}{a}}])/a\text{-power torsion}\] as the transition maps may not satisfy the assumptions of the lemma. But now, since flatness implies torsion free (Lemma 0538) and since \(M_{I, a}\) is the quotient of \(M'_{I, a}\) (because we arranged it so the elements \(k_1, \ldots, k_t\) map to zero in \(M_{I, a}\)) by the \(a\)-power torsion submodule we also conclude that \(M'_{I, a} = M_{I, a}\) for such a pair and we win.

Lemma

Let \(R\) be a ring. Let \(M\) be a finite \(R\)-module. Let \(k \geq 0\) and \(I = \text{Fit}_k(M)\). For every \(a \in I\) with \(R' = R[\frac{I}{a}]\) the strict transform \[M' = (M \otimes_R R')/a\text{-power torsion}\] has \(\text{Fit}_k(M') = R'\).

Proof

First observe that \(\text{Fit}_k(M \otimes_R R') = IR' = aR'\). The first equality by Lemma 07ZA part (3) and the second equality by Algebra, Lemma 07Z3. From Lemma 080Z and exactness of localization we see that \(M'_{\mathfrak p'}\) can be generated by \(\leq k\) elements for every prime \(\mathfrak p'\) of \(R'\). Then \(\text{Fit}_k(M') = R'\) for example by Lemma 07ZC.

Lemma

Let \(R\) be a ring. Let \(M\) be a finite \(R\)-module. Let \(k \geq 0\) and \(I = \text{Fit}_k(M)\). Assume that \(M_\mathfrak p\) is free of rank \(k\) for every \(\mathfrak p \not \in V(I)\). Then for every \(a \in I\) with \(R' = R[\frac{I}{a}]\) the strict transform \[M' = (M \otimes_R R')/a\text{-power torsion}\] is locally free of rank \(k\).

Proof

By Lemma 0CZM we have \(\text{Fit}_k(M') = R'\). By Lemma 07ZD it suffices to show that \(\text{Fit}_{k - 1}(M') = 0\). Recall that \(R' \subset R'_a = R_a\), see Algebra, Lemma 07Z3. Hence it suffices to prove that \(\text{Fit}_{k - 1}(M')\) maps to zero in \(R'_a = R_a\). Since clearly \((M')_a = M_a\) this reduces us to showing that \(\text{Fit}_{k - 1}(M_a) = 0\) because formation of Fitting ideals commutes with base change according to Lemma 07ZA part (3). This is true by our assumption that \(M_a\) is finite locally free of rank \(k\) (see Algebra, Lemma 00NX) and the already cited Lemma 07ZD.

Lemma

Let \(R\) be a ring. Let \(M\) be a finite \(R\)-module. Let \(f \in R\) be an element such that \(M_f\) is finite locally free of rank \(r\). Then there exists a finitely generated ideal \(I \subset R\) with \(V(f) = V(I)\) such that for all \(a \in I\) with \(R' = R[\frac{I}{a}]\) the strict transform \[M' = (M \otimes_R R')/a\text{-power torsion}\] is locally free of rank \(r\).

Proof

Choose a surjection \(R^{\oplus n} \to M\). Choose a finite submodule \(K \subset \Ker(R^{\oplus n} \to M)\) such that \(R^{\oplus n}/K \to M\) becomes an isomorphism after inverting \(f\). This is possible because \(M_f\) is of finite presentation for example by Algebra, Lemma 00NX. Set \(M_1 = R^{\oplus n}/K\) and suppose we can prove the lemma for \(M_1\). Say \(I \subset R\) is the corresponding ideal. Then for \(a \in I\) the map \[M_1' = (M_1 \otimes_R R')/a\text{-power torsion} \longrightarrow M' = (M \otimes_R R')/a\text{-power torsion}\] is surjective. It is also an isomorphism after inverting \(a\) in \(R'\) as \(R'_a = R_f\), see Algebra, Lemma 080U. But \(a\) is a nonzerodivisor on \(M'_1\), whence the displayed map is an isomorphism. Thus it suffices to prove the lemma in case \(M\) is a finitely presented \(R\)-module.

Assume \(M\) is a finitely presented \(R\)-module. Then \(J = \text{Fit}_r(M) \subset S\) is a finitely generated ideal. We claim that \(I = fJ\) works.

We first check that \(V(f) = V(I)\). The inclusion \(V(f) \subset V(I)\) is clear. Conversely, if \(f \not \in \mathfrak p\), then \(\mathfrak p\) is not an element of \(V(J)\) by Lemma 07ZC. Thus \(\mathfrak p \not \in V(fJ) = V(I)\).

Let \(a \in I\) and set \(R' = R[\frac{I}{a}]\). We may write \(a = fb\) for some \(b \in J\). By Algebra, Lemmas 07Z3 and 0BBI we see that \(J R' = b R'\) and \(b\) is a nonzerodivisor in \(R'\). Let \(\mathfrak p' \subset R' = R[\frac{I}{a}]\) be a prime ideal. Then \(JR'_{\mathfrak p'}\) is generated by \(b\). It follows from Lemma 080Z that \(M'_{\mathfrak p'}\) can be generated by \(r\) elements. Since \(M'\) is finite, there exist \(m_1, \ldots, m_r \in M'\) and \(g \in R'\), \(g \not \in \mathfrak p'\) such that the corresponding map \((R')^{\oplus r} \to M'\) becomes surjective after inverting \(g\).

Finally, consider the ideal \(J' = \text{Fit}_{k - 1}(M')\). Note that \(J' R'_g\) is generated by the coefficients of relations between \(m_1, \ldots, m_r\) (compatibility of Fitting ideal with base change). Thus it suffices to show that \(J' = 0\), see Lemma 07ZD. Since \(R'_a = R_f\) (Algebra, Lemma 080U) and \(M'_a = M_f\) is free of rank \(r\) we see that \(J'_a = 0\). Since \(a\) is a nonzerodivisor in \(R'\) we conclude that \(J' = 0\) and we win.

Completion and flatness

In this section we discuss when the completion of a “big” flat module is flat.

Lemma

Let \(R\) be a ring. Let \(I \subset R\) be an ideal. Let \(A\) be a set. Assume \(R\) is Noetherian and complete with respect to \(I\). There is a canonical map \[\left(\bigoplus\nolimits_{\alpha \in A} R\right)^\wedge \longrightarrow \prod\nolimits_{\alpha \in A} R\] from the \(I\)-adic completion of the direct sum into the product which is universally injective.

Proof

By definition an element \(x\) of the left hand side is \(x = (x_n)\) where \(x_n = (x_{n, \alpha}) \in \bigoplus\nolimits_{\alpha \in A} R/I^n\) such that \(x_{n, \alpha} = x_{n + 1, \alpha} \bmod I^n\). As \(R = R^\wedge\) we see that for any \(\alpha\) there exists a \(y_\alpha \in R\) such that \(x_{n, \alpha} = y_\alpha \bmod I^n\). Note that for each \(n\) there are only finitely many \(\alpha\) such that the elements \(x_{n, \alpha}\) are nonzero. Conversely, given \((y_\alpha) \in \prod_\alpha R\) such that for each \(n\) there are only finitely many \(\alpha\) such that \(y_{\alpha} \bmod I^n\) is nonzero, then this defines an element of the left hand side. Hence we can think of an element of the left hand side as infinite “convergent sums” \(\sum_\alpha y_\alpha\) with \(y_\alpha \in R\) such that for each \(n\) there are only finitely many \(y_\alpha\) which are nonzero modulo \(I^n\). The displayed map maps this element to the element to \((y_\alpha)\) in the product. In particular the map is injective.

Let \(Q\) be a finite \(R\)-module. We have to show that the map \[Q \otimes_R \left(\bigoplus\nolimits_{\alpha \in A} R\right)^\wedge \longrightarrow Q \otimes_R \left(\prod\nolimits_{\alpha \in A} R\right)\] is injective, see Algebra, Theorem 058K. Choose a presentation \(R^{\oplus k} \to R^{\oplus m} \to Q \to 0\) and denote \(q_1, \ldots, q_m \in Q\) the corresponding generators for \(Q\). By Artin-Rees (Algebra, Lemma 00IN) there exists a constant \(c\) such that \(\Im(R^{\oplus k} \to R^{\oplus m}) \cap (I^N)^{\oplus m} \subset \Im((I^{N - c})^{\oplus k} \to R^{\oplus m})\). Let us contemplate the diagram \[\xymatrix{ \bigoplus_{l = 1}^k \left(\bigoplus\nolimits_{\alpha \in A} R\right)^\wedge \ar[r] \ar[d] & \bigoplus_{j = 1}^m \left(\bigoplus\nolimits_{\alpha \in A} R\right)^\wedge \ar[r] \ar[d] & Q \otimes_R \left(\bigoplus\nolimits_{\alpha \in A} R\right)^\wedge \ar[r] \ar[d] & 0 \\ \bigoplus_{l = 1}^k \left(\prod\nolimits_{\alpha \in A} R\right) \ar[r] & \bigoplus_{j = 1}^m \left(\prod\nolimits_{\alpha \in A} R\right) \ar[r] & Q \otimes_R \left(\prod\nolimits_{\alpha \in A} R\right) \ar[r] & 0 }\] with exact rows. Pick an element \(\sum_j \sum_\alpha y_{j, \alpha}\) of \(\bigoplus_{j = 1, \ldots, m} \left(\bigoplus\nolimits_{\alpha \in A} R\right)^\wedge\). If this element maps to zero in the module \(Q \otimes_R \left(\prod\nolimits_{\alpha \in A} R\right)\), then we see in particular that \(\sum_j q_j \otimes y_{j, \alpha} = 0\) in \(Q\) for each \(\alpha\). Thus we can find an element \((z_{1, \alpha}, \ldots, z_{k, \alpha}) \in \bigoplus_{l = 1, \ldots, k} R\) which maps to \((y_{1, \alpha}, \ldots, y_{m, \alpha}) \in \bigoplus_{j = 1, \ldots, m} R\). Moreover, if \(y_{j, \alpha} \in I^{N_\alpha}\) for \(j = 1, \ldots, m\), then we may assume that \(z_{l, \alpha} \in I^{N_\alpha - c}\) for \(l = 1, \ldots, k\). Hence the sum \(\sum_l \sum_\alpha z_{l, \alpha}\) is “convergent” and defines an element of \(\bigoplus_{l = 1, \ldots, k} \left(\bigoplus\nolimits_{\alpha \in A} R\right)^\wedge\) which maps to the element \(\sum_j \sum_\alpha y_{j, \alpha}\) we started out with. Thus the right vertical arrow is injective and we win.

The following lemma can also be deduced from Lemma 0912 below.

Lemma

Let \(R\) be a ring. Let \(I \subset R\) be an ideal. Let \(A\) be a set. Assume \(R\) is Noetherian. The completion \((\bigoplus\nolimits_{\alpha \in A} R)^\wedge\) is a flat \(R\)-module.

Proof

Denote \(R^\wedge\) the completion of \(R\) with respect to \(I\). As \(R \to R^\wedge\) is flat by Algebra, Lemma 00MB it suffices to prove that \((\bigoplus\nolimits_{\alpha \in A} R)^\wedge\) is a flat \(R^\wedge\)-module (use Algebra, Lemma 00HC). Since \[(\bigoplus\nolimits_{\alpha \in A} R)^\wedge = (\bigoplus\nolimits_{\alpha \in A} R^\wedge)^\wedge\] we may replace \(R\) by \(R^\wedge\) and assume that \(R\) is complete with respect to \(I\) (see Algebra, Lemma 031C). In this case Lemma 05BC tells us the map \((\bigoplus\nolimits_{\alpha \in A} R)^\wedge \to \prod_{\alpha \in A} R\) is universally injective. Thus, by Algebra, Lemma 058P it suffices to show that \(\prod_{\alpha \in A} R\) is flat. By Algebra, Proposition 05CZ (and Algebra, Lemma 05CY) we see that \(\prod_{\alpha \in A} R\) is flat.

Lemma

Let \(A\) be a Noetherian ring. Let \(I\) be an ideal of \(A\). Let \(M\) be a finite \(A\)-module. For every \(p > 0\) there exists a \(c > 0\) such that \(\text{Tor}_p^A(M, A/I^n) \to \text{Tor}_p^A(M, A/I^{n - c})\) is zero for all \(n \geq c\).

Proof

Proof for \(p = 1\). Choose a short exact sequence \(0 \to K \to A^{\oplus t} \to M \to 0\). Then \(\text{Tor}_1^A(M, A/I^n) = K \cap (I^n)^{\oplus t}/I^nK\). By Artin-Rees (Algebra, Lemma 00IN) there is a constant \(c \geq 0\) such that \(K \cap (I^n)^{\oplus t} \subset I^{n - c}K\) for \(n \geq c\). Thus the result for \(p = 1\). For \(p > 1\) we have \(\text{Tor}_p^A(M, A/I^n) = \text{Tor}^A_{p - 1}(K, A/I^n)\). Thus the lemma follows by induction.