Algebraic Curves

Introduction

In this chapter we develop some of the theory of algebraic curves. A reference covering algebraic curves over the complex numbers is the book [ACGH].

What we already know. Besides general algebraic geometry, we have already proved some specific results on algebraic curves. Here is a list.

1. We have discussed affine opens of and ample invertible sheaves on \(1\) dimensional Noetherian schemes in Varieties, Section 09N7.

2. We have seen a curve is either affine or projective in Varieties, Section 0A22.

3. We have discussed degrees of locally free modules on proper curves in Varieties, Section 0AYQ.

4. We have discussed the Picard scheme of a nonsingular projective curve over an algebraically closed field in Picard Schemes of Curves, Section 0B93.

Curves and function fields

In this section we elaborate on the results of Varieties, Section 0BXM in the case of curves.

Lemma

Let \(k\) be a field. Let \(X\) be a curve and \(Y\) a proper variety. Let \(U \subset X\) be a nonempty open and let \(f : U \to Y\) be a morphism. If \(x \in X\) is a closed point such that \(\mathcal{O}_{X, x}\) is a discrete valuation ring, then there exist an open \(U \subset U' \subset X\) containing \(x\) and a morphism of varieties \(f' : U' \to Y\) extending \(f\).

Proof

This is a special case of Morphisms, Lemma 0BX7.

Lemma

Let \(k\) be a field. Let \(X\) be a normal curve and \(Y\) a proper variety. The set of rational maps from \(X\) to \(Y\) is the same as the set of morphisms \(X \to Y\).

Proof

A rational map from \(X\) to \(Y\) can be extended to a morphism \(X \to Y\) by Lemma 0BXY as every local ring is a discrete valuation ring (for example by Varieties, Lemma 0B8Y). Conversely, if two morphisms \(f,g: X \to Y\) are equivalent as rational maps, then \(f = g\) by Morphisms, Lemma 01RH.

Lemma

Let \(k\) be a field. Let \(f : X \to Y\) be a nonconstant morphism of curves over \(k\). If \(Y\) is normal, then \(f\) is flat.

Proof

Pick \(x \in X\) mapping to \(y \in Y\). Then \(\mathcal{O}_{Y, y}\) is either a field or a discrete valuation ring (Varieties, Lemma 0B8Y). Since \(f\) is nonconstant it is dominant (as it must map the generic point of \(X\) to the generic point of \(Y\)). This implies that \(\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}\) is injective (Morphisms, Lemma 0CC1). Hence \(\mathcal{O}_{X, x}\) is torsion free as a \(\mathcal{O}_{Y, y}\)-module and therefore \(\mathcal{O}_{X, x}\) is flat as a \(\mathcal{O}_{Y, y}\)-module by More on Algebra, Lemma 0539.

Lemma

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

1. \(Y\) is separated over \(k\),

2. \(X\) is proper of dimension \(\leq 1\) over \(k\),

3. \(f(Z)\) has at least two points for every irreducible component \(Z \subset X\) of dimension \(1\).

Then \(f\) is finite.

Proof

The morphism \(f\) is proper by Morphisms, Lemma 01W6. Thus \(f(X)\) is closed and images of closed points are closed. Let \(y \in Y\) be the image of a closed point in \(X\). Then \(f^{-1}(\{y\})\) is a closed subset of \(X\) not containing any of the generic points of irreducible components of dimension \(1\) by condition (3). It follows that \(f^{-1}(\{y\})\) is finite. Hence \(f\) is finite over an open neighbourhood of \(y\) by More on Morphisms, Lemma 02UP (if \(Y\) is Noetherian, then you can use the easier Cohomology of Schemes, Lemma 02OH). Since we’ve seen above that there are enough of these points \(y\), the proof is complete.

Lemma

Let \(k\) be a field. Let \(X \to Y\) be a morphism of varieties with \(Y\) proper and \(X\) a curve. There exists a factorization \(X \to \overline{X} \to Y\) where \(X \to \overline{X}\) is an open immersion and \(\overline{X}\) is a projective curve.

Proof

This is clear from Lemma 0BXY and Varieties, Lemma 0BXW.

Here is the main theorem of this section. We will say a morphism \(f : X \to Y\) of varieties is constant if the image \(f(X)\) consists of a single point \(y\) of \(Y\). If this happens then \(y\) is a closed point of \(Y\) (since the image of a closed point of \(X\) will be a closed point of \(Y\)).

Theorem

Let \(k\) be a field. The following categories are canonically equivalent

1. The category of finitely generated field extensions \(K/k\) of transcendence degree \(1\).

2. The category of curves and dominant rational maps.

3. The category of normal projective curves and nonconstant morphisms.

4. The category of nonsingular projective curves and nonconstant morphisms.

5. The category of regular projective curves and nonconstant morphisms.

6. The category of normal proper curves and nonconstant morphisms.

Proof

The equivalence between categories (1) and (2) is the restriction of the equivalence of Varieties, Theorem 0BXN. Namely, a variety is a curve if and only if its function field has transcendence degree \(1\), see for example Varieties, Lemma 0A21.

The categories in (3), (4), (5), and (6) are the same. First of all, the terms “regular” and “nonsingular” are synonyms, see Properties, Definition 02IS. Being normal and regular are the same thing for Noetherian \(1\)-dimensional schemes (Properties, Lemmas 0569 and 0BX2). See Varieties, Lemma 0B8Y for the case of curves. Thus (3) is the same as (5). Finally, (6) is the same as (3) by Varieties, Lemma 0A26.

If \(f : X \to Y\) is a nonconstant morphism of nonsingular projective curves, then \(f\) sends the generic point \(\eta\) of \(X\) to the generic point \(\xi\) of \(Y\). Hence we obtain a morphism \(k(Y) = \mathcal{O}_{Y, \xi} \to \mathcal{O}_{X, \eta} = k(X)\) in the category (1). If two morphisms \(f,g: X \to Y\) gives the same morphism \(k(Y) \to k(X)\), then by the equivalence between (1) and (2), \(f\) and \(g\) are equivalent as rational maps, so \(f=g\) by Lemma 0BXZ. Conversely, suppose that we have a map \(k(Y) \to k(X)\) in the category (1). Then we obtain a morphism \(U \to Y\) for some nonempty open \(U \subset X\). By Lemma 0BXY this extends to all of \(X\) and we obtain a morphism in the category (5). Thus we see that there is a fully faithful functor (5)\(\to\)(1).

To finish the proof we have to show that every \(K/k\) in (1) is the function field of a normal projective curve. We already know that \(K = k(X)\) for some curve \(X\). After replacing \(X\) by its normalization (which is a variety birational to \(X\)) we may assume \(X\) is normal (Varieties, Lemma 0BXR). Then we choose \(X \to \overline{X}\) with \(\overline{X} \setminus X = \{x_1, \ldots, x_n\}\) as in Varieties, Lemma 0BXW. Since \(X\) is normal and since each of the local rings \(\mathcal{O}_{\overline{X}, x_i}\) is normal we conclude that \(\overline{X}\) is a normal projective curve as desired. (Remark: We can also first compactify using Varieties, Lemma 0BXV and then normalize using Varieties, Lemma 0BXR. Doing it this way we avoid using the somewhat tricky Morphisms, Lemma 0BXB.)

Definition

Let \(k\) be a field. Let \(X\) be a curve. A nonsingular projective model of \(X\) is a pair \((Y, \varphi)\) where \(Y\) is a nonsingular projective curve and \(\varphi : k(X) \to k(Y)\) is an isomorphism of function fields.

A nonsingular projective model is determined up to unique isomorphism by Theorem 0BY1. Thus we often say “the nonsingular projective model”. We usually drop \(\varphi\) from the notation. Warning: it needn’t be the case that \(Y\) is smooth over \(k\) but Lemma 0BY3 shows this can only happen in positive characteristic.

Lemma

Let \(k\) be a field. Let \(X\) be a curve and let \(Y\) be the nonsingular projective model of \(X\). If \(k\) is perfect, then \(Y\) is a smooth projective curve.

Proof

See Varieties, Lemma 0B8Y for example.

Lemma

Let \(k\) be a field. Let \(X\) be a geometrically irreducible curve over \(k\). For a field extension \(K/k\) denote \(Y_K\) a nonsingular projective model of \((X_K)_{red}\).

1. If \(X\) is proper, then \(Y_K\) is the normalization of \(X_K\).

2. There exists \(K/k\) finite purely inseparable such that \(Y_K\) is smooth.

3. Whenever \(Y_K\) is smooth¹ we have \(H^0(Y_K, \mathcal{O}_{Y_K}) = K\).

4. Given a commutative diagram \[\xymatrix{ \Omega & K' \ar[l] \\ K \ar[u] & k \ar[l] \ar[u] }\] of fields such that \(Y_K\) and \(Y_{K'}\) are smooth, then \(Y_\Omega = (Y_K)_\Omega = (Y_{K'})_\Omega\).

Proof

Let \(X'\) be a nonsingular projective model of \(X\). Then \(X'\) and \(X\) have isomorphic nonempty open subschemes. In particular \(X'\) is geometrically irreducible as \(X\) is (some details omitted). Thus we may assume that \(X\) is projective.

Assume \(X\) is proper. Then \(X_K\) is proper and hence the normalization \((X_K)^\nu\) is proper as a scheme finite over a proper scheme (Varieties, Lemma 0BXR and Morphisms, Lemmas 01WN and 01W3). On the other hand, \(X_K\) is irreducible as \(X\) is geometrically irreducible. Hence \(X_K^\nu\) is proper, normal, irreducible, and birational to \((X_K)_{red}\). This proves (1) because a proper curve is projective (Varieties, Lemma 0A26).

Proof of (2). As \(X\) is proper and we have (1), we can apply Varieties, Lemma 0BXT to find \(K/k\) finite purely inseparable such that \(Y_K\) is geometrically normal. Then \(Y_K\) is geometrically regular as normal and regular are the same for curves (Properties, Lemma 0BX2). Then \(Y\) is a smooth variety by Varieties, Lemma 038X.

If \(Y_K\) is geometrically reduced, then \(Y_K\) is geometrically integral (Varieties, Lemma 038K) and we see that \(H^0(Y_K, \mathcal{O}_{Y_K}) = K\) by Varieties, Lemma 04L2. This proves (3) because a smooth variety is geometrically reduced (even geometrically regular, see Varieties, Lemma 038X).

If \(Y_K\) is smooth, then for every extension \(\Omega/K\) the base change \((Y_K)_\Omega\) is smooth over \(\Omega\) (Morphisms, Lemma 01VB). Hence it is clear that \(Y_\Omega = (Y_K)_\Omega\). This proves (4).

Linear series

We deviate from the classical story (see Remark 0CCT) by defining linear series in the following manner.

Definition

Let \(k\) be a field. Let \(X\) be a proper scheme of dimension \(\leq 1\) over \(k\). Let \(d \geq 0\) and \(r \geq 0\). A linear series of degree \(d\) and dimension \(r\) is a pair \((\mathcal{L}, V)\) where \(\mathcal{L}\) is an invertible \(\mathcal{O}_X\)-module of degree \(d\) (Varieties, Definition 0AYR) and \(V \subset H^0(X, \mathcal{L})\) is a \(k\)-subvector space of dimension \(r + 1\). We will abbreviate this by saying \((\mathcal{L}, V)\) is a \(\mathfrak g^r_d\) on \(X\).

We will mostly use this when \(X\) is a nonsingular proper curve. In fact, the definition above is just one way to generalize the classical definition of a \(\mathfrak g^r_d\). For example, if \(X\) is a proper curve, then one can generalize linear series by allowing \(\mathcal{L}\) to be a torsion free coherent \(\mathcal{O}_X\)-module of rank \(1\). On a nonsingular curve every torsion free coherent module is locally free, so this agrees with our notion for nonsingular proper curves.

The following lemma explains the geometric meaning of linear series for proper nonsingular curves.

Lemma

Let \(k\) be a field. Let \(X\) be a nonsingular proper curve over \(k\). Let \((\mathcal{L}, V)\) be a \(\mathfrak g^r_d\) on \(X\). Then there exists a morphism \[\varphi : X \longrightarrow \mathbf{P}^r_k = \text{Proj}(k[T_0, \ldots, T_r])\] of varieties over \(k\) and a map \(\alpha : \varphi^*\mathcal{O}_{\mathbf{P}^r_k}(1) \to \mathcal{L}\) such that \(\varphi^*T_0, \ldots, \varphi^*T_r\) are sent to a basis of \(V\) by \(\alpha\).

Proof

Let \(s_0, \ldots, s_r \in V\) be a \(k\)-basis. Since \(X\) is nonsingular the image \(\mathcal{L}' \subset \mathcal{L}\) of the map \(s_0, \ldots, s_r : \mathcal{O}_X^{\oplus r + 1} \to \mathcal{L}\) is an invertible \(\mathcal{O}_X\)-module for example by Divisors, Lemma 0CC4. Then we use Constructions, Lemma 01NE to get a morphism \[\varphi = \varphi_{(\mathcal{L}', (s_0, \ldots, s_r))} : X \longrightarrow \mathbf{P}^r_k\] as in the statement of the lemma.

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme of dimension \(\leq 1\) over \(k\). If \(X\) has a \(\mathfrak g^r_d\), then \(X\) has a \(\mathfrak g^s_d\) for all \(0 \leq s \leq r\).

Proof

This is true because a vector space \(V\) of dimension \(r + 1\) over \(k\) has a linear subspace of dimension \(s + 1\) for all \(0 \leq s \leq r\).

Lemma

Let \(k\) be a field. Let \(X\) be a nonsingular proper curve over \(k\). Let \((\mathcal{L}, V)\) be a \(\mathfrak g^1_d\) on \(X\). Then the morphism \(\varphi : X \to \mathbf{P}^1_k\) of Lemma 0CCP either

1. is nonconstant and has degree \(\leq d\), or

2. factors through a closed point of \(\mathbf{P}^1_k\) and in this case \(H^0(X, \mathcal{O}_X) \not = k\).

Proof

By Lemma 0CCP we see that \(\mathcal{L}' = \varphi^*\mathcal{O}_{\mathbf{P}^1_k}(1)\) has a nonzero map \(\mathcal{L}' \to \mathcal{L}\). Hence by Varieties, Lemma 0B40 we see that \(0 \leq \deg(\mathcal{L}') \leq d\). If \(\deg(\mathcal{L}') = 0\), then the same lemma tells us \(\mathcal{L}' \cong \mathcal{O}_X\) and since we have two linearly independent sections we find we are in case (2). If \(\deg(\mathcal{L}') > 0\) then \(\varphi\) is nonconstant (since the pullback of an invertible module by a constant morphism is trivial). Hence \[\deg(\mathcal{L}') = \deg(X/\mathbf{P}^1_k) \deg(\mathcal{O}_{\mathbf{P}^1_k}(1))\] by Varieties, Lemma 0AYZ. This finishes the proof as the degree of \(\mathcal{O}_{\mathbf{P}^1_k}(1)\) is \(1\).

Lemma

Let \(k\) be a field. Let \(X\) be a proper curve over \(k\) with \(H^0(X, \mathcal{O}_X) = k\). If \(X\) has a \(\mathfrak g^r_d\), then \(r \leq d\). If equality holds, then \(H^1(X, \mathcal{O}_X) = 0\), i.e., the genus of \(X\) (Definition 0BY7) is \(0\).

Proof

Let \((\mathcal{L}, V)\) be a \(\mathfrak g^r_d\). Since this will only increase \(r\), we may assume \(V = H^0(X, \mathcal{L})\). Choose a nonzero element \(s \in V\). Then the zero scheme of \(s\) is an effective Cartier divisor \(D \subset X\), we have \(\mathcal{L} = \mathcal{O}_X(D)\), and we have a short exact sequence \[0 \to \mathcal{O}_X \to \mathcal{L} \to \mathcal{L}|_D \to 0\] see Divisors, Lemma 01X0 and Remark 0C6K. By Varieties, Lemma 0AYY we have \(\deg(D) = \deg(\mathcal{L}) = d\). Since \(D\) is an Artinian scheme we have \(\mathcal{L}|_D \cong \mathcal{O}_D\)². Thus \[\dim_k H^0(D, \mathcal{L}|_D) = \dim_k H^0(D, \mathcal{O}_D) = \deg(D) = d\] On the other hand, by assumption \(\dim_k H^0(X, \mathcal{O}_X) = 1\) and \(\dim H^0(X, \mathcal{L}) = r + 1\). We conclude that \(r + 1 \leq 1 + d\), i.e., \(r \leq d\) as in the lemma.

Assume equality holds. Then \(H^0(X, \mathcal{L}) \to H^0(X, \mathcal{L}|_D)\) is surjective. If we knew that \(H^1(X, \mathcal{L})\) was zero, then we would conclude that \(H^1(X, \mathcal{O}_X)\) is zero by the long exact cohomology sequence and the proof would be complete. Our strategy will be to replace \(\mathcal{L}\) by a large power which has vanishing. As \(\mathcal{L}|_D\) is the trivial invertible module (see above), we can find a section \(t\) of \(\mathcal{L}\) whose restriction of \(D\) generates \(\mathcal{L}|_D\). Consider the multiplication map \[\mu : H^0(X, \mathcal{L}) \otimes_k H^0(X, \mathcal{L}) \longrightarrow H^0(X, \mathcal{L}^{\otimes 2})\] and consider the short exact sequence \[0 \to \mathcal{L} \xrightarrow{s} \mathcal{L}^{\otimes 2} \to \mathcal{L}^{\otimes 2}|_D \to 0\] Since \(H^0(\mathcal{L}) \to H^0(\mathcal{L}|_D)\) is surjective and since \(t\) maps to a trivialization of \(\mathcal{L}|_D\) we see that \(\mu(H^0(X, \mathcal{L}) \otimes t)\) gives a subspace of \(H^0(X, \mathcal{L}^{\otimes 2})\) surjecting onto the global sections of \(\mathcal{L}^{\otimes 2}|_D\). Thus we see that \[\dim H^0(X, \mathcal{L}^{\otimes 2}) = r + 1 + d = 2r + 1 = \deg(\mathcal{L}^{\otimes 2}) + 1\] Ok, so \(\mathcal{L}^{\otimes 2}\) has the same property as \(\mathcal{L}\), i.e., that the dimension of the space of global sections is equal to the degree plus one. Since \(\mathcal{L}\) is ample (Varieties, Lemma 0B5X) there exists some \(n_0\) such that \(\mathcal{L}^{\otimes n}\) has vanishing \(H^1\) for all \(n \geq n_0\) (Cohomology of Schemes, Lemma 0B5T). Thus applying the argument above to \(\mathcal{L}^{\otimes n}\) with \(n = 2^m\) for some sufficiently large \(m\) we conclude the lemma is true.

Remark

Let \(X\) be a smooth projective curve over an algebraically closed field \(k\). We say two effective Cartier divisors \(D, D' \subset X\) are linearly equivalent if and only if \(\mathcal{O}_X(D) \cong \mathcal{O}_X(D')\) as \(\mathcal{O}_X\)-modules. Since \(\Pic(X) = \text{Cl}(X)\) (Divisors, Lemma 0BE9) we see that \(D\) and \(D'\) are linearly equivalent if and only if the Weil divisors associated to \(D\) and \(D'\) define the same element of \(\text{Cl}(X)\). Given an effective Cartier divisor \(D \subset X\) of degree \(d\) the complete linear system or complete linear series \(|D|\) of \(D\) is the set of effective Cartier divisors \(E \subset X\) which are linearly equivalent to \(D\). Another way to say it is that \(|D|\) is the set of closed points of the fibre of the morphism \[\gamma_d : \underline{\Hilbfunctor}^d_{X/k} \longrightarrow \underline{\Picardfunctor}^d_{X/k}\] (Picard Schemes of Curves, Lemma 0BA0) over the closed point corresponding to \(\mathcal{O}_X(D)\). This gives \(|D|\) a natural scheme structure and it turns out that \(|D| \cong \mathbf{P}^m_k\) with \(m + 1 = h^0(\mathcal{O}_X(D))\). In fact, more canonically we have \[|D| = \mathbf{P}(H^0(X, \mathcal{O}_X(D))^\vee)\] where \((-)^\vee\) indicates \(k\)-linear dual and \(\mathbf{P}\) is as in Constructions, Example 0FCY. In this language a linear system or a linear series on \(X\) is a closed subvariety \(L \subset |D|\) which can be cut out by linear equations. If \(L\) has dimension \(r\), then \(L = \mathbf{P}(V^\vee)\) where \(V \subset H^0(X, \mathcal{O}_X(D))\) is a linear subspace of dimension \(r + 1\). Thus the classical linear series \(L \subset |D|\) corresponds to the linear series \((\mathcal{O}_X(D), V)\) as defined above.

Duality

In this section we work out the consequences of the very general material on dualizing complexes and duality for proper \(1\)-dimensional schemes over fields. If you are interested in the analogous discussion for higher dimension proper schemes over fields, see Duality for Schemes, Section 0FVU.

Lemma

Let \(X\) be a proper scheme of dimension \(\leq 1\) over a field \(k\). There exists a dualizing complex \(\omega_X^\bullet\) with the following properties

1. \(H^i(\omega_X^\bullet)\) is nonzero only for \(i = -1, 0\),

2. \(\omega_X = H^{-1}(\omega_X^\bullet)\) is a coherent Cohen-Macaulay module whose support is the irreducible components of dimension \(1\),

3. for \(x \in X\) closed, the module \(H^0(\omega_{X, x}^\bullet)\) is nonzero if and only if either

   1. \(\dim(\mathcal{O}_{X, x}) = 0\) or

   2. \(\dim(\mathcal{O}_{X, x}) = 1\) and \(\mathcal{O}_{X, x}\) is not Cohen-Macaulay,

4. for \(K \in D_\QCoh(\mathcal{O}_X)\) there are functorial isomorphisms³ \[\Ext^i_X(K, \omega_X^\bullet) = \Hom_k(H^{-i}(X, K), k)\] compatible with shifts and distinguished triangles,

5. there are functorial isomorphisms \(\Hom(\mathcal{F}, \omega_X) = \Hom_k(H^1(X, \mathcal{F}), k)\) for \(\mathcal{F}\) quasi-coherent on \(X\),

6. if \(X \to \Spec(k)\) is smooth of relative dimension \(1\), then \(\omega_X \cong \Omega_{X/k}\).

Proof

Denote \(f : X \to \Spec(k)\) the structure morphism. We start with the relative dualizing complex \[\omega_X^\bullet = \omega_{X/k}^\bullet = a(\mathcal{O}_{\Spec(k)})\] as described in Duality for Schemes, Remark 0B6S. Then property (4) holds by construction as \(a\) is the right adjoint for \(f_* : D_\QCoh(\mathcal{O}_X) \to D(\mathcal{O}_{\Spec(k)})\). Since \(f\) is proper we have \(f^!(\mathcal{O}_{\Spec(k)}) = a(\mathcal{O}_{\Spec(k)})\) by definition, see Duality for Schemes, Section 0A9Y. Hence \(\omega_X^\bullet\) and \(\omega_X\) are as in Duality for Schemes, Example 0AWI and as in Duality for Schemes, Example 0AWJ. Parts (1) and (2) follow from Duality for Schemes, Lemma 0AWN. For a closed point \(x \in X\) we see that \(\omega_{X, x}^\bullet\) is a normalized dualizing complex over \(\mathcal{O}_{X, x}\), see Duality for Schemes, Lemma 0AWL. Assertion (3) then follows from Dualizing Complexes, Lemma 0AWS. Assertion (5) follows from Duality for Schemes, Lemma 0AWP for coherent \(\mathcal{F}\) and in general by unwinding (4) for \(K = \mathcal{F}[0]\) and \(i = -1\). Assertion (6) follows from Duality for Schemes, Lemma 0BRT.

Lemma

Let \(X\) be a proper scheme over a field \(k\) which is Cohen-Macaulay and equidimensional of dimension \(1\). The module \(\omega_X\) of Lemma 0BS2 has the following properties

1. \(\omega_X\) is a dualizing module on \(X\) (Duality for Schemes, Section 0AWH),

2. \(\omega_X\) is a coherent Cohen-Macaulay module whose support is \(X\),

3. there are functorial isomorphisms \(\Ext^i_X(K, \omega_X[1]) = \Hom_k(H^{-i}(X, K), k)\) compatible with shifts for \(K \in D_\QCoh(X)\),

4. there are functorial isomorphisms \(\Ext^{1 + i}(\mathcal{F}, \omega_X) = \Hom_k(H^{-i}(X, \mathcal{F}), k)\) for \(\mathcal{F}\) quasi-coherent on \(X\).

Proof

Recall from the proof of Lemma 0BS2 that \(\omega_X\) is as in Duality for Schemes, Example 0AWI and hence is a dualizing module. The other statements follow from Lemma 0BS2 and the fact that \(\omega_X^\bullet = \omega_X[1]\) as \(X\) is Cohen-Macualay (Duality for Schemes, Lemma 0AWT).

Remark

Let \(X\) be a proper scheme of dimension \(\leq 1\) over a field \(k\). Let \(\omega_X^\bullet\) and \(\omega_X\) be as in Lemma 0BS2. If \(\mathcal{E}\) is a finite locally free \(\mathcal{O}_X\)-module with dual \(\mathcal{E}^\vee\) then we have canonical isomorphisms \[\Hom_k(H^{-i}(X, \mathcal{E}), k) = H^i(X, \mathcal{E}^\vee \otimes_{\mathcal{O}_X}^\mathbf{L} \omega_X^\bullet)\] This follows from the lemma and Cohomology, Lemma 08DQ. If \(X\) is Cohen-Macaulay and equidimensional of dimension \(1\), then we have canonical isomorphisms \[\Hom_k(H^{-i}(X, \mathcal{E}), k) = H^{1 + i}(X, \mathcal{E}^\vee \otimes_{\mathcal{O}_X} \omega_X)\] by Lemma 0BS3. In particular if \(\mathcal{L}\) is an invertible \(\mathcal{O}_X\)-module, then we have \[\dim_k H^0(X, \mathcal{L}) = \dim_k H^1(X, \mathcal{L}^{\otimes -1} \otimes_{\mathcal{O}_X} \omega_X)\] and \[\dim_k H^1(X, \mathcal{L}) = \dim_k H^0(X, \mathcal{L}^{\otimes -1} \otimes_{\mathcal{O}_X} \omega_X)\]

Here is a sanity check for the dualizing complex.

Lemma

Let \(X\) be a proper scheme of dimension \(\leq 1\) over a field \(k\). Let \(\omega_X^\bullet\) and \(\omega_X\) be as in Lemma 0BS2.

1. If \(X \to \Spec(k)\) factors as \(X \to \Spec(k') \to \Spec(k)\) for some field \(k'\), then \(\omega_X^\bullet\) and \(\omega_X\) satisfy properties (4), (5), (6) with \(k\) replaced with \(k'\).

2. If \(K/k\) is a field extension, then the pullback of \(\omega_X^\bullet\) and \(\omega_X\) to the base change \(X_K\) are as in Lemma 0BS2 for the morphism \(X_K \to \Spec(K)\).

Proof

Denote \(f : X \to \Spec(k)\) the structure morphism. Assertion (1) really means that \(\omega_X^\bullet\) and \(\omega_X\) are as in Lemma 0BS2 for the morphism \(f' : X \to \Spec(k')\). In the proof of Lemma 0BS2 we took \(\omega_X^\bullet = a(\mathcal{O}_{\Spec(k)})\) where \(a\) be is the right adjoint of Duality for Schemes, Lemma 0A9E for \(f\). Thus we have to show \(a(\mathcal{O}_{\Spec(k)}) \cong a'(\mathcal{O}_{\Spec(k)})\) where \(a'\) be is the right adjoint of Duality for Schemes, Lemma 0A9E for \(f'\). Since \(k' \subset H^0(X, \mathcal{O}_X)\) we see that \(k'/k\) is a finite extension (Cohomology of Schemes, Lemma 02O6). By uniqueness of adjoints we have \(a = a' \circ b\) where \(b\) is the right adjoint of Duality for Schemes, Lemma 0A9E for \(g : \Spec(k') \to \Spec(k)\). Another way to say this: we have \(f^! = (f')^! \circ g^!\). Thus it suffices to show that \(\Hom_k(k', k) \cong k'\) as \(k'\)-modules, see Duality for Schemes, Example 0A9F. This holds because these are \(k'\)-vector spaces of the same dimension (namely dimension \(1\)).

Proof of (2). This holds because we have base change for \(a\) by Duality for Schemes, Lemma 0AA8. See discussion in Duality for Schemes, Remark 0B6S.

Lemma

Let \(X\) be a proper scheme of dimension \(\leq 1\) over a field \(k\). Let \(i : Y \to X\) be a closed immersion. Let \(\omega_X^\bullet\), \(\omega_X\), \(\omega_Y^\bullet\), \(\omega_Y\) be as in Lemma 0BS2. Then

1. \(\omega_Y^\bullet = R\SheafHom(\mathcal{O}_Y, \omega_X^\bullet)\),

2. \(\omega_Y = \SheafHom(\mathcal{O}_Y, \omega_X)\) and \(i_*\omega_Y = \SheafHom_{\mathcal{O}_X}(i_*\mathcal{O}_Y, \omega_X)\).

Proof

Denote \(g : Y \to \Spec(k)\) and \(f : X \to \Spec(k)\) the structure morphisms. Then \(g = f \circ i\). Denote \(a, b, c\) the right adjoint of Duality for Schemes, Lemma 0A9E for \(f, g, i\). Then \(b = c \circ a\) by uniqueness of right adjoints and because \(Rg_* = Rf_* \circ Ri_*\). In the proof of Lemma 0BS2 we set \(\omega_X^\bullet = a(\mathcal{O}_{\Spec(k)})\) and \(\omega_Y^\bullet = b(\mathcal{O}_{\Spec(k)})\). Hence \(\omega_Y^\bullet = c(\omega_X^\bullet)\) which implies (1) by Duality for Schemes, Lemma 0A9X. Since \(\omega_X = H^{-1}(\omega_X^\bullet)\) and \(\omega_Y = H^{-1}(\omega_Y^\bullet)\) we conclude that \(\omega_Y = \SheafHom(\mathcal{O}_Y, \omega_X)\). This implies \(i_*\omega_Y = \SheafHom_{\mathcal{O}_X}(i_*\mathcal{O}_Y, \omega_X)\) by Duality for Schemes, Lemma 0A77.

Lemma

Let \(X\) be a proper scheme over a field \(k\) which is Gorenstein, reduced, and equidimensional of dimension \(1\). Let \(i : Y \to X\) be a reduced closed subscheme equidimensional of dimension \(1\). Let \(j : Z \to X\) be the scheme theoretic closure of \(X \setminus Y\). Then

1. \(Y\) and \(Z\) are Cohen-Macaulay,

2. if \(\mathcal{I} \subset \mathcal{O}_X\), resp. \(\mathcal{J} \subset \mathcal{O}_X\) is the ideal sheaf of \(Y\), resp. \(Z\) in \(X\), then \[\mathcal{I} = i_*\mathcal{I}' \quad\text{and}\quad \mathcal{J} = j_*\mathcal{J}'\] where \(\mathcal{I}' \subset \mathcal{O}_Z\), resp. \(\mathcal{J}' \subset \mathcal{O}_Y\) is the ideal sheaf of \(Y \cap Z\) in \(Z\), resp. \(Y\),

3. \(\omega_Y = \mathcal{J}'(i^*\omega_X)\) and \(i_*(\omega_Y) = \mathcal{J}\omega_X\),

4. \(\omega_Z = \mathcal{I}'(i^*\omega_X)\) and \(i_*(\omega_Z) = \mathcal{I}\omega_X\),

5. we have the following short exact sequences \[\begin{align*} 0 \to \omega_X \to i_*i^*\omega_X \oplus j_*j^*\omega_X \to \mathcal{O}_{Y \cap Z} \to 0 \\ 0 \to i_*\omega_Y \to \omega_X \to j_*j^*\omega_X \to 0 \\ 0 \to j_*\omega_Z \to \omega_X \to i_*i^*\omega_X \to 0 \\ 0 \to i_*\omega_Y \oplus j_*\omega_Z \to \omega_X \to \mathcal{O}_{Y \cap Z} \to 0 \\ 0 \to \omega_Y \to i^*\omega_X \to \mathcal{O}_{Y \cap Z} \to 0 \\ 0 \to \omega_Z \to j^*\omega_X \to \mathcal{O}_{Y \cap Z} \to 0 \end{align*}\]

Here \(\omega_X\), \(\omega_Y\), \(\omega_Z\) are as in Lemma 0BS2.

Proof

A reduced \(1\)-dimensional Noetherian scheme is Cohen-Macaulay, so (1) is true. Since \(X\) is reduced, we see that \(X = Y \cup Z\) scheme theoretically. With notation as in Morphisms, Lemma 0C4J and by the statement of that lemma we have a short exact sequence \[0 \to \mathcal{O}_X \to \mathcal{O}_Y \oplus \mathcal{O}_Z \to \mathcal{O}_{Y \cap Z} \to 0\] Since \(\mathcal{J} = \Ker(\mathcal{O}_X \to \mathcal{O}_Z)\), \(\mathcal{J}' = \Ker(\mathcal{O}_Y \to \mathcal{O}_{Y \cap Z})\), \(\mathcal{I} = \Ker(\mathcal{O}_X \to \mathcal{O}_Y)\), and \(\mathcal{I}' = \Ker(\mathcal{O}_Z \to \mathcal{O}_{Y \cap Z})\) a diagram chase implies (2). Observe that \(\mathcal{I} + \mathcal{J}\) is the ideal sheaf of \(Y \cap Z\) and that \(\mathcal{I} \cap \mathcal{J} = 0\). Hence we have the following exact sequences \[\begin{align*} 0 \to \mathcal{O}_X \to \mathcal{O}_Y \oplus \mathcal{O}_Z \to \mathcal{O}_{Y \cap Z} \to 0 \\ 0 \to \mathcal{J} \to \mathcal{O}_X \to \mathcal{O}_Z \to 0 \\ 0 \to \mathcal{I} \to \mathcal{O}_X \to \mathcal{O}_Y \to 0 \\ 0 \to \mathcal{J} \oplus \mathcal{I} \to \mathcal{O}_X \to \mathcal{O}_{Y \cap Z} \to 0 \\ 0 \to \mathcal{J}' \to \mathcal{O}_Y \to \mathcal{O}_{Y \cap Z} \to 0 \\ 0 \to \mathcal{I}' \to \mathcal{O}_Z \to \mathcal{O}_{Y \cap Z} \to 0 \end{align*}\] Since \(X\) is Gorenstein \(\omega_X\) is an invertible \(\mathcal{O}_X\)-module (Duality for Schemes, Lemma 0BFQ). Since \(Y \cap Z\) has dimension \(0\) we have \(\omega_X|_{Y \cap Z} \cong \mathcal{O}_{Y \cap Z}\). Thus if we prove (3) and (4), then we obtain the short exact sequences of the lemma by tensoring the above short exact sequence with the invertible module \(\omega_X\). By symmetry it suffices to prove (3) and by (2) it suffices to prove \(i_*(\omega_Y) = \mathcal{J}\omega_X\).

We have \(i_*\omega_Y = \SheafHom_{\mathcal{O}_X}(i_*\mathcal{O}_Y, \omega_X)\) by Lemma 0E33. Again using that \(\omega_X\) is invertible we finally conclude that it suffices to show \(\SheafHom_{\mathcal{O}_X}(\mathcal{O}_X/\mathcal{I}, \mathcal{O}_X)\) maps isomorphically to \(\mathcal{J}\) by evaluation at \(1\). In other words, that \(\mathcal{J}\) is the annihilator of \(\mathcal{I}\). This follows from the above.

Riemann-Roch

Let \(k\) be a field. Let \(X\) be a proper scheme of dimension \(\leq 1\) over \(k\). In Varieties, Section 0AYQ we have defined the degree of a locally free \(\mathcal{O}_X\)-module \(\mathcal{E}\) of constant rank by the formula [0BRX]\[\begin{equation} \deg(\mathcal{E}) = \chi(X, \mathcal{E}) - \text{rank}(\mathcal{E})\chi(X, \mathcal{O}_X) \end{equation}\] see Varieties, Definition 0AYR. In the chapter on Chow Homology we defined the first Chern class of \(\mathcal{E}\) as an operation on cycles (Chow Homology, Section 02U4) and we proved that [0BRY]\[\begin{equation} \deg(\mathcal{E}) = \deg(c_1(\mathcal{E}) \cap [X]_1) \end{equation}\] see Chow Homology, Lemma 0AZ3. Combining (0BRX) and (0BRY) we obtain our first version of the Riemann-Roch formula [0BRZ]\[\begin{equation} \chi(X, \mathcal{E}) = \deg(c_1(\mathcal{E}) \cap [X]_1) + \text{rank}(\mathcal{E})\chi(X, \mathcal{O}_X) \end{equation}\] If \(\mathcal{L}\) is an invertible \(\mathcal{O}_X\)-module, then we can also consider the numerical intersection \((\mathcal{L} \cdot X)\) as defined in Varieties, Definition 0BEP. However, this does not give anything new as [0BS0]\[\begin{equation} (\mathcal{L} \cdot X) = \deg(\mathcal{L}) \end{equation}\] by Varieties, Lemma 0BEY. If \(\mathcal{L}\) is ample, then this integer is positive and is called the degree [0BS1]\[\begin{equation} \deg_\mathcal{L}(X) = (\mathcal{L} \cdot X) = \deg(\mathcal{L}) \end{equation}\] of \(X\) with respect to \(\mathcal{L}\), see Varieties, Definition 0BEW.

To obtain a true Riemann-Roch theorem we would like to write \(\chi(X, \mathcal{O}_X)\) as the degree of a canonical zero cycle on \(X\). We refer to [F] for a fully general version of this. We will use duality to get a formula in the case where \(X\) is Gorenstein; however, in some sense this is a cheat (for example because this method cannot work in higher dimension).

We first use Lemmas 0BS2 and 0BS3 to get a relation between the euler characteristic of \(\mathcal{O}_X\) and the euler characteristic of the dualizing complex or the dualizing module.

Lemma

Let \(X\) be a proper scheme of dimension \(\leq 1\) over a field \(k\). With \(\omega_X^\bullet\) and \(\omega_X\) as in Lemma 0BS2 we have \[\chi(X, \mathcal{O}_X) = \chi(X, \omega_X^\bullet)\] If \(X\) is Cohen-Macaulay and equidimensional of dimension \(1\), then \[\chi(X, \mathcal{O}_X) = - \chi(X, \omega_X)\]

Proof

We define the right hand side of the first formula as follows: \[\chi(X, \omega_X^\bullet) = \sum\nolimits_{i \in \mathbf{Z}} (-1)^i\dim_k H^i(X, \omega_X^\bullet)\] This is well defined because \(\omega_X^\bullet\) is in \(D^b_{\textit{Coh}}(\mathcal{O}_X)\), but also because \[H^i(X, \omega_X^\bullet) = \Ext^i(\mathcal{O}_X, \omega_X^\bullet) = H^{-i}(X, \mathcal{O}_X)\] which is always finite dimensional and nonzero only if \(i = 0, -1\). This of course also proves the first formula. The second is a consequence of the first because \(\omega_X^\bullet = \omega_X[1]\) in the CM case, see Lemma 0BS3.

We will use Lemma 0BS5 to get the desired formula for \(\chi(X, \mathcal{O}_X)\) in the case that \(\omega_X\) is invertible, i.e., that \(X\) is Gorenstein. The statement is that \(-1/2\) of the first Chern class of \(\omega_X\) capped with the cycle \([X]_1\) associated to \(X\) is a natural zero cycle on \(X\) with half-integer coefficients whose degree is \(\chi(X, \mathcal{O}_X)\). The occurrence of fractions in the statement of Riemann-Roch cannot be avoided.

Lemma

Let \(X\) be a proper scheme over a field \(k\) which is Gorenstein and equidimensional of dimension \(1\). Let \(\omega_X\) be as in Lemma 0BS2. Then

1. \(\omega_X\) is an invertible \(\mathcal{O}_X\)-module,

2. \(\deg(\omega_X) = -2\chi(X, \mathcal{O}_X)\),

3. for a locally free \(\mathcal{O}_X\)-module \(\mathcal{E}\) of constant rank we have \[\chi(X, \mathcal{E}) = \deg(\mathcal{E}) - \textstyle{\frac{1}{2}} \text{rank}(\mathcal{E}) \deg(\omega_X)\] and \(\dim_k(H^i(X, \mathcal{E})) = \dim_k(H^{1 - i}(X, \mathcal{E}^\vee \otimes_{\mathcal{O}_X} \omega_X))\) for all \(i \in \mathbf{Z}\).

Nonsingular (normal) curves are Gorenstein, see Duality for Schemes, Lemma 0DWG.

Proof

Recall that Gorenstein schemes are Cohen-Macaulay (Duality for Schemes, Lemma 0C00) and hence \(\omega_X\) is a dualizing module on \(X\), see Lemma 0BS3. It follows more or less from the definition of the Gorenstein property that the dualizing sheaf is invertible, see Duality for Schemes, Section 0AWV. By (0BRZ) applied to \(\omega_X\) we have \[\chi(X, \omega_X) = \deg(c_1(\omega_X) \cap [X]_1) + \chi(X, \mathcal{O}_X)\] Combined with Lemma 0BS5 this gives \[2\chi(X, \mathcal{O}_X) = - \deg(c_1(\omega_X) \cap [X]_1) = - \deg(\omega_X)\] the second equality by (0BRY). Putting this back into (0BRZ) for \(\mathcal{E}\) gives the displayed formula of the lemma. The symmetry in dimensions is a consequence of duality for \(X\), see Remark 0BS4.

Some vanishing results

This section contains some very weak vanishing results. Please see Section 0E39 for a few more and more interesting results.

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) having dimension \(1\) and \(H^0(X, \mathcal{O}_X) = k\). Then \(X\) is connected, Cohen-Macaulay, and equidimensional of dimension \(1\).

Proof

Since \(\Gamma(X, \mathcal{O}_X) = k\) has no nontrivial idempotents, we see that \(X\) is connected. This already shows that \(X\) is equidimensional of dimension \(1\) (any irreducible component of dimension \(0\) would be a connected component). Let \(\mathcal{I} \subset \mathcal{O}_X\) be the maximal coherent submodule supported in closed points. Then \(\mathcal{I}\) exists (Divisors, Lemma 02OL) and is globally generated (Varieties, Lemma 0AYT). Since \(1 \in \Gamma(X, \mathcal{O}_X)\) is not a section of \(\mathcal{I}\) we conclude that \(\mathcal{I} = 0\). Thus \(X\) does not have embedded points (Divisors, Lemma 02OL). Thus \(X\) has \((S_1)\) by Divisors, Lemma 0346. Hence \(X\) is Cohen-Macaulay.

In this section we work in the following situation.

Situation

Here \(k\) is a field, \(X\) is a proper scheme over \(k\) having dimension \(1\) and \(H^0(X, \mathcal{O}_X) = k\).

By Lemma 0BY5 the scheme \(X\) is Cohen-Macaulay and equidimensional of dimension \(1\). The dualizing module \(\omega_X\) discussed in Lemmas 0BS2 and 0BS3 has nonvanishing \(H^1\) because in fact \(\dim_k H^1(X, \omega_X) = \dim_k H^0(X, \mathcal{O}_X) = 1\). It turns out that anything slightly more “positive” than \(\omega_X\) has vanishing \(H^1\).

Lemma

In Situation 0B5D. Given an exact sequence \[\omega_X \to \mathcal{F} \to \mathcal{Q} \to 0\] of coherent \(\mathcal{O}_X\)-modules with \(H^1(X, \mathcal{Q}) = 0\) (for example if \(\dim(\text{Supp}(\mathcal{Q})) = 0\)), then either \(H^1(X, \mathcal{F}) = 0\) or \(\mathcal{F} = \omega_X \oplus \mathcal{Q}\).

Proof

(The parenthetical statement follows from Cohomology of Schemes, Lemma 0B3J.) Since \(H^0(X, \mathcal{O}_X) = k\) is dual to \(H^1(X, \omega_X)\) (see Section 0B5B) we see that \(\dim H^1(X, \omega_X) = 1\). The sheaf \(\omega_X\) represents the functor \(\mathcal{F} \mapsto \Hom_k(H^1(X, \mathcal{F}), k)\) on the category of coherent \(\mathcal{O}_X\)-modules (Duality for Schemes, Lemma 0AWP). Consider an exact sequence as in the statement of the lemma and assume that \(H^1(X, \mathcal{F}) \not = 0\). Since \(H^1(X, \mathcal{Q}) = 0\) we see that \(H^1(X, \omega_X) \to H^1(X, \mathcal{F})\) is an isomorphism. By the universal property of \(\omega_X\) stated above, we conclude there is a map \(\mathcal{F} \to \omega_X\) whose action on \(H^1\) is the inverse of this isomorphism. The composition \(\omega_X \to \mathcal{F} \to \omega_X\) is the identity (by the universal property) and the lemma is proved.

Lemma

In Situation 0B5D. Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module which is globally generated and not isomorphic to \(\mathcal{O}_X\). Then \(H^1(X, \omega_X \otimes \mathcal{L}) = 0\).

Proof

By duality as discussed in Section 0B5B we have to show that \(H^0(X, \mathcal{L}^{\otimes - 1}) = 0\). If not, then we can choose a global section \(t\) of \(\mathcal{L}^{\otimes - 1}\) and a global section \(s\) of \(\mathcal{L}\) such that \(st \not = 0\). However, then \(st\) is a constant multiple of \(1\), by our assumption that \(H^0(X, \mathcal{O}_X) = k\). It follows that \(\mathcal{L} \cong \mathcal{O}_X\), which is a contradiction.

Lemma

In Situation 0B5D. Given an exact sequence \[\omega_X \to \mathcal{F} \to \mathcal{Q} \to 0\] of coherent \(\mathcal{O}_X\)-modules with \(\dim(\text{Supp}(\mathcal{Q})) = 0\) and \(\dim_k H^0(X, \mathcal{Q}) \geq 2\) and such that there is no nonzero submodule \(\mathcal{Q}' \subset \mathcal{F}\) such that \(\mathcal{Q}' \to \mathcal{Q}\) is injective. Then the submodule of \(\mathcal{F}\) generated by global sections surjects onto \(\mathcal{Q}\).

Proof

Let \(\mathcal{F}' \subset \mathcal{F}\) be the submodule generated by global sections and the image of \(\omega_X \to \mathcal{F}\). Since \(\dim_k H^0(X, \mathcal{Q}) \geq 2\) and \(\dim_k H^1(X, \omega_X) = \dim_k H^0(X, \mathcal{O}_X) = 1\), we see that \(\mathcal{F}' \to \mathcal{Q}\) is not zero and \(\omega_X \to \mathcal{F}'\) is not an isomorphism. Hence \(H^1(X, \mathcal{F}') = 0\) by Lemma 0B5E and our assumption on \(\mathcal{F}\). Consider the short exact sequence \[0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{Q}/\Im(\mathcal{F}' \to \mathcal{Q}) \to 0\] If the quotient on the right is nonzero, then we obtain a contradiction because then \(H^0(X, \mathcal{F})\) is bigger than \(H^0(X, \mathcal{F}')\).

Here is an example global generation statement.

Lemma

In Situation 0B5D assume that \(X\) is integral. Let \(0 \to \omega_X \to \mathcal{F} \to \mathcal{Q} \to 0\) be a short exact sequence of coherent \(\mathcal{O}_X\)-modules with \(\mathcal{F}\) torsion free, \(\dim(\text{Supp}(\mathcal{Q})) = 0\), and \(\dim_k H^0(X, \mathcal{Q}) \geq 2\). Then \(\mathcal{F}\) is globally generated.

Proof

Consider the submodule \(\mathcal{F}'\) generated by the global sections. By Lemma 0B5F we see that \(\mathcal{F}' \to \mathcal{Q}\) is surjective, in particular \(\mathcal{F}' \not = 0\). Since \(X\) is a curve, we see that \(\mathcal{F}' \subset \mathcal{F}\) is an inclusion of rank \(1\) sheaves, hence \(\mathcal{Q}' = \mathcal{F}/\mathcal{F}'\) is supported in finitely many points. To get a contradiction, assume that \(\mathcal{Q}'\) is nonzero. Then we see that \(H^1(X, \mathcal{F}') \not = 0\). Then we get a nonzero map \(\mathcal{F}' \to \omega_X\) by the universal property (Duality for Schemes, Lemma 0AWP). The image of the composition \(\mathcal{F}' \to \omega_X \to \mathcal{F}\) is generated by global sections, hence is inside of \(\mathcal{F}'\). Thus we get a nonzero self map \(\mathcal{F}' \to \mathcal{F}'\). Since \(\mathcal{F}'\) is torsion free of rank \(1\) on a proper curve this has to be an automorphism (details omitted). But then this implies that \(\mathcal{F}'\) is contained in \(\omega_X \subset \mathcal{F}\) contradicting the surjectivity of \(\mathcal{F}' \to \mathcal{Q}\).

Lemma

In Situation 0B5D. Let \(\mathcal{L}\) be a very ample invertible \(\mathcal{O}_X\)-module with \(\deg(\mathcal{L}) \geq 2\). Then \(\omega_X \otimes_{\mathcal{O}_X} \mathcal{L}\) is globally generated.

Proof

Assume \(k\) is algebraically closed. Let \(x \in X\) be a closed point. Let \(C_i \subset X\) be the irreducible components and for each \(i\) let \(x_i \in C_i\) be the generic point. By Varieties, Lemma 0B58 we can choose a section \(s \in H^0(X, \mathcal{L})\) such that \(s\) vanishes at \(x\) but not at \(x_i\) for all \(i\). The corresponding module map \(s : \mathcal{O}_X \to \mathcal{L}\) is injective with cokernel \(\mathcal{Q}\) supported in finitely many points and with \(H^0(X, \mathcal{Q}) \geq 2\). Consider the corresponding exact sequence \[0 \to \omega_X \to \omega_X \otimes \mathcal{L} \to \omega_X \otimes \mathcal{Q} \to 0\] By Lemma 0B5F we see that the module generated by global sections surjects onto \(\omega_X \otimes \mathcal{Q}\). Since \(x\) was arbitrary this proves the lemma. Some details omitted.

We will reduce the case where \(k\) is not algebraically closed, to the algebraically closed field case. We suggest the reader skip the rest of the proof. Choose an algebraic closure \(\overline{k}\) of \(k\) and consider the base change \(X_{\overline{k}}\). Let us check that \(X_{\overline{k}} \to \Spec(\overline{k})\) is an example of Situation 0B5D. By flat base change (Cohomology of Schemes, Lemma 02KH) we see that \(H^0(X_{\overline{k}}, \mathcal{O}) = \overline{k}\). The scheme \(X_{\overline{k}}\) is proper over \(\overline{k}\) (Morphisms, Lemma 01W4) and equidimensional of dimension \(1\) (Morphisms, Lemma 02FY). The pullback of \(\omega_X\) to \(X_{\overline{k}}\) is the dualizing module of \(X_{\overline{k}}\) by Lemma 0E32. The pullback of \(\mathcal{L}\) to \(X_{\overline{k}}\) is very ample (Morphisms, Lemma 0B3F). The degree of the pullback of \(\mathcal{L}\) to \(X_{\overline{k}}\) is equal to the degree of \(\mathcal{L}\) on \(X\) (Varieties, Lemma 0B59). Finally, we see that \(\omega_X \otimes \mathcal{L}\) is globally generated if and only if its base change is so (Varieties, Lemma 0B57). In this way we see that the result follows from the result in the case of an algebraically closed ground field.

Very ample invertible sheaves

An often used criterion for very ampleness of an invertible module \(\mathcal{L}\) on a scheme \(X\) of finite type over an algebraically closed field is: sections of \(\mathcal{L}\) separate points and tangent vectors (Varieties, Section 0E8R). Here is another criterion for curves; please compare with Varieties, Subsection 08A2.

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) having dimension \(1\) and \(H^0(X, \mathcal{O}_X) = k\). Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. Assume

1. \(\mathcal{L}\) has a regular global section,

2. \(H^1(X, \mathcal{L}) = 0\), and

3. \(\mathcal{L}\) is ample.

Then \(\mathcal{L}^{\otimes 6}\) is very ample on \(X\) over \(k\).

Proof

Let \(s\) be a regular global section of \(\mathcal{L}\). Let \(i : Z = Z(s) \to X\) be the zero scheme of \(s\), see Divisors, Section 0C4S. By condition (3) we see that \(Z \not = \emptyset\) (small detail omitted). Consider the short exact sequence \[0 \to \mathcal{O}_X \xrightarrow{s} \mathcal{L} \to i_*(\mathcal{L}|_Z) \to 0\] Tensoring with \(\mathcal{L}\) we obtain \[0 \to \mathcal{L} \to \mathcal{L}^{\otimes 2} \to i_*(\mathcal{L}^{\otimes 2}|_Z) \to 0\] Observe that \(Z\) has dimension \(0\) (Divisors, Lemma 056N) and hence is the spectrum of an Artinian ring (Varieties, Lemma 06LH) hence \(\mathcal{L}|_Z \cong \mathcal{O}_Z\) (Algebra, Lemma 02M9). The short exact sequence also shows that \(H^1(X, \mathcal{L}^{\otimes 2}) = 0\) (for example using Varieties, Lemma 0AYT to see vanishing in the spot on the right). Using induction on \(n \geq 1\) and the sequence \[0 \to \mathcal{L}^{\otimes n} \xrightarrow{s} \mathcal{L}^{\otimes n + 1} \to i_*(\mathcal{L}^{\otimes n + 1}|_Z) \to 0\] we see that \(H^1(X, \mathcal{L}^{\otimes n}) = 0\) for \(n > 0\) and that there exists a global section \(t_{n + 1}\) of \(\mathcal{L}^{\otimes n + 1}\) which gives a trivialization of \(\mathcal{L}^{\otimes n + 1}|_Z \cong \mathcal{O}_Z\).

Consider the multiplication map \[\mu_n : H^0(X, \mathcal{L}) \otimes_k H^0(X, \mathcal{L}^{\otimes n}) \oplus H^0(X, \mathcal{L}^{\otimes 2}) \otimes_k H^0(X, \mathcal{L}^{\otimes n - 1}) \longrightarrow H^0(X, \mathcal{L}^{\otimes n + 1})\] We claim this is surjective for \(n \geq 3\). To see this we consider the short exact sequence \[0 \to \mathcal{L}^{\otimes n} \xrightarrow{s} \mathcal{L}^{\otimes n + 1} \to i_*(\mathcal{L}^{\otimes n + 1}|_Z) \to 0\] The sections of \(\mathcal{L}^{\otimes n + 1}\) coming from the left in this sequence are in the image of \(\mu_n\). On the other hand, since \(H^0(\mathcal{L}^{\otimes 2}) \to H^0(\mathcal{L}^{\otimes 2}|_Z)\) is surjective (see above) and since \(t_{n - 1}\) maps to a trivialization of \(\mathcal{L}^{\otimes n - 1}|_Z\) we see that \(\mu_n(H^0(X, \mathcal{L}^{\otimes 2}) \otimes t_{n - 1})\) gives a subspace of \(H^0(X, \mathcal{L}^{\otimes n + 1})\) surjecting onto the global sections of \(\mathcal{L}^{\otimes n + 1}|_Z\). This proves the claim.

From the claim in the previous paragraph we conclude that the graded \(k\)-algebra \[S = \bigoplus\nolimits_{n \geq 0} H^0(X, \mathcal{L}^{\otimes n})\] is generated in degrees \(0, 1, 2, 3\) over \(k\). Recall that \(X = \text{Proj}(S)\), see Morphisms, Lemma 0C6J. Thus \(S^{(6)} = \bigoplus_{n} S_{6n}\) is generated in degree \(1\). This means that \(\mathcal{L}^{\otimes 6}\) is very ample as desired.

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) having dimension \(1\) and \(H^0(X, \mathcal{O}_X) = k\). Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. Assume

1. \(\mathcal{L}\) is globally generated,

2. \(H^1(X, \mathcal{L}) = 0\), and

3. \(\mathcal{L}\) is ample.

Then \(\mathcal{L}^{\otimes 2}\) is very ample on \(X\) over \(k\).

Proof

Choose basis \(s_0, \ldots, s_n\) of \(H^0(X, \mathcal{L}^{\otimes 2})\) over \(k\). By property (1) we see that \(\mathcal{L}^{\otimes 2}\) is globally generated and we get a morphism \[\varphi_{\mathcal{L}^{\otimes 2}, (s_0, \ldots, s_n)} : X \longrightarrow \mathbf{P}^n_k\] See Constructions, Section 01ND. The lemma asserts that this morphism is a closed immersion. To check this we may replace \(k\) by its algebraic closure, see Descent, Lemma 02L6. Thus we may assume \(k\) is algebraically closed.

Assume \(k\) is algebraically closed. For each generic point \(\eta_i \in X\) let \(V_i \subset H^0(X, \mathcal{L})\) be the \(k\)-subvector space of sections vanishing at \(\eta_i\). Since \(\mathcal{L}\) is globally generated, we see that \(V_i \not = H^0(X, \mathcal{L})\). Since \(X\) has only a finite number of irreducible components and \(k\) is infinite, we can find \(s \in H^0(X, \mathcal{L})\) nonvanishing at \(\eta_i\) for all \(i\). Then \(s\) is a regular section of \(\mathcal{L}\) (because \(X\) is Cohen-Macaulay by Lemma 0BY5 and hence \(\mathcal{L}\) has no embedded associated points).

In particular, all of the statements given in the proof of Lemma 0E8V hold with this \(s\). Moreover, as \(\mathcal{L}\) is globally generated, we can find a global section \(t \in H^0(X, \mathcal{L})\) such that \(t|_Z\) is nonvanishing (argue as above using the finite number of points of \(Z\)). Then in the proof of Lemma 0E8V we can use \(t\) to see that additionally the multiplication map \[\mu_n : H^0(X, \mathcal{L}) \otimes_k H^0(X, \mathcal{L}^{\otimes 2}) \longrightarrow H^0(X, \mathcal{L}^{\otimes 3})\] is surjective. Thus \[S = \bigoplus\nolimits_{n \geq 0} H^0(X, \mathcal{L}^{\otimes n})\] is generated in degrees \(0, 1, 2\) over \(k\). Arguing as in the proof of Lemma 0E8V we find that \(S^{(2)} = \bigoplus_{n} S_{2n}\) is generated in degree \(1\). This means that \(\mathcal{L}^{\otimes 2}\) is very ample as desired. Some details omitted.

The genus of a curve

If \(X\) is a smooth projective geometrically irreducible curve over a field \(k\), then we’ve previously defined the genus of \(X\) as the dimension of \(H^1(X, \mathcal{O}_X)\), see Picard Schemes of Curves, Definition 0B9W. Observe that \(H^0(X, \mathcal{O}_X) = k\) in this case, see Varieties, Lemma 04L2. Let us generalize this as follows.

Definition

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) having dimension \(1\) and \(H^0(X, \mathcal{O}_X) = k\). Then the genus of \(X\) is \(g = \dim_k H^1(X, \mathcal{O}_X)\).

This is sometimes called the arithmetic genus of \(X\). In the literature the arithmetic genus of a proper curve \(X\) over \(k\) is sometimes defined as \[p_a(X) = 1 - \chi(X, \mathcal{O}_X) = 1 - \dim_k H^0(X, \mathcal{O}_X) + \dim_k H^1(X, \mathcal{O}_X)\] This agrees with our definition when it applies because we assume \(H^0(X, \mathcal{O}_X) = k\). But note that

1. \(p_a(X)\) can be negative, and

2. \(p_a(X)\) depends on the base field \(k\) and should be written \(p_a(X/k)\).

For example if \(k = \mathbf{Q}\) and \(X = \mathbf{P}^1_{\mathbf{Q}(i)}\) then \(p_a(X/\mathbf{Q}) = -1\) and \(p_a(X/\mathbf{Q}(i)) = 0\).

The assumption that \(H^0(X, \mathcal{O}_X) = k\) in our definition has two consequences. On the one hand, it means there is no confusion about the base field. On the other hand, it implies the scheme \(X\) is Cohen-Macaulay and equidimensional of dimension \(1\) (Lemma 0BY5). If \(\omega_X\) denotes the dualizing module as in Lemmas 0BS2 and 0BS3 we see that [0BY8]\[\begin{equation} g = \dim_k H^1(X, \mathcal{O}_X) = \dim_k H^0(X, \omega_X) \end{equation}\] by duality, see Remark 0BS4.

If \(X\) is proper over \(k\) of dimension \(\leq 1\) and \(H^0(X, \mathcal{O}_X)\) is not equal to the ground field \(k\), instead of using the arithmetic genus \(p_a(X)\) given by the displayed formula above we shall use the invariant \(\chi(X, \mathcal{O}_X)\). In fact, it is advocated in [FAC, page 276] and [Hirzebruch, Introduction] that we should call \(\chi(X, \mathcal{O}_X)\) the arithmetic genus.

Lemma

Let \(k'/k\) be a field extension. Let \(X\) be a proper scheme over \(k\) having dimension \(1\) and \(H^0(X, \mathcal{O}_X) = k\). Then \(X_{k'}\) is a proper scheme over \(k'\) having dimension \(1\) and \(H^0(X_{k'}, \mathcal{O}_{X_{k'}}) = k'\). Moreover the genus of \(X_{k'}\) is equal to the genus of \(X\).

Proof

The dimension of \(X_{k'}\) is \(1\) for example by Morphisms, Lemma 02FY. The morphism \(X_{k'} \to \Spec(k')\) is proper by Morphisms, Lemma 01W4. The equality \(H^0(X_{k'}, \mathcal{O}_{X_{k'}}) = k'\) follows from Cohomology of Schemes, Lemma 02KH. The equality of the genus follows from the same lemma.

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) having dimension \(1\) and \(H^0(X, \mathcal{O}_X) = k\). If \(X\) is Gorenstein, then \[\deg(\omega_X) = 2g - 2\] where \(g\) is the genus of \(X\) and \(\omega_X\) is as in Lemma 0BS2.

Proof

Immediate from Lemma 0BS6.

Lemma

Let \(X\) be a smooth proper curve over a field \(k\) with \(H^0(X, \mathcal{O}_X) = k\). Then \[\dim_k H^0(X, \Omega_{X/k}) = g \quad\text{and}\quad \deg(\Omega_{X/k}) = 2g - 2\] where \(g\) is the genus of \(X\).

Proof

By Lemma 0BS2 we have \(\Omega_{X/k} = \omega_X\). Hence the formulas hold by (0BY8) and Lemma 0C19.

Plane curves

Let \(k\) be a field. A plane curve will be a curve \(X\) which is isomorphic to a closed subscheme of \(\mathbf{P}^2_k\). Often the embedding \(X \to \mathbf{P}^2_k\) will be considered given. By Divisors, Example 0BXK a curve is determined by the corresponding homogeneous ideal \[I(X) = \Ker\left( k[T_0, T_2, T_2] \longrightarrow \bigoplus \Gamma(X, \mathcal{O}_X(n)) \right)\] Recall that in this situation we have \[X = \text{Proj}(k[T_0, T_2, T_2]/I)\] as closed subschemes of \(\mathbf{P}^2_k\). For more general information on these constructions we refer the reader to Divisors, Example 0BXK and the references therein. It turns out that \(I(X) = (F)\) for some homogeneous polynomial \(F \in k[T_0, T_1, T_2]\), see Lemma 0BYB. Since \(X\) is irreducible, it follows that \(F\) is irreducible, see Lemma 0BYC. Moreover, looking at the short exact sequence \[0 \to \mathcal{O}_{\mathbf{P}^2_k}(-d) \xrightarrow{F} \mathcal{O}_{\mathbf{P}^2_k} \to \mathcal{O}_X \to 0\] where \(d = \deg(F)\) we find that \(H^0(X, \mathcal{O}_X) = k\) and that \(X\) has genus \((d - 1)(d - 2)/2\), see proof of Lemma 0BYD.

To find smooth plane curves it is easiest to write explicit equations. Let \(p\) denote the characteristic of \(k\). If \(p\) does not divide \(d\), then we can take \[F = T_0^d + T_1^d + T_2^d\] The corresponding curve \(X = V_+(F)\) is called the Fermat curve of degree \(d\). It is smooth because on each standard affine piece \(D_+(T_i)\) we obtain a curve isomorphic to the affine curve \[\Spec(k[x, y]/(x^d + y^d + 1))\] The ring map \(k \to k[x, y]/(x^d + y^d + 1)\) is smooth by Algebra, Lemma 00TE as \(d x^{d - 1}\) and \(d y^{d - 1}\) generate the unit ideal in \(k[x, y]/(x^d + y^d + 1)\). If \(p | d\) but \(p \not = 3\) then you can use the equation \[F = T_0^{d - 1}T_1 + T_1^{d - 1}T_2 + T_2^{d - 1}T_0\] Namely, on the affine pieces you get \(x + x^{d - 1}y + y^{d - 1}\) with derivatives \(1 - x^{d - 2}y\) and \(x^{d - 1} - y^{d - 2}\) whose common zero set (of all three) is empty⁴. We leave it to the reader to make examples in characteristic \(3\).

More generally for any field \(k\) and any \(n\) and \(d\) there exists a smooth hypersurface of degree \(d\) in \(\mathbf{P}^n_k\), see for example [Poonen].

Of course, in this way we only find smooth curves whose genus is a triangular number. To get smooth curves of an arbitrary genus one can look for smooth curves lying on \(\mathbf{P}^1 \times \mathbf{P}^1\) (insert future reference here).

Lemma

Let \(Z \subset \mathbf{P}^2_k\) be a closed subscheme which is equidimensional of dimension \(1\) and has no embedded points (equivalently \(Z\) is Cohen-Macaulay). Then the ideal \(I(Z) \subset k[T_0, T_1, T_2]\) corresponding to \(Z\) is principal.

Proof

This is a special case of Divisors, Lemma 0BXL (see also Varieties, Lemma 0BXU). The parenthetical statement follows from the fact that a \(1\) dimensional Noetherian scheme is Cohen-Macaulay if and only if it has no embedded points, see Divisors, Lemma 0BXG.

Lemma

Let \(Z \subset \mathbf{P}^2_k\) be as in Lemma 0BYB and let \(I(Z) = (F)\) for some \(F \in k[T_0, T_1, T_2]\). Then \(Z\) is a curve if and only if \(F\) is irreducible.

Proof

If \(F\) is reducible, say \(F = F' F''\) then let \(Z'\) be the closed subscheme of \(\mathbf{P}^2_k\) defined by \(F'\). It is clear that \(Z' \subset Z\) and that \(Z' \not = Z\). Since \(Z'\) has dimension \(1\) as well, we conclude that either \(Z\) is not reduced, or that \(Z\) is not irreducible. Conversely, write \(Z = \sum a_i D_i\) where \(D_i\) are the irreducible components of \(Z\), see Divisors, Lemmas 0AGB and 0BXH. Let \(F_i \in k[T_0, T_1, T_2]\) be the homogeneous polynomial generating the ideal of \(D_i\). Then it is clear that \(F\) and \(\prod F_i^{a_i}\) cut out the same closed subscheme of \(\mathbf{P}^2_k\). Hence \(F = \lambda \prod F_i^{a_i}\) for some \(\lambda \in k^*\) because both generate the ideal of \(Z\). Thus we see that if \(F\) is irreducible, then \(Z\) is a prime divisor, i.e., a curve.

Lemma

Let \(Z \subset \mathbf{P}^2_k\) be as in Lemma 0BYB and let \(I(Z) = (F)\) for some \(F \in k[T_0, T_1, T_2]\). Then \(H^0(Z, \mathcal{O}_Z) = k\) and the genus of \(Z\) is \((d - 1)(d - 2)/2\) where \(d = \deg(F)\).

Proof

Let \(S = k[T_0, T_1, T_2]\). There is an exact sequence of graded modules \[0 \to S(-d) \xrightarrow{F} S \to S/(F) \to 0\] Denote \(i : Z \to \mathbf{P}^2_k\) the given closed immersion. Applying the exact functor \(\widetilde{\ }\) (Constructions, Lemma 01M7) we obtain \[0 \to \mathcal{O}_{\mathbf{P}^2_k}(-d) \to \mathcal{O}_{\mathbf{P}^2_k} \to i_*\mathcal{O}_Z \to 0\] because \(F\) generates the ideal of \(Z\). Note that the cohomology groups of \(\mathcal{O}_{\mathbf{P}^2_k}(-d)\) and \(\mathcal{O}_{\mathbf{P}^2_k}\) are given in Cohomology of Schemes, Lemma 01XT. On the other hand, we have \(H^q(Z, \mathcal{O}_Z) = H^q(\mathbf{P}^2_k, i_*\mathcal{O}_Z)\) by Cohomology of Schemes, Lemma 089W. Applying the long exact cohomology sequence we first obtain that \[k = H^0(\mathbf{P}^2_k, \mathcal{O}_{\mathbf{P}^2_k}) \longrightarrow H^0(Z, \mathcal{O}_Z)\] is an isomorphism and next that the boundary map \[H^1(Z, \mathcal{O}_Z) \longrightarrow H^2(\mathbf{P}^2_k, \mathcal{O}_{\mathbf{P}^2_k}(-d)) \cong k[T_0, T_1, T_2]_{d - 3}\] is an isomorphism. Since it is easy to see that the dimension of this is \((d - 1)(d - 2)/2\) the proof is finished.

Lemma

Let \(Z \subset \mathbf{P}^2_k\) be as in Lemma 0BYB and let \(I(Z) = (F)\) for some \(F \in k[T_0, T_1, T_2]\). If \(Z \to \Spec(k)\) is smooth in at least one point and \(k\) is infinite, then there exists a closed point \(z \in Z\) contained in the smooth locus such that \(\kappa(z)/k\) is finite separable of degree at most \(d\).

Proof

Suppose that \(z' \in Z\) is a point where \(Z \to \Spec(k)\) is smooth. After renumbering the coordinates if necessary we may assume \(z'\) is contained in \(D_+(T_0)\). Set \(f = F(1, x, y) \in k[x, y]\). Then \(Z \cap D_+(X_0)\) is isomorphic to the spectrum of \(k[x, y]/(f)\). Let \(f_x, f_y\) be the partial derivatives of \(f\) with respect to \(x, y\). Since \(z'\) is a smooth point of \(Z/k\) we see that either \(f_x\) or \(f_y\) is nonzero in \(z'\) (see discussion in Algebra, Section 00T1). After renumbering the coordinates we may assume \(f_y\) is not zero at \(z'\). Hence there is a nonempty open subscheme \(V \subset Z \cap D_{+}(X_0)\) such that the projection \[p : V \longrightarrow \Spec(k[x])\] is étale. Because the degree of \(f\) as a polynomial in \(y\) is at most \(d\), we see that the degrees of the fibres of the projection \(p\) are at most \(d\) (see discussion in Morphisms, Section 03J3). Moreover, as \(p\) is étale the image of \(p\) is an open \(U \subset \Spec(k[x])\). Finally, since \(k\) is infinite, the set of \(k\)-rational points \(U(k)\) of \(U\) is infinite, in particular not empty. Pick any \(t \in U(k)\) and let \(z \in V\) be a point mapping to \(t\). Then \(z\) works.

Curves of genus zero

Later we will need to know what a proper genus zero curve looks like. It turns out that a Gorenstein proper genus zero curve is a plane curve of degree \(2\), i.e., a conic, see Lemma 0C6N. A general proper genus zero curve is obtained from a nonsingular one (over a bigger field) by a pushout procedure, see Lemma 0DJB. Since a nonsingular curve is Gorenstein, these two results cover all possible cases.

Lemma

Let \(X\) be a proper curve over a field \(k\) with \(H^0(X, \mathcal{O}_X) = k\). If \(X\) has genus \(0\), then every invertible \(\mathcal{O}_X\)-module \(\mathcal{L}\) of degree \(0\) is trivial.

Proof

Namely, we have \(\dim_k H^0(X, \mathcal{L}) \geq 0 + 1 - 0 = 1\) by Riemann-Roch (Lemma 0BS6), hence \(\mathcal{L}\) has a nonzero section, hence \(\mathcal{L} \cong \mathcal{O}_X\) by Varieties, Lemma 0B40.

Lemma

Let \(X\) be a proper curve over a field \(k\) with \(H^0(X, \mathcal{O}_X) = k\). Assume \(X\) has genus \(0\). Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module of degree \(d > 0\). Then we have

1. \(\dim_k H^0(X, \mathcal{L}) = d + 1\) and \(\dim_k H^1(X, \mathcal{L}) = 0\),

2. \(\mathcal{L}\) is very ample and defines a closed immersion into \(\mathbf{P}^d_k\).

Proof

By definition of degree and genus we have \[\dim_k H^0(X, \mathcal{L}) - \dim_k H^1(X, \mathcal{L}) = d + 1\] Let \(s\) be a nonzero section of \(\mathcal{L}\). Then the zero scheme of \(s\) is an effective Cartier divisor \(D \subset X\), we have \(\mathcal{L} = \mathcal{O}_X(D)\) and we have a short exact sequence \[0 \to \mathcal{O}_X \to \mathcal{L} \to \mathcal{L}|_D \to 0\] see Divisors, Lemma 01X0 and Remark 0C6K. Since \(H^1(X, \mathcal{O}_X) = 0\) by assumption, we see that \(H^0(X, \mathcal{L}) \to H^0(X, \mathcal{L}|_D)\) is surjective. As \(\mathcal{L}|_D\) is generated by global sections (because \(\dim(D) = 0\), see Varieties, Lemma 0AYT) we conclude that the invertible module \(\mathcal{L}\) is generated by global sections. In fact, since \(D\) is an Artinian scheme we have \(\mathcal{L}|_D \cong \mathcal{O}_D\)⁵ and hence we can find a section \(t\) of \(\mathcal{L}\) whose restriction of \(D\) generates \(\mathcal{L}|_D\). The short exact sequence also shows that \(H^1(X, \mathcal{L}) = 0\).

For \(n \geq 1\) consider the multiplication map \[\mu_n : H^0(X, \mathcal{L}) \otimes_k H^0(X, \mathcal{L}^{\otimes n}) \longrightarrow H^0(X, \mathcal{L}^{\otimes n + 1})\] We claim this is surjective. To see this we consider the short exact sequence \[0 \to \mathcal{L}^{\otimes n} \xrightarrow{s} \mathcal{L}^{\otimes n + 1} \to \mathcal{L}^{\otimes n + 1}|_D \to 0\] The sections of \(\mathcal{L}^{\otimes n + 1}\) coming from the left in this sequence are in the image of \(\mu_n\). On the other hand, since \(H^0(\mathcal{L}) \to H^0(\mathcal{L}|_D)\) is surjective and since \(t^n\) maps to a trivialization of \(\mathcal{L}^{\otimes n}|_D\) we see that \(\mu_n(H^0(X, \mathcal{L}) \otimes t^n)\) gives a subspace of \(H^0(X, \mathcal{L}^{\otimes n + 1})\) surjecting onto the global sections of \(\mathcal{L}^{\otimes n + 1}|_D\). This proves the claim.

Observe that \(\mathcal{L}\) is ample by Varieties, Lemma 0B5X. Hence Morphisms, Lemma 0C6J gives an isomorphism \[X \longrightarrow \text{Proj}\left( \bigoplus\nolimits_{n \geq 0} H^0(X, \mathcal{L}^{\otimes n})\right)\] Since the maps \(\mu_n\) are surjective for all \(n \geq 1\) we see that the graded algebra on the right hand side is a quotient of the symmetric algebra on \(H^0(X, \mathcal{L})\). Choosing a \(k\)-basis \(s_0, \ldots, s_d\) of \(H^0(X, \mathcal{L})\) we see that it is a quotient of a polynomial algebra in \(d + 1\) variables. Since quotients of graded rings correspond to closed immersions of \(\text{Proj}\) (Constructions, Lemma 01N1) we find a closed immersion \(X \to \mathbf{P}^d_k\). We omit the verification that this morphism is the morphism of Constructions, Lemma 01NE associated to the sections \(s_0, \ldots, s_d\) of \(\mathcal{L}\).

Lemma

Let \(X\) be a proper curve over a field \(k\) with \(H^0(X, \mathcal{O}_X) = k\). If \(X\) is Gorenstein and has genus \(0\), then \(X\) is isomorphic to a plane curve of degree \(2\).

Proof

Consider the invertible sheaf \(\mathcal{L} = \omega_X^{\otimes -1}\) where \(\omega_X\) is as in Lemma 0BS2. Then \(\deg(\omega_X) = -2\) by Lemma 0C19 and hence \(\deg(\mathcal{L}) = 2\). By Lemma 0C6T we conclude that choosing a basis \(s_0, s_1, s_2\) of the \(k\)-vector space of global sections of \(\mathcal{L}\) we obtain a closed immersion \[\varphi_{(\mathcal{L}, (s_0, s_1, s_2))} : X \longrightarrow \mathbf{P}^2_k\] Thus \(X\) is a plane curve of some degree \(d\). Let \(F \in k[T_0, T_1, T_2]_d\) be its equation (Lemma 0BYB). Because the genus of \(X\) is \(0\) we see that \(d\) is \(1\) or \(2\) (Lemma 0BYD). Observe that \(F\) restricts to the zero section on \(\varphi(X)\) and hence \(F(s_0, s_1, s_2)\) is the zero section of \(\mathcal{L}^{\otimes 2}\). Because \(s_0, s_1, s_2\) are linearly independent we see that \(F\) cannot be linear, i.e., \(d = \deg(F) \geq 2\). Thus \(d = 2\) and the proof is complete.

Proposition

Let \(k\) be a field. Let \(X\) be a proper curve over \(k\). The following are equivalent

1. \(X \cong \mathbf{P}^1_k\),

2. \(X\) is smooth and geometrically irreducible over \(k\), \(X\) has genus \(0\), and \(X\) has an invertible module of odd degree,

3. \(X\) is geometrically integral over \(k\), \(X\) has genus \(0\), \(X\) is Gorenstein, and \(X\) has an invertible sheaf of odd degree,

4. \(H^0(X, \mathcal{O}_X) = k\), \(X\) has genus \(0\), \(X\) is Gorenstein, and \(X\) has an invertible sheaf of odd degree,

5. \(X\) is geometrically integral over \(k\), \(X\) has genus \(0\), and \(X\) has an invertible \(\mathcal{O}_X\)-module of degree \(1\),

6. \(H^0(X, \mathcal{O}_X) = k\), \(X\) has genus \(0\), and \(X\) has an invertible \(\mathcal{O}_X\)-module of degree \(1\),

7. \(H^1(X, \mathcal{O}_X) = 0\) and \(X\) has an invertible \(\mathcal{O}_X\)-module of degree \(1\),

8. \(H^1(X, \mathcal{O}_X) = 0\) and \(X\) has closed points \(x_1, \ldots, x_n\) such that \(\mathcal{O}_{X, x_i}\) is normal and \(\gcd([\kappa(x_i) : k]) = 1\), and

9. add more here.

Proof

We will prove that each condition (2) – (8) implies (1) and we omit the verification that (1) implies (2) – (8).

Assume (2). A smooth scheme over \(k\) is geometrically reduced (Varieties, Lemma 056T) and regular (Varieties, Lemma 056S). Hence \(X\) is Gorenstein (Duality for Schemes, Lemma 0DWG). Thus we reduce to (3).

Assume (3). Since \(X\) is geometrically integral over \(k\) we have \(H^0(X, \mathcal{O}_X) = k\) by Varieties, Lemma 04L2. and we reduce to (4).

Assume (4). Since \(X\) is Gorenstein the dualizing module \(\omega_X\) as in Lemma 0BS2 has degree \(\deg(\omega_X) = -2\) by Lemma 0C19. Combined with the assumed existence of an odd degree invertible module, we conclude there exists an invertible module of degree \(1\). In this way we reduce to (6).

Assume (5). Since \(X\) is geometrically integral over \(k\) we have \(H^0(X, \mathcal{O}_X) = k\) by Varieties, Lemma 04L2. and we reduce to (6).

Assume (6). Then \(X \cong \mathbf{P}^1_k\) by Lemma 0C6T.

Assume (7). Observe that \(\kappa = H^0(X, \mathcal{O}_X)\) is a field finite over \(k\) by Varieties, Lemma 04L2. If \(d = [\kappa : k] > 1\), then every invertible sheaf has degree divisible by \(d\) and there cannot be an invertible sheaf of degree \(1\). Hence \(d = 1\) and we reduce to case (6).

Assume (8). Observe that \(\kappa = H^0(X, \mathcal{O}_X)\) is a field finite over \(k\) by Varieties, Lemma 04L2. Since \(\kappa \subset \kappa(x_i)\) we see that \(k = \kappa\) by the assumption on the gcd of the degrees. The same condition allows us to find integers \(a_i\) such that \(1 = \sum a_i[\kappa(x_i) : k]\). Because \(x_i\) defines an effective Cartier divisor on \(X\) by Varieties, Lemma 0B8Y we can consider the invertible module \(\mathcal{O}_X(\sum a_i x_i)\). By our choice of \(a_i\) the degree of \(\mathcal{L}\) is \(1\). Thus \(X \cong \mathbf{P}^1_k\) by Lemma 0C6T.

Lemma

Let \(X\) be a proper curve over a field \(k\) with \(H^0(X, \mathcal{O}_X) = k\). Assume \(X\) is singular and has genus \(0\). Then there exists a diagram \[\xymatrix{ x' \ar[d] \ar[r] & X' \ar[d]^\nu \ar[r] & \Spec(k') \ar[d] \\ x \ar[r] & X \ar[r] & \Spec(k) }\] where

1. \(k'/k\) is a nontrivial finite extension,

2. \(X' \cong \mathbf{P}^1_{k'}\),

3. \(x'\) is a \(k'\)-rational point of \(X'\),

4. \(x\) is a \(k\)-rational point of \(X\),

5. \(X' \setminus \{x'\} \to X \setminus \{x\}\) is an isomorphism,

6. \(0 \to \mathcal{O}_X \to \nu_*\mathcal{O}_{X'} \to k'/k \to 0\) is a short exact sequence where \(k'/k = \kappa(x')/\kappa(x)\) indicates the skyscraper sheaf on the point \(x\).

Proof

Let \(\nu : X' \to X\) be the normalization of \(X\), see Varieties, Sections 0BXQ and 0C44. Since \(X\) is singular \(\nu\) is not an isomorphism. Then \(k' = H^0(X', \mathcal{O}_{X'})\) is a finite extension of \(k\) (Varieties, Lemma 04L2). The short exact sequence \[0 \to \mathcal{O}_X \to \nu_*\mathcal{O}_{X'} \to \mathcal{Q} \to 0\] and the fact that \(\mathcal{Q}\) is supported in finitely many closed points give us that

1. \(H^1(X', \mathcal{O}_{X'}) = 0\), i.e., \(X'\) has genus \(0\) as a curve over \(k'\),

2. there is a short exact sequence \(0 \to k \to k' \to H^0(X, \mathcal{Q}) \to 0\).

In particular \(k'/k\) is a nontrivial extension.

Next, we consider what is often called the conductor ideal \[\mathcal{I} = \SheafHom_{\mathcal{O}_X}(\nu_*\mathcal{O}_{X'}, \mathcal{O}_X)\] This is a quasi-coherent \(\mathcal{O}_X\)-module. We view \(\mathcal{I}\) as an ideal in \(\mathcal{O}_X\) via the map \(\varphi \mapsto \varphi(1)\). Thus \(\mathcal{I}(U)\) is the set of \(f \in \mathcal{O}_X(U)\) such that \(f \left(\nu_*\mathcal{O}_{X'}(U)\right) \subset \mathcal{O}_X(U)\). In other words, the condition is that \(f\) annihilates \(\mathcal{Q}\). In other words, there is a defining exact sequence \[0 \to \mathcal{I} \to \mathcal{O}_X \to \SheafHom_{\mathcal{O}_X}(\mathcal{Q}, \mathcal{Q})\] Let \(U \subset X\) be an affine open containing the support of \(\mathcal{Q}\). Then \(V = \mathcal{Q}(U) = H^0(X, \mathcal{Q})\) is a \(k\)-vector space of dimension \(n - 1\). The image of \(\mathcal{O}_X(U) \to \Hom_k(V, V)\) is a commutative subalgebra, hence has dimension \(\leq n - 1\) over \(k\) (this is a property of commutative subalgebras of matrix algebras; details omitted). We conclude that we have a short exact sequence \[0 \to \mathcal{I} \to \mathcal{O}_X \to \mathcal{A} \to 0\] where \(\text{Supp}(\mathcal{A}) = \text{Supp}(\mathcal{Q})\) and \(\dim_k H^0(X, \mathcal{A}) \leq n - 1\). On the other hand, the description \(\mathcal{I} = \SheafHom_{\mathcal{O}_X}(\nu_*\mathcal{O}_{X'}, \mathcal{O}_X)\) provides \(\mathcal{I}\) with a \(\nu_*\mathcal{O}_{X'}\)-module structure such that the inclusion map \(\mathcal{I} \to \nu_*\mathcal{O}_{X'}\) is a \(\nu_*\mathcal{O}_{X'}\)-module map. We conclude that \(\mathcal{I} = \nu_*\mathcal{I}'\) for some quasi-coherent sheaf of ideals \(\mathcal{I}' \subset \mathcal{O}_{X'}\), see Morphisms, Lemma 01SB. Define \(\mathcal{A}'\) as the cokernel: \[0 \to \mathcal{I}' \to \mathcal{O}_{X'} \to \mathcal{A}' \to 0\] Combining the exact sequences so far we obtain a short exact sequence \(0 \to \mathcal{A} \to \nu_*\mathcal{A}' \to \mathcal{Q} \to 0\). Using the estimate above, combined with \(\dim_k H^0(X, \mathcal{Q}) = n - 1\), gives \[\dim_k H^0(X', \mathcal{A}') = \dim_k H^0(X, \mathcal{A}) + \dim_k H^0(X, \mathcal{Q}) \leq 2 n - 2\] However, since \(X'\) is a curve over \(k'\) we see that the left hand side is divisible by \(n\) (Varieties, Lemma 0C6P). As \(\mathcal{A}\) and \(\mathcal{A}'\) cannot be zero, we conclude that \(\dim_k H^0(X', \mathcal{A}') = n\) which means that \(\mathcal{I}'\) is the ideal sheaf of a \(k'\)-rational point \(x'\). By Proposition 0C6U we find \(X' \cong \mathbf{P}^1_{k'}\). Going back to the equalities above, we conclude that \(\dim_k H^0(X, \mathcal{A}) = 1\). This means that \(\mathcal{I}\) is the ideal sheaf of a \(k\)-rational point \(x\). Then \(\mathcal{A} = \kappa(x) = k\) and \(\mathcal{A}' = \kappa(x') = k'\) as skyscraper sheaves. Comparing the exact sequences given above, this immediately implies the result on structure sheaves as stated in the lemma.

Example

In fact, the situation described in Lemma 0DJB occurs for any nontrivial finite extension \(k'/k\). Namely, we can consider \[A = \{f \in k'[x] \mid f(0) \in k \}\] The spectrum of \(A\) is an affine curve, which we can glue to the spectrum of \(B = k'[y]\) using the isomorphism \(A_x \cong B_y\) sending \(x^{-1}\) to \(y\). The result is a proper curve \(X\) with \(H^0(X, \mathcal{O}_X) = k\) and singular point \(x\) corresponding to the maximal ideal \(A \cap (x)\). The normalization of \(X\) is \(\mathbf{P}^1_{k'}\) exactly as in the lemma.

Geometric genus

If \(X\) is a proper and smooth curve over \(k\) with \(H^0(X, \mathcal{O}_X) = k\), then \[p_g(X) = \dim_k H^0(X, \Omega_{X/k})\] is called the geometric genus of \(X\). By Lemma 0C1A the geometric genus of \(X\) agrees with the (arithmetic) genus. However, in higher dimensions there is a difference between the geometric genus and the arithmetic genus, see Remark 0BYG.

For singular curves, we will define the geometric genus as follows.

Definition

Let \(k\) be a field. Let \(X\) be a geometrically irreducible curve over \(k\). The geometric genus of \(X\) is the genus of a smooth projective model of \(X\) possibly defined over an extension field of \(k\) as in Lemma 0BY4.

If \(k\) is perfect, then the nonsingular projective model \(Y\) of \(X\) is smooth (Lemma 0BY3) and the geometric genus of \(X\) is just the genus of \(Y\). But if \(k\) is not perfect, this may not be true. In this case we choose an extension \(K/k\) such that the nonsingular projective model \(Y_K\) of \((X_K)_{red}\) is a smooth projective curve and we define the geometric genus of \(X\) to be the genus of \(Y_K\). This is well defined by Lemmas 0BY4 and 0BY9.

Remark

Suppose that \(X\) is a \(d\)-dimensional proper smooth variety over an algebraically closed field \(k\). Then the arithmetic genus is often defined as \(p_a(X) = (-1)^d(\chi(X, \mathcal{O}_X) - 1)\) and the geometric genus as \(p_g(X) = \dim_k H^0(X, \Omega^d_{X/k})\). In this situation the arithmetic genus and the geometric genus no longer agree even though it is still true that \(\omega_X \cong \Omega_{X/k}^d\). For example, if \(d = 2\), then we have \[\begin{align*} p_a(X) - p_g(X) & = h^0(X, \mathcal{O}_X) - h^1(X, \mathcal{O}_X) + h^2(X, \mathcal{O}_X) - 1 - h^0(X, \Omega^2_{X/k}) \\ & = - h^1(X, \mathcal{O}_X) + h^2(X, \mathcal{O}_X) - h^0(X, \omega_X) \\ & = - h^1(X, \mathcal{O}_X) \end{align*}\] where \(h^i(X, \mathcal{F}) = \dim_k H^i(X, \mathcal{F})\) and where the last equality follows from duality. Hence for a surface the difference \(p_g(X) - p_a(X)\) is always nonnegative; it is sometimes called the irregularity of the surface. If \(X = C_1 \times C_2\) is a product of smooth projective curves of genus \(g_1\) and \(g_2\), then the irregularity is \(g_1 + g_2\).

Riemann-Hurwitz

Let \(k\) be a field. Let \(f : X \to Y\) be a morphism of smooth curves over \(k\). Then we obtain a canonical exact sequence \[f^*\Omega_{Y/k} \xrightarrow{\text{d}f} \Omega_{X/k} \longrightarrow \Omega_{X/Y} \longrightarrow 0\] by Morphisms, Lemma 01UX. Since \(X\) and \(Y\) are smooth, the sheaves \(\Omega_{X/k}\) and \(\Omega_{Y/k}\) are invertible modules, see Morphisms, Lemma 02G1. Assume the first map is nonzero, i.e., assume \(f\) is generically étale, see Lemma 0C1C. Let \(R \subset X\) be the closed subscheme cut out by the different \(\mathfrak{D}_f\) of \(f\). By Discriminants, Lemma 0BWJ this is the same as the vanishing locus of \(\text{d}f\), it is an effective Cartier divisor, and we get \[f^*\Omega_{Y/k} \otimes_{\mathcal{O}_X} \mathcal{O}_X(R) = \Omega_{X/k}\] In particular, if \(X\), \(Y\) are projective with \(k = H^0(Y, \mathcal{O}_Y) = H^0(X, \mathcal{O}_X)\) and \(X\), \(Y\) have genus \(g_X\), \(g_Y\), then we get the Riemann-Hurwitz formula \[\begin{align*} 2g_X - 2 & = \deg(\Omega_{X/k}) \\ & = \deg(f^*\Omega_{Y/k} \otimes_{\mathcal{O}_X} \mathcal{O}_X(R)) \\ & = \deg(f) \deg(\Omega_{Y/k}) + \deg(R) \\ & = \deg(f) (2g_Y - 2) + \deg(R) \end{align*}\] The first and last equality by Lemma 0C1A. The second equality by the isomorphism of invertible sheaves given above. The third equality by additivity of degrees (Varieties, Lemma 0AYX), the formula for the degree of a pullback (Varieties, Lemma 0AYZ), and finally the formula for the degree of \(\mathcal{O}_X(R)\) (Varieties, Lemma 0AYY).

To use the Riemann-Hurwitz formula we need to compute \(\deg(R) = \dim_k \Gamma(R, \mathcal{O}_R)\). By the structure of zero dimensional schemes over \(k\) (see for example Varieties, Lemma 06LH), we see that \(R\) is a finite disjoint union of spectra of Artinian local rings \(R = \coprod_{x \in R} \Spec(\mathcal{O}_{R, x})\) with each \(\mathcal{O}_{R, x}\) of finite dimension over \(k\). Thus \[\deg(R) = \sum\nolimits_{x \in R} \dim_k \mathcal{O}_{R, x} = \sum\nolimits_{x \in R} d_x [\kappa(x) : k]\] with \[d_x = \text{length}_{\mathcal{O}_{R, x}} \mathcal{O}_{R, x} = \text{length}_{\mathcal{O}_{X, x}} \mathcal{O}_{R, x}\] the multiplicity of \(x\) in \(R\) (see Algebra, Lemma 02M0). Let \(x \in X\) be a closed point with image \(y \in Y\). Looking at stalks we obtain an exact sequence \[\Omega_{Y/k, y} \to \Omega_{X/k, x} \to \Omega_{X/Y, x} \to 0\] Choosing local generators \(\eta_x\) and \(\eta_y\) of the (free rank \(1\)) modules \(\Omega_{X/k, x}\) and \(\Omega_{Y/k, y}\) we see that \(\eta_y \mapsto h \eta_x\) for some nonzero \(h \in \mathcal{O}_{X, x}\). By the exact sequence we see that \(\Omega_{X/Y, x} \cong \mathcal{O}_{X, x}/h\mathcal{O}_{X, x}\) as \(\mathcal{O}_{X, x}\)-modules. Since the divisor \(R\) is cut out by \(h\) (see above) we have \(\mathcal{O}_{R, x} = \mathcal{O}_{X, x}/h\mathcal{O}_{X, x}\). Thus we find the following equalities \[\begin{align*} d_x & = \text{length}_{\mathcal{O}_{X, x}}(\mathcal{O}_{R, x}) \\ & = \text{length}_{\mathcal{O}_{X, x}}(\mathcal{O}_{X, x}/h\mathcal{O}_{X, x}) \\ & = \text{length}_{\mathcal{O}_{X, x}}(\Omega_{X/Y, x}) \\ & = \text{ord}_{\mathcal{O}_{X, x}}(h) \\ & = \text{ord}_{\mathcal{O}_{X, x}}(``\eta_y/\eta_x") \end{align*}\] The first equality by our definition of \(d_x\). The second and third we saw above. The fourth equality is the definition of \(\text{ord}\), see Algebra, Definition 02MD. Note that since \(\mathcal{O}_{X, x}\) is a discrete valuation ring, the integer \(\text{ord}_{\mathcal{O}_{X, x}}(h)\) just the valuation of \(h\). The fifth equality is a mnemonic.

Here is a case where one can “calculate” the multiplicity \(d_x\) in terms of other invariants. Namely, if \(\kappa(x)\) is separable over \(k\), then we may choose \(\eta_x = \text{d}s\) and \(\eta_y = \text{d}t\) where \(s\) and \(t\) are uniformizers in \(\mathcal{O}_{X, x}\) and \(\mathcal{O}_{Y, y}\) (Lemma 0C1E). Then \(t \mapsto u s^{e_x}\) for some unit \(u \in \mathcal{O}_{X, x}\) where \(e_x\) is the ramification index of the extension \(\mathcal{O}_{Y, y} \subset \mathcal{O}_{X, x}\). Hence we get \[\eta_y = \text{d}t = \text{d}(u s^{e_x}) = e s^{e_x - 1} u \text{d}s + s^{e_x} \text{d}u\] Writing \(\text{d}u = w \text{d}s\) for some \(w \in \mathcal{O}_{X, x}\) we see that \[``\eta_y/\eta_x" = e s^{e_x - 1} u + s^{e_x} w = (e_x u + s w)s^{e_x - 1}\] We conclude that the order of vanishing of this is \(e_x - 1\) unless the characteristic of \(\kappa(x)\) is \(p > 0\) and \(p\) divides \(e_x\) in which case the order of vanishing is \(> e_x - 1\).

Combining all of the above we find that if \(k\) has characteristic zero, then \[2g_X - 2 = (2g_Y - 2)\deg(f) + \sum\nolimits_{x \in X} (e_x - 1)[\kappa(x) : k]\] where \(e_x\) is the ramification index of \(\mathcal{O}_{X, x}\) over \(\mathcal{O}_{Y, f(x)}\). This precise formula will hold if and only if all the ramification is tame, i.e., when the residue field extensions \(\kappa(x)/\kappa(y)\) are separable and \(e_x\) is prime to the characteristic of \(k\), although the arguments above are insufficient to prove this. We refer the reader to Lemma 0C1F and its proof.

Lemma

Let \(k\) be a field. Let \(f : X \to Y\) be a morphism of smooth curves over \(k\). The following are equivalent

1. \(\text{d}f : f^*\Omega_{Y/k} \to \Omega_{X/k}\) is nonzero,

2. \(\Omega_{X/Y}\) is supported on a proper closed subset of \(X\),

3. there exists a nonempty open \(U \subset X\) such that \(f|_U : U \to Y\) is unramified,

4. there exists a nonempty open \(U \subset X\) such that \(f|_U : U \to Y\) is étale,

5. the extension \(k(X)/k(Y)\) of function fields is finite separable.

Proof

Since \(X\) and \(Y\) are smooth, the sheaves \(\Omega_{X/k}\) and \(\Omega_{Y/k}\) are invertible modules, see Morphisms, Lemma 02G1. Using the exact sequence \[f^*\Omega_{Y/k} \longrightarrow \Omega_{X/k} \longrightarrow \Omega_{X/Y} \longrightarrow 0\] of Morphisms, Lemma 01UX we see that (1) and (2) are equivalent and equivalent to the condition that \(f^*\Omega_{Y/k} \to \Omega_{X/k}\) is nonzero in the generic point. The equivalence of (2) and (3) follows from Morphisms, Lemma 02G5. The equivalence between (3) and (4) follows from Morphisms, Lemma 02GV and the fact that flatness is automatic (Lemma 0CCK). To see the equivalence of (5) and (4) use Algebra, Lemma 07ND. Some details omitted.

Lemma

Let \(f : X \to Y\) be a morphism of smooth proper curves over a field \(k\) which satisfies the equivalent conditions of Lemma 0C1C. If \(k = H^0(Y, \mathcal{O}_Y) = H^0(X, \mathcal{O}_X)\) and \(X\) and \(Y\) have genus \(g_X\) and \(g_Y\), then \[2g_X - 2 = (2g_Y - 2) \deg(f) + \deg(R)\] where \(R \subset X\) is the effective Cartier divisor cut out by the different of \(f\).

Proof

See discussion above; we used Discriminants, Lemma 0BWJ, Lemma 0C1A, and Varieties, Lemmas 0AYX and 0AYZ.

Lemma

Let \(X \to \Spec(k)\) be smooth of relative dimension \(1\) at a closed point \(x \in X\). If \(\kappa(x)\) is separable over \(k\), then for any uniformizer \(s\) in the discrete valuation ring \(\mathcal{O}_{X, x}\) the element \(\text{d}s\) freely generates \(\Omega_{X/k, x}\) over \(\mathcal{O}_{X, x}\).

Proof

The ring \(\mathcal{O}_{X, x}\) is a discrete valuation ring by Algebra, Lemma 00TT. Since \(x\) is closed \(\kappa(x)\) is finite over \(k\). Hence if \(\kappa(x)/k\) is separable, then any uniformizer \(s\) maps to a nonzero element of \(\Omega_{X/k, x} \otimes_{\mathcal{O}_{X, x}} \kappa(x)\) by Algebra, Lemma 00TU. Since \(\Omega_{X/k, x}\) is free of rank \(1\) over \(\mathcal{O}_{X, x}\) the result follows.

Lemma

Notation and assumptions as in Lemma 0C1D. For a closed point \(x \in X\) let \(d_x\) be the multiplicity of \(x\) in \(R\). Then \[2g_X - 2 = (2g_Y - 2) \deg(f) + \sum\nolimits d_x [\kappa(x) : k]\] Moreover, we have the following results

1. \(d_x = \text{length}_{\mathcal{O}_{X, x}}(\Omega_{X/Y, x})\),

2. \(d_x \geq e_x - 1\) where \(e_x\) is the ramification index of \(\mathcal{O}_{X, x}\) over \(\mathcal{O}_{Y, y}\),

3. \(d_x = e_x - 1\) if and only if \(\mathcal{O}_{X, x}\) is tamely ramified over \(\mathcal{O}_{Y, y}\).

Proof

By Lemma 0C1D and the discussion above (which used Varieties, Lemma 06LH and Algebra, Lemma 02M0) it suffices to prove the results on the multiplicity \(d_x\) of \(x\) in \(R\). Part (1) was proved in the discussion above. In the discussion above we proved (2) and (3) only in the case where \(\kappa(x)\) is separable over \(k\). In the rest of the proof we give a uniform treatment of (2) and (3) using material on differents of quasi-finite Gorenstein morphisms.

First, observe that \(f\) is a quasi-finite Gorenstein morphism. This is true for example because \(f\) is a flat quasi-finite morphism and \(X\) is Gorenstein (see Duality for Schemes, Lemma 0C12) or because it was shown in the proof of Discriminants, Lemma 0BWJ (which we used above). Thus \(\omega_{X/Y}\) is invertible by Discriminants, Lemma 0C16 and the same remains true after replacing \(X\) by opens and after performing a base change by some \(Y' \to Y\). We will use this below without further mention.

Choose affine opens \(U \subset X\) and \(V \subset Y\) such that \(x \in U\), \(y \in V\), \(f(U) \subset V\), and \(x\) is the only point of \(U\) lying over \(y\). Write \(U = \Spec(A)\) and \(V = \Spec(B)\). Then \(R \cap U\) is the different of \(f|_U : U \to V\). By Discriminants, Lemma 0BW7 formation of the different commutes with arbitrary base change in our case. By our choice of \(U\) and \(V\) we have \[A \otimes_B \kappa(y) = \mathcal{O}_{X, x} \otimes_{\mathcal{O}_{Y, y}} \kappa(y) = \mathcal{O}_{X, x}/(s^{e_x})\] where \(e_x\) is the ramification index as in the statement of the lemma. Let \(C = \mathcal{O}_{X, x}/(s^{e_x})\) viewed as a finite algebra over \(\kappa(y)\). Let \(\mathfrak{D}_{C/\kappa(y)}\) be the different of \(C\) over \(\kappa(y)\) in the sense of Discriminants, Definition 0BW4. It suffices to show: \(\mathfrak{D}_{C/\kappa(y)}\) is nonzero if and only if the extension \(\mathcal{O}_{Y, y} \subset \mathcal{O}_{X, x}\) is tamely ramified and in the tamely ramified case \(\mathfrak{D}_{C/\kappa(y)}\) is equal to the ideal generated by \(s^{e_x - 1}\) in \(C\). Recall that tame ramification means exactly that \(\kappa(x)/\kappa(y)\) is separable and that the characteristic of \(\kappa(y)\) does not divide \(e_x\). On the other hand, the different of \(C/\kappa(y)\) is nonzero if and only if \(\tau_{C/\kappa(y)} \in \omega_{C/\kappa(y)}\) is nonzero. Namely, since \(\omega_{C/\kappa(y)}\) is an invertible \(C\)-module (as the base change of \(\omega_{A/B}\)) it is free of rank \(1\), say with generator \(\lambda\). Write \(\tau_{C/\kappa(y)} = h\lambda\) for some \(h \in C\). Then \(\mathfrak{D}_{C/\kappa(y)} = (h) \subset C\) whence the claim. By Discriminants, Lemma 0C13 we have \(\tau_{C/\kappa(y)} \not = 0\) if and only if \(\kappa(x)/\kappa(y)\) is separable and \(e_x\) is prime to the characteristic. Finally, even if \(\tau_{C/\kappa(y)}\) is nonzero, then it is still the case that \(s \tau_{C/\kappa(y)} = 0\) because \(s\tau_{C/\kappa(y)} : C \to \kappa(y)\) sends \(c\) to the trace of the nilpotent operator \(sc\) which is zero. Hence \(sh = 0\), hence \(h \in (s^{e_x - 1})\) which proves that \(\mathfrak{D}_{C/\kappa(y)} \subset (s^{e_x - 1})\) always. Since \((s^{e_x - 1}) \subset C\) is the smallest nonzero ideal, we have proved the final assertion.

Inseparable maps

Some remarks on the behaviour of the genus under inseparable maps.

Lemma

Let \(k\) be a field. Let \(f : X \to Y\) be a surjective morphism of curves over \(k\). If \(X\) is smooth over \(k\) and \(Y\) is normal, then \(Y\) is smooth over \(k\).

Proof

Let \(y \in Y\). Pick \(x \in X\) mapping to \(y\). By Varieties, Lemma 05AX it suffices to show that \(f\) is flat at \(x\). This follows from Lemma 0CCK.

Lemma

Let \(k\) be a field of characteristic \(p > 0\). Let \(f : X \to Y\) be a nonconstant morphism of proper nonsingular curves over \(k\). If the extension \(k(X)/k(Y)\) of function fields is purely inseparable, then there exists a factorization \[X = X_0 \to X_1 \to \ldots \to X_n = Y\] such that each \(X_i\) is a proper nonsingular curve and \(X_i \to X_{i + 1}\) is a degree \(p\) morphism with \(k(X_{i + 1}) \subset k(X_i)\) inseparable.

Proof

This follows from Theorem 0BY1 and the fact that a finite purely inseparable extension of fields can always be gotten as a sequence of (inseparable) extensions of degree \(p\), see Fields, Lemma 09HI.

Lemma

Let \(k\) be a field of characteristic \(p > 0\). Let \(f : X \to Y\) be a nonconstant morphism of proper nonsingular curves over \(k\). If \(X\) is smooth and \(k(Y) \subset k(X)\) is inseparable of degree \(p\), then there is a unique isomorphism \(Y = X^{(p)}\) such that \(f\) is \(F_{X/k}\).

Proof

The relative frobenius morphism \(F_{X/k} : X \to X^{(p)}\) is constructed in Varieties, Section 0CC6. Observe that \(X^{(p)}\) is a smooth proper curve over \(k\) as a base change of \(X\). The morphism \(F_{X/k}\) has degree \(p\) by Varieties, Lemma 0CCF. Thus \(k(X^{(p)})\) and \(k(Y)\) are both subfields of \(k(X)\) with \([k(X) : k(Y)] = [k(X) : k(X^{(p)})] = p\). To prove the lemma it suffices to show that \(k(Y) = k(X^{(p)})\) inside \(k(X)\). See Theorem 0BY1.

Write \(K = k(X)\). Consider the map \(\text{d} : K \to \Omega_{K/k}\). It follows from Lemma 0C1C that both \(k(Y)\) is contained in the kernel of \(\text{d}\). By Varieties, Lemma 0CCC we see that \(k(X^{(p)})\) is in the kernel of \(\text{d}\). Since \(X\) is a smooth curve we know that \(\Omega_{K/k}\) is a vector space of dimension \(1\) over \(K\). Then More on Algebra, Lemma 07P2. implies that \(\Ker(\text{d}) = kK^p\) and that \([K : kK^p] = p\). Thus \(k(Y) = kK^p = k(X^{(p)})\) for reasons of degree.

Lemma

Let \(k\) be a field of characteristic \(p > 0\). Let \(f : X \to Y\) be a nonconstant morphism of proper nonsingular curves over \(k\). If \(X\) is smooth and \(k(Y) \subset k(X)\) is purely inseparable, then there is a unique \(n \geq 0\) and a unique isomorphism \(Y = X^{(p^n)}\) such that \(f\) is the \(n\)-fold relative Frobenius of \(X/k\).

Proof

The \(n\)-fold relative Frobenius of \(X/k\) is defined in Varieties, Remark 0CCG. The lemma follows by combining Lemmas 0CCY and 0CCX.

Lemma

Let \(k\) be a field of characteristic \(p > 0\). Let \(f : X \to Y\) be a nonconstant morphism of proper nonsingular curves over \(k\). Assume

1. \(X\) is smooth,

2. \(H^0(X, \mathcal{O}_X) = k\),

3. \(k(X)/k(Y)\) is purely inseparable.

Then \(Y\) is smooth, \(H^0(Y, \mathcal{O}_Y) = k\), and the genus of \(Y\) is equal to the genus of \(X\).

Proof

By Lemma 0CCZ we see that \(Y = X^{(p^n)}\) is the base change of \(X\) by \(F_{\Spec(k)}^n\). Thus \(Y\) is smooth and the result on the cohomology and genus follows from Lemma 0BY9.

Example

This example will show that the genus can change under a purely inseparable morphism of nonsingular projective curves. Let \(k\) be a field of characteristic \(3\). Assume there exists an element \(a \in k\) which is not a \(3\)rd power. For example \(k = \mathbf{F}_3(a)\) would work. Let \(X\) be the plane curve with homogeneous equation \[F = T_1^2T_0 - T_2^3 + aT_0^3\] as in Section 0BYA. On the affine piece \(D_+(T_0)\) using coordinates \(x = T_1/T_0\) and \(y = T_2/T_0\) we obtain \(x^2 - y^3 + a = 0\) which defines a nonsingular affine curve. Moreover, the point at infinity \((0 : 1: 0)\) is a smooth point. Hence \(X\) is a nonsingular projective curve of genus \(1\) (Lemma 0BYD). On the other hand, consider the morphism \(f : X \to \mathbf{P}^1_k\) which on \(D_+(T_0)\) sends \((x, y)\) to \(x \in \mathbf{A}^1_k \subset \mathbf{P}^1_k\). Then \(f\) is a morphism of proper nonsingular curves over \(k\) inducing an inseparable function field extension of degree \(p = 3\) but the genus of \(X\) is \(1\) and the genus of \(\mathbf{P}^1_k\) is \(0\).

Proposition

Let \(k\) be a field of characteristic \(p > 0\). Let \(f : X \to Y\) be a nonconstant morphism of proper smooth curves over \(k\). Then we can factor \(f\) as \[X \longrightarrow X^{(p^n)} \longrightarrow Y\] where \(X^{(p^n)} \to Y\) is a nonconstant morphism of proper smooth curves inducing a separable field extension \(k(X^{(p^n)})/k(Y)\), we have \[X^{(p^n)} = X \times_{\Spec(k), F_{\Spec(k)}^n} \Spec(k),\] and \(X \to X^{(p^n)}\) is the \(n\)-fold relative frobenius of \(X\).

Proof

By Fields, Lemma 030K there is a subextension \(k(X)/E/k(Y)\) such that \(k(X)/E\) is purely inseparable and \(E/k(Y)\) is separable. By Theorem 0BY1 this corresponds to a factorization \(X \to Z \to Y\) of \(f\) with \(Z\) a nonsingular proper curve. Apply Lemma 0CCZ to the morphism \(X \to Z\) to conclude.

Lemma

Let \(k\) be a field of characteristic \(p > 0\). Let \(X\) be a smooth proper curve over \(k\). Let \((\mathcal{L}, V)\) be a \(\mathfrak g^r_d\) with \(r \geq 1\). Then one of the following two is true

1. there exists a \(\mathfrak g^1_d\) whose corresponding morphism \(X \to \mathbf{P}^1_k\) (Lemma 0CCP) is generically étale (i.e., is as in Lemma 0C1C), or

2. there exists a \(\mathfrak g^r_{d'}\) on \(X^{(p)}\) where \(d' \leq d/p\).

Proof

Pick two \(k\)-linearly independent elements \(s, t \in V\). Then \(f = s/t\) is the rational function defining the morphism \(X \to \mathbf{P}^1_k\) corresponding to the linear series \((\mathcal{L}, ks + kt)\). If this morphism is not generically étale, then \(f \in k(X^{(p)})\) by Proposition 0CD2. Now choose a basis \(s_0, \ldots, s_r\) of \(V\) and let \(\mathcal{L}' \subset \mathcal{L}\) be the invertible sheaf generated by \(s_0, \ldots, s_r\). Set \(f_i = s_i/s_0\) in \(k(X)\). If for each pair \((s_0, s_i)\) we have \(f_i \in k(X^{(p)})\), then the morphism \[\varphi = \varphi_{(\mathcal{L}', (s_0, \ldots, s_r)} : X \longrightarrow \mathbf{P}^r_k = \text{Proj}(k[T_0, \ldots, T_r])\] factors through \(X^{(p)}\) as this is true over the affine open \(D_+(T_0)\) and we can extend the morphism over the affine part to the whole of the smooth curve \(X^{(p)}\) by Lemma 0BXZ. Introducing notation, say we have the factorization \[X \xrightarrow{F_{X/k}} X^{(p)} \xrightarrow{\psi} \mathbf{P}^r_k\] of \(\varphi\). Then \(\mathcal{N} = \psi^*\mathcal{O}_{\mathbf{P}^1_k}(1)\) is an invertible \(\mathcal{O}_{X^{(p)}}\)-module with \(\mathcal{L}' = F_{X/k}^*\mathcal{N}\) and with \(\psi^*T_0, \ldots, \psi^*T_r\) \(k\)-linearly independent (as they pullback to \(s_0, \ldots, s_r\) on \(X\)). Finally, we have \[d = \deg(\mathcal{L}) \geq \deg(\mathcal{L}') = \deg(F_{X/k}) \deg(\mathcal{N}) = p \deg(\mathcal{N})\] as desired. Here we used Varieties, Lemmas 0B40, 0AYZ, and 0CCF.

Lemma

Let \(k\) be a field. Let \(X\) be a smooth proper curve over \(k\) with \(H^0(X, \mathcal{O}_X) = k\) and genus \(g \geq 2\). Then there exists a closed point \(x \in X\) with \(\kappa(x)/k\) separable of degree \(\leq 2g - 2\).

Proof

Set \(\omega = \Omega_{X/k}\). By Lemma 0C1A this has degree \(2g - 2\) and has \(g\) global sections. Thus we have a \(\mathfrak g^{g - 1}_{2g - 2}\). By the trivial Lemma 0CCQ there exists a \(\mathfrak g^1_{2g - 2}\) and by Lemma 0CCR we obtain a morphism \[\varphi : X \longrightarrow \mathbf{P}^1_k\] of some degree \(d \leq 2g - 2\). Since \(\varphi\) is flat (Lemma 0CCK) and finite (Lemma 0CCL) it is finite locally free of degree \(d\) (Morphisms, Lemma 02KB). Pick any rational point \(t \in \mathbf{P}^1_k\) and any point \(x \in X\) with \(\varphi(x) = t\). Then \[d \geq [\kappa(x) : \kappa(t)] = [\kappa(x) : k]\] for example by Morphisms, Lemmas 0CC2 and 03J5. Thus if \(k\) is perfect (for example has characteristic zero or is finite) then the lemma is proved. Thus we reduce to the case discussed in the next paragraph.

Assume that \(k\) is an infinite field of characteristic \(p > 0\). As above we will use that \(X\) has a \(\mathfrak g^{g - 1}_{2g - 2}\). The smooth proper curve \(X^{(p)}\) has the same genus as \(X\). Hence its genus is \(> 0\). We conclude that \(X^{(p)}\) does not have a \(\mathfrak g^{g - 1}_d\) for any \(d \leq g - 1\) by Lemma 0CCS. Applying Lemma 0CD3 to our \(\mathfrak g^{g - 1}_{2g - 2}\) (and noting that \(2g - 2/p \leq g - 1\)) we conclude that possibility (2) does not occur. Hence we obtain a morphism \[\varphi : X \longrightarrow \mathbf{P}^1_k\] which is generically étale (in the sense of the lemma) and has degree \(\leq 2g - 2\). Let \(U \subset X\) be the nonempty open subscheme where \(\varphi\) is étale. Then \(\varphi(U) \subset \mathbf{P}^1_k\) is a nonempty Zariski open and we can pick a \(k\)-rational point \(t \in \varphi(U)\) as \(k\) is infinite. Let \(u \in U\) be a point with \(\varphi(u) = t\). Then \(\kappa(u)/\kappa(t)\) is separable (Morphisms, Lemma 02GL), \(\kappa(t) = k\), and \([\kappa(u) : k] \leq 2g - 2\) as before.

The following lemma does not really belong in this section but we don’t know a good place for it elsewhere.

Lemma

Let \(X\) be a smooth curve over a field \(k\). Let \(\overline{x} \in X_{\overline{k}}\) be a closed point with image \(x \in X\). The ramification index of \(\mathcal{O}_{X, x} \subset \mathcal{O}_{X_{\overline{k}}, \overline{x}}\) is the inseparable degree of \(\kappa(x)/k\).

Proof

After shrinking \(X\) we may assume there is an étale morphism \(\pi : X \to \mathbf{A}^1_k\), see Morphisms, Lemma 054L. Then we can consider the diagram of local rings \[\xymatrix{ \mathcal{O}_{X_{\overline{k}}, \overline{x}} & \mathcal{O}_{\mathbf{A}^1_{\overline{k}}, \pi(\overline{x})} \ar[l] \\ \mathcal{O}_{X, x} \ar[u] & \mathcal{O}_{\mathbf{A}^1_k, \pi(x)} \ar[l] \ar[u] }\] The horizontal arrows have ramification index \(1\) as they correspond to étale morphisms. Moreover, the extension \(\kappa(x)/\kappa(\pi(x))\) is separable hence \(\kappa(x)\) and \(\kappa(\pi(x))\) have the same inseparable degree over \(k\). By multiplicativity of ramification indices it suffices to prove the result when \(x\) is a point of the affine line.

Assume \(X = \mathbf{A}^1_k\). In this case, the local ring of \(X\) at \(x\) looks like \[\mathcal{O}_{X, x} = k[t]_{(P)}\] where \(P\) is an irreducible monic polynomial over \(k\). Then \(P(t) = Q(t^q)\) for some separable polynomial \(Q \in k[t]\), see Fields, Lemma 09H0. Observe that \(\kappa(x) = k[t]/(P)\) has inseparable degree \(q\) over \(k\). On the other hand, over \(\overline{k}\) we can factor \(Q(t) = \prod (t - \alpha_i)\) with \(\alpha_i\) pairwise distinct. Write \(\alpha_i = \beta_i^q\) for some unique \(\beta_i \in \overline{k}\). Then our point \(\overline{x}\) corresponds to one of the \(\beta_i\) and we conclude because the ramification index of \[k[t]_{(P)} \longrightarrow \overline{k}[t]_{(t - \beta_i)}\] is indeed equal to \(q\) as the uniformizer \(P\) maps to \((t - \beta_i)^q\) times a unit.

Pushouts

Let \(k\) be a field. Consider a solid diagram \[\xymatrix{ Z' \ar[d] \ar[r]_{i'} & X' \ar@{..>}[d]^a \\ Z \ar@{..>}[r]^i & X }\] of schemes over \(k\) satisfying

1. \(X'\) is separated of finite type over \(k\) of dimension \(\leq 1\),

2. \(i : Z' \to X'\) is a closed immersion,

3. \(Z'\) and \(Z\) are finite over \(\Spec(k)\), and

4. \(Z' \to Z\) is surjective.

In this situation every finite set of points of \(X'\) are contained in an affine open, see Varieties, Proposition 09NN. Thus the assumptions of More on Morphisms, Proposition 0E25 are satisfied and we obtain the following

1. the pushout \(X = Z \amalg_{Z'} X'\) exists in the category of schemes,

2. \(i : Z \to X\) is a closed immersion,

3. \(a : X' \to X\) is integral surjective,

4. \(X \to \Spec(k)\) is separated by More on Morphisms, Lemma 0E26

5. \(X \to \Spec(k)\) is of finite type by More on Morphisms, Lemmas 0E27,

6. thus \(a : X' \to X\) is finite by Morphisms, Lemmas 01WJ and 01T8,

7. if \(X' \to \Spec(k)\) is proper, then \(X \to \Spec(k)\) is proper by Morphisms, Lemma 03GN.

The following lemma can be generalized significantly.

Lemma

In the situation above, let \(Z = \Spec(k')\) where \(k'\) is a field and \(Z' = \Spec(k'_1 \times \ldots \times k'_n)\) with \(k'_i/k'\) finite extensions of fields. Let \(x \in X\) be the image of \(Z \to X\) and \(x'_i \in X'\) the image of \(\Spec(k'_i) \to X'\). Then we have a fibre product diagram \[\xymatrix{ \prod\nolimits_{i = 1, \ldots, n} k'_i & \prod\nolimits_{i = 1, \ldots, n} \mathcal{O}_{X', x'_i}^\wedge \ar[l] \\ k' \ar[u] & \mathcal{O}_{X, x}^\wedge \ar[u] \ar[l] }\] where the horizontal arrows are given by the maps to the residue fields.

Proof

Choose an affine open neighbourhood \(\Spec(A)\) of \(x\) in \(X\). Let \(\Spec(A') \subset X'\) be the inverse image. By construction we have a fibre product diagram \[\xymatrix{ \prod\nolimits_{i = 1, \ldots, n} k'_i & A' \ar[l] \\ k' \ar[u] & A \ar[u] \ar[l] }\] Since everything is finite over \(A\) we see that the diagram remains a fibre product diagram after completion with respect to the maximal ideal \(\mathfrak m \subset A\) corresponding to \(x\) (Algebra, Lemma 00MB). Finally, apply Algebra, Lemma 07N9 to identify the completion of \(A'\).

Glueing and squishing

Below we will indicate \(k[\epsilon]\) the algebra of dual numbers over \(k\) as defined in Varieties, Definition 0B29.

Lemma

Let \(k\) be an algebraically closed field. Let \(k \subset A\) be a ring extension such that \(A\) has exactly two \(k\)-sub algebras, then either \(A = k \times k\) or \(A = k[\epsilon]\).

Proof

The assumption means \(k \not = A\) and any subring \(k \subset C \subset A\) is equal to either \(k\) or \(A\). Let \(t \in A\), \(t \not \in k\). Then \(A\) is generated by \(t\) over \(k\). Hence \(A = k[x]/I\) for some ideal \(I\). If \(I = (0)\), then we have the subalgebra \(k[x^2]\) which is not allowed. Otherwise \(I\) is generated by a monic polynomial \(P\). Write \(P = \prod_{i = 1}^d (t - a_i)\). If \(d > 2\), then the subalgebra generated by \((t - a_1)(t - a_2)\) gives a contradiction. Thus \(d = 2\). If \(a_1 \not = a_2\), then \(A = k \times k\), if \(a_1 = a_2\), then \(A = k[\epsilon]\).

Example

Let \(k\) be an algebraically closed field. Let \(f : X' \to X\) be a morphism of algebraic \(k\)-schemes. We say \(X\) is obtained by glueing \(a\) and \(b\) in \(X'\) if the following are true:

1. \(a, b \in X'(k)\) are distinct points which map to the same point \(x \in X(k)\),

2. \(f\) is finite and \(f^{-1}(X \setminus \{x\}) \to X \setminus \{x\}\) is an isomorphism,

3. there is a short exact sequence \[0 \to \mathcal{O}_X \to f_*\mathcal{O}_{X'} \xrightarrow{a - b} x_*k \to 0\] where arrow on the right sends a local section \(h\) of \(f_*\mathcal{O}_{X'}\) to the difference \(h(a) - h(b) \in k\).

If this is the case, then there also is a short exact sequence \[0 \to \mathcal{O}_X^* \to f_*\mathcal{O}_{X'}^* \xrightarrow{ab^{-1}} x_*k^* \to 0\] where arrow on the right sends a local section \(h\) of \(f_*\mathcal{O}_{X'}^*\) to the multiplicative difference \(h(a)h(b)^{-1} \in k^*\).

Example

Let \(k\) be an algebraically closed field. Let \(f : X' \to X\) be a morphism of algebraic \(k\)-schemes. We say \(X\) is obtained by squishing the tangent vector \(\vartheta\) in \(X'\) if the following are true:

1. \(\vartheta : \Spec(k[\epsilon]) \to X'\) is a closed immersion over \(k\) such that \(f \circ \vartheta\) factors through a point \(x \in X(k)\),

2. \(f\) is finite and \(f^{-1}(X \setminus \{x\}) \to X \setminus \{x\}\) is an isomorphism,

3. there is a short exact sequence \[0 \to \mathcal{O}_X \to f_*\mathcal{O}_{X'} \xrightarrow{\vartheta} x_*k \to 0\] where arrow on the right sends a local section \(h\) of \(f_*\mathcal{O}_{X'}\) to the coefficient of \(\epsilon\) in \(\vartheta^\sharp(h) \in k[\epsilon]\).

If this is the case, then there also is a short exact sequence \[0 \to \mathcal{O}_X^* \to f_*\mathcal{O}_{X'}^* \xrightarrow{\vartheta} x_*k \to 0\] where arrow on the right sends a local section \(h\) of \(f_*\mathcal{O}_{X'}^*\) to \(\text{d}\log(\vartheta^\sharp(h))\) where \(\text{d}\log : k[\epsilon]^* \to k\) is the homomorphism of abelian groups sending \(a + b\epsilon\) to \(b/a \in k\).

Lemma

Let \(k\) be an algebraically closed field. Let \(f : X' \to X\) be a finite morphism of algebraic \(k\)-schemes such that \(\mathcal{O}_X \subset f_*\mathcal{O}_{X'}\) and such that \(f\) is an isomorphism away from a finite set of points. Then there is a factorization \[X' = X_n \to X_{n - 1} \to \ldots \to X_1 \to X_0 = X\] such that each \(X_i \to X_{i - 1}\) is either the glueing of two points or the squishing of a tangent vector (see Examples 0C1J and 0C1K).

Proof

Let \(U \subset X\) be the maximal open set over which \(f\) is an isomorphism. Then \(X \setminus U = \{x_1, \ldots, x_n\}\) with \(x_i \in X(k)\). We will consider factorizations \(X' \to Y \to X\) of \(f\) such that both morphisms are finite and \[\mathcal{O}_X \subset g_*\mathcal{O}_Y \subset f_*\mathcal{O}_{X'}\] where \(g : Y \to X\) is the given morphism. By assumption \(\mathcal{O}_{X, x} \to (f_*\mathcal{O}_{X'})_x\) is an isomorphism onless \(x = x_i\) for some \(i\). Hence the cokernel \[f_*\mathcal{O}_{X'}/\mathcal{O}_X = \bigoplus \mathcal{Q}_i\] is a direct sum of skyscraper sheaves \(\mathcal{Q}_i\) supported at \(x_1, \ldots, x_n\). Because the displayed quotient is a coherent \(\mathcal{O}_X\)-module, we conclude that \(\mathcal{Q}_i\) has finite length over \(\mathcal{O}_{X, x_i}\). Hence we can argue by induction on the sum of these lengths, i.e., the length of the whole cokernel.

If \(n > 1\), then we can define an \(\mathcal{O}_X\)-subalgebra \(\mathcal{A} \subset f_*\mathcal{O}_{X'}\) by taking the inverse image of \(\mathcal{Q}_1\). This will give a nontrivial factorization and we win by induction.

Assume \(n = 1\). We abbreviate \(x = x_1\). Consider the finite \(k\)-algebra extension \[A = \mathcal{O}_{X, x} \subset (f_*\mathcal{O}_{X'})_x = B\] Note that \(\mathcal{Q} = \mathcal{Q}_1\) is the skyscraper sheaf with value \(B/A\). We have a \(k\)-subalgebra \(A \subset A + \mathfrak m_A B \subset B\). If both inclusions are strict, then we obtain a nontrivial factorization and we win by induction as above. If \(A + \mathfrak m_A B = B\), then \(A = B\) by Nakayama, then \(f\) is an isomorphism and there is nothing to prove. We conclude that we may assume \(B = A + \mathfrak m_A B\). Set \(C = B/\mathfrak m_A B\). If \(C\) has more than \(2\) \(k\)-subalgebras, then we obtain a subalgebra between \(A\) and \(B\) by taking the inverse image in \(B\). Thus we may assume \(C\) has exactly \(2\) \(k\)-subalgebras. Thus \(C = k \times k\) or \(C = k[\epsilon]\) by Lemma 0C1I. In this case \(f\) is correspondingly the glueing two points or the squishing of a tangent vector.

Lemma

Let \(k\) be an algebraically closed field. If \(f : X' \to X\) is the glueing of two points \(a, b\) as in Example 0C1J, then there is an exact sequence \[k^* \to \Pic(X) \to \Pic(X') \to 0\] The first map is zero if \(a\) and \(b\) are on different connected components of \(X'\) and injective if \(X'\) is proper and \(a\) and \(b\) are on the same connected component of \(X'\).

Proof

The map \(\Pic(X) \to \Pic(X')\) is surjective by Varieties, Lemma 0C0T. Using the short exact sequence \[0 \to \mathcal{O}_X^* \to f_*\mathcal{O}_{X'}^* \xrightarrow{ab^{-1}} x_*k^* \to 0\] we obtain \[H^0(X', \mathcal{O}_{X'}^*) \xrightarrow{ab^{-1}} k^* \to H^1(X, \mathcal{O}_X^*) \to H^1(X, f_*\mathcal{O}_{X'}^*)\] We have \(H^1(X, f_*\mathcal{O}_{X'}^*) \subset H^1(X', \mathcal{O}_{X'}^*)\) (for example by the Leray spectral sequence, see Cohomology, Lemma 01F2). Hence the kernel of \(\Pic(X) \to \Pic(X')\) is the cokernel of \(ab^{-1} : H^0(X', \mathcal{O}_{X'}^*) \to k^*\). If \(a\) and \(b\) are on different connected components of \(X'\), then \(ab^{-1}\) is surjective. Because \(k\) is algebraically closed any regular function on a reduced connected proper scheme over \(k\) comes from an element of \(k\), see Varieties, Lemma 0BUG. Thus \(ab^{-1}\) is zero if \(X'\) is proper and \(a\) and \(b\) are on the same connected component.

Lemma

Let \(k\) be an algebraically closed field. If \(f : X' \to X\) is the squishing of a tangent vector \(\vartheta\) as in Example 0C1K, then there is an exact sequence \[(k, +) \to \Pic(X) \to \Pic(X') \to 0\] and the first map is injective if \(X'\) is proper and reduced.

Proof

The map \(\Pic(X) \to \Pic(X')\) is surjective by Varieties, Lemma 0C0T. Using the short exact sequence \[0 \to \mathcal{O}_X^* \to f_*\mathcal{O}_{X'}^* \xrightarrow{\vartheta} x_*k \to 0\] of Example 0C1K we obtain \[H^0(X', \mathcal{O}_{X'}^*) \xrightarrow{\vartheta} k \to H^1(X, \mathcal{O}_X^*) \to H^1(X, f_*\mathcal{O}_{X'}^*)\] We have \(H^1(X, f_*\mathcal{O}_{X'}^*) \subset H^1(X', \mathcal{O}_{X'}^*)\) (for example by the Leray spectral sequence, see Cohomology, Lemma 01F2). Hence the kernel of \(\Pic(X) \to \Pic(X')\) is the cokernel of the map \(\vartheta : H^0(X', \mathcal{O}_{X'}^*) \to k\). Because \(k\) is algebraically closed any regular function on a reduced connected proper scheme over \(k\) comes from an element of \(k\), see Varieties, Lemma 0BUG. Thus the final statement of the lemma.

Multicross and nodal singularities

In this section we discuss the simplest possible curve singularities.

Let \(k\) be a field. Consider the complete local \(k\)-algebra [0C1U]\[\begin{equation} A = \{(f_1, \ldots, f_n) \in k[[t]] \times \ldots \times k[[t]] \mid f_1(0) = \ldots = f_n(0)\} \end{equation}\] In the language introduced in Varieties, Definition 0C41 we see that \(A\) is a wedge of \(n\) copies of the power series ring in \(1\) variable over \(k\). Observe that \(k[[t]] \times \ldots \times k[[t]]\) is the integral closure of \(A\) in its total ring of fractions. Hence the \(\delta\)-invariant of \(A\) is \(n - 1\). There is an isomorphism \[k[[x_1, \ldots, x_n]]/(\{x_ix_j\}_{i \not = j}) \longrightarrow A\] obtained by sending \(x_i\) to \((0, \ldots, 0, t, 0, \ldots, 0)\) in \(A\). It follows that \(\dim(A) = 1\) and \(\dim_k \mathfrak m/\mathfrak m^2 = n\). In particular, \(A\) is regular if and only if \(n = 1\).

Lemma

Let \(k\) be a separably closed field. Let \(A\) be a \(1\)-dimensional reduced Nagata local \(k\)-algebra with residue field \(k\). Then \[\delta\text{-invariant }A \geq \text{number of branches of }A - 1\] If equality holds, then \(A^\wedge\) is as in (0C1U).

Proof

Since the residue field of \(A\) is separably closed, the number of branches of \(A\) is equal to the number of geometric branches of \(A\), see More on Algebra, Definition 0C26. The inequality holds by Varieties, Lemma 0C43. Assume equality holds. We may replace \(A\) by the completion of \(A\); this does not change the number of branches or the \(\delta\)-invariant, see More on Algebra, Lemma 0C2D and Varieties, Lemma 0C3W. Then \(A\) is strictly henselian, see Algebra, Lemma 04GM. By Varieties, Lemma 0C42 we see that \(A\) is a wedge of complete discrete valuation rings. Each of these is isomorphic to \(k[[t]]\) by Algebra, Lemma 0C0S. Hence \(A\) is as in (0C1U).

Definition

Let \(k\) be an algebraically closed field. Let \(X\) be an algebraic \(1\)-dimensional \(k\)-scheme. Let \(x \in X\) be a closed point. We say \(x\) defines a multicross singularity if the completion \(\mathcal{O}_{X, x}^\wedge\) is isomorphic to (0C1U) for some \(n \geq 2\). We say \(x\) is a node, or an ordinary double point, or defines a nodal singularity if \(n = 2\).

These singularities are in some sense the simplest kind of singularities one can have on a curve over an algebraically closed field.

Lemma

Let \(k\) be an algebraically closed field. Let \(X\) be a reduced algebraic \(1\)-dimensional \(k\)-scheme. Let \(x \in X\). The following are equivalent

1. \(x\) defines a multicross singularity,

2. the \(\delta\)-invariant of \(X\) at \(x\) is the number of branches of \(X\) at \(x\) minus \(1\),

3. there is a sequence of morphisms \(U_n \to U_{n - 1} \to \ldots \to U_0 = U \subset X\) where \(U\) is an open neighbourhood of \(x\), where \(U_n\) is nonsingular, and where each \(U_i \to U_{i - 1}\) is the glueing of two points as in Example 0C1J.

Proof

The equivalence of (1) and (2) is Lemma 0C1V.

Assume (3). We will argue by descending induction on \(i\) that all singularities of \(U_i\) are multicross. This is true for \(U_n\) as \(U_n\) has no singular points. If \(U_i\) is gotten from \(U_{i + 1}\) by glueing \(a, b \in U_{i + 1}\) to a point \(c \in U_i\), then we see that \[\mathcal{O}_{U_i, c}^\wedge \subset \mathcal{O}_{U_{i + 1}, a}^\wedge \times \mathcal{O}_{U_{i + 1}, b}^\wedge\] is the set of elements having the same residue classes in \(k\). Thus the number of branches at \(c\) is the sum of the number of branches at \(a\) and \(b\), and the \(\delta\)-invariant at \(c\) is the sum of the \(\delta\)-invariants at \(a\) and \(b\) plus \(1\) (because the displayed inclusion has codimension \(1\)). This proves that (2) holds as desired.

Assume the equivalent conditions (1) and (2). We may choose an open \(U \subset X\) such that \(x\) is the only singular point of \(U\). Then we apply Lemma 0C1L to the normalization morphism \[U^\nu = U_n \to U_{n - 1} \to \ldots \to U_1 \to U_0 = U\] All we have to do is show that in none of the steps we are squishing a tangent vector. Suppose \(U_{i + 1} \to U_i\) is the smallest \(i\) such that this is the squishing of a tangent vector \(\theta\) at \(u' \in U_{i + 1}\) lying over \(u \in U_i\). Arguing as above, we see that \(u_i\) is a multicross singularity (because the maps \(U_i \to \ldots \to U_0\) are glueing of pairs of points). But now the number of branches at \(u'\) and \(u\) is the same and the \(\delta\)-invariant of \(U_i\) at \(u\) is \(1\) bigger than the \(\delta\)-invariant of \(U_{i + 1}\) at \(u'\). By Lemma 0C1V this implies that \(u\) cannot be a multicross singularity which is a contradiction.

Lemma

Let \(k\) be an algebraically closed field. Let \(X\) be a reduced algebraic \(1\)-dimensional \(k\)-scheme. Let \(x \in X\) be a multicross singularity (Definition 0C1W). If \(X\) is Gorenstein, then \(x\) is a node.

Proof

The map \(\mathcal{O}_{X, x} \to \mathcal{O}_{X, x}^\wedge\) is flat and unramified in the sense that \(\kappa(x) = \mathcal{O}_{X, x}^\wedge/\mathfrak m_x \mathcal{O}_{X, x}^\wedge\). (See More on Algebra, Section 07NU.) Thus \(X\) is Gorenstein implies \(\mathcal{O}_{X, x}\) is Gorenstein, implies \(\mathcal{O}_{X, x}^\wedge\) is Gorenstein by Dualizing Complexes, Lemma 0BJL. Thus it suffices to show that the ring \(A\) in (0C1U) with \(n \geq 2\) is Gorenstein if and only if \(n = 2\).

If \(n = 2\), then \(A = k[[x, y]]/(xy)\) is a complete intersection and hence Gorenstein. For example this follows from Duality for Schemes, Lemma 0BVA applied to \(k[[x, y]] \to A\) and the fact that the regular local ring \(k[[x, y]]\) is Gorenstein by Dualizing Complexes, Lemma 0AWX.

Assume \(n > 2\). If \(A\) where Gorenstein, then \(A\) would be a dualizing complex over \(A\) (Duality for Schemes, Definition 0AWW). Then \(R\Hom(k, A)\) would be equal to \(k[n]\) for some \(n \in \mathbf{Z}\), see Dualizing Complexes, Lemma 0A7L. It would follow that \(\Ext^1_A(k, A) \cong k\) or \(\Ext^1_A(k, A) = 0\) (depending on the value of \(n\); in fact \(n\) has to be \(-1\) but it doesn’t matter to us here). Using the exact sequence \[0 \to \mathfrak m_A \to A \to k \to 0\] we find that \[\Ext^1_A(k, A) = \Hom_A(\mathfrak m_A, A)/A\] where \(A \to \Hom_A(\mathfrak m_A, A)\) is given by \(a \mapsto (a' \mapsto aa')\). Let \(e_i \in \Hom_A(\mathfrak m_A, A)\) be the element that sends \((f_1, \ldots, f_n) \in \mathfrak m_A\) to \((0, \ldots, 0, f_i, 0, \ldots, 0)\). The reader verifies easily that \(e_1, \ldots, e_{n - 1}\) are \(k\)-linearly independent in \(\Hom_A(\mathfrak m_A, A)/A\). Thus \(\dim_k \Ext^1_A(k, A) \geq n - 1 \geq 2\) which finishes the proof. (Observe that \(e_1 + \ldots + e_n\) is the image of \(1\) under the map \(A \to \Hom_A(\mathfrak m_A, A)\).)

Torsion in the Picard group

In this section we bound the torsion in the Picard group of a \(1\)-dimensional proper scheme over a field. We will use this in our study of semistable reduction for curves.

There does not seem to be an elementary way to obtain the result of Lemma 0C1Z. Analyzing the proof there are two key ingredients: (1) there is an abelian variety classifying degree zero invertible sheaves on a smooth projective curve and (2) the structure of torsion points on an abelian variety can be determined.

Lemma

Let \(k\) be an algebraically closed field. Let \(X\) be a smooth projective curve of genus \(g\) over \(k\).

1. If \(n \geq 1\) is invertible in \(k\), then \(\Pic(X)[n] \cong (\mathbf{Z}/n\mathbf{Z})^{\oplus 2g}\).

2. If the characteristic of \(k\) is \(p > 0\), then there exists an integer \(0 \leq f \leq g\) such that \(\Pic(X)[p^m] \cong (\mathbf{Z}/p^m\mathbf{Z})^{\oplus f}\) for all \(m \geq 1\).

Proof

Let \(\Pic^0(X) \subset \Pic(X)\) denote the subgroup of invertible sheaves of degree \(0\). In other words, there is a short exact sequence \[0 \to \Pic^0(X) \to \Pic(X) \xrightarrow{\deg} \mathbf{Z} \to 0.\] The group \(\Pic^0(X)\) is the \(k\)-points of the group scheme \(\underline{\Picardfunctor}^0_{X/k}\), see Picard Schemes of Curves, Lemma 0BA0. The same lemma tells us that \(\underline{\Picardfunctor}^0_{X/k}\) is a \(g\)-dimensional abelian variety over \(k\) as defined in Groupoids, Definition 03RO. Thus we conclude by the results of Groupoids, Proposition 03RP.

Lemma

Let \(k\) be a field. Let \(n\) be prime to the characteristic of \(k\). Let \(X\) be a smooth proper curve over \(k\) with \(H^0(X, \mathcal{O}_X) = k\) and of genus \(g\).

1. If \(g = 1\) then there exists a finite separable extension \(k'/k\) such that \(X_{k'}\) has a \(k'\)-rational point and \(\Pic(X_{k'})[n] \cong (\mathbf{Z}/n\mathbf{Z})^{\oplus 2}\).

2. If \(g \geq 2\) then there exists a finite separable extension \(k'/k\) with \([k' : k] \leq (2g - 2)(n^{2g})!\) such that \(X_{k'}\) has a \(k'\)-rational point and \(\Pic(X_{k'})[n] \cong (\mathbf{Z}/n\mathbf{Z})^{\oplus 2g}\).

Proof

Assume \(g \geq 2\). First we may choose a finite separable extension of degree at most \(2g - 2\) such that \(X\) acquires a rational point, see Lemma 0CD4. Thus we may assume \(X\) has a \(k\)-rational point \(x \in X(k)\) but now we have to prove the lemma with \([k' : k] \leq (n^{2g})!\). Let \(k \subset k^{sep} \subset \overline{k}\) be a separable algebraic closure inside an algebraic closure. By Lemma 0C1Z we have \[\Pic(X_{\overline{k}})[n] \cong (\mathbf{Z}/n\mathbf{Z})^{\oplus 2g}\] By Picard Schemes of Curves, Lemma 0CD5 we conclude that \[\Pic(X_{k^{sep}})[n] \cong (\mathbf{Z}/n\mathbf{Z})^{\oplus 2g}\] By Picard Schemes of Curves, Lemma 0CD5 there is a continuous action \[\text{Gal}(k^{sep}/k) \longrightarrow \text{Aut}(\Pic(X_{k^{sep}})[n]\] and the lemma is true for the fixed field \(k'\) of the kernel of this map. The kernel is open because the action is continuous which implies that \(k'/k\) is finite. By Galois theory \(\text{Gal}(k'/k)\) is the image of the displayed arrow. Since the permutation group of a set of cardinality \(n^{2g}\) has cardinality \((n^{2g})!\) we conclude by Galois theory that \([k' : k] \leq (n^{2g})!\). (Of course this proves the lemma with the bound \(|\text{GL}_{2g}(\mathbf{Z}/n\mathbf{Z})|\), but all we want here is that there is some bound.)

If the genus is \(1\), then there is no upper bound on the degree of a finite separable field extension over which \(X\) acquires a rational point (details omitted). Still, there is such an extension for example by Varieties, Lemma 056U. The rest of the proof is the same as in the case of \(g \geq 2\).

Proposition

Let \(k\) be an algebraically closed field. Let \(X\) be a proper scheme over \(k\) which is reduced, connected, and has dimension \(1\). Let \(g\) be the genus of \(X\) and let \(g_{geom}\) be the sum of the geometric genera of the irreducible components of \(X\). For any prime \(\ell\) different from the characteristic of \(k\) we have \[\dim_{\mathbf{F}_\ell} \Pic(X)[\ell] \leq g + g_{geom}\] and equality holds if and only if all the singularities of \(X\) are multicross.

Proof

Let \(\nu : X^\nu \to X\) be the normalization (Varieties, Lemma 0C1R). Choose a factorization \[X^\nu = X_n \to X_{n - 1} \to \ldots \to X_1 \to X_0 = X\] as in Lemma 0C1L. Let us denote \(h^0_i = \dim_k H^0(X_i, \mathcal{O}_{X_i})\) and \(h^1_i = \dim_k H^1(X_i, \mathcal{O}_{X_i})\). By Lemmas 0C1M and 0C1N for each \(n > i \geq 0\) we have one of the following there possibilities

1. \(X_i\) is obtained by glueing \(a, b \in X_{i + 1}\) which are on different connected components: in this case \(\Pic(X_i) = \Pic(X_{i + 1})\), \(h^0_{i + 1} = h^0_i + 1\), \(h^1_{i + 1} = h^1_i\),

2. \(X_i\) is obtained by glueing \(a, b \in X_{i + 1}\) which are on the same connected component: in this case there is a short exact sequence \[0 \to k^* \to \Pic(X_i) \to \Pic(X_{i + 1}) \to 0,\] and \(h^0_{i + 1} = h^0_i\), \(h^1_{i + 1} = h^1_i - 1\),

3. \(X_i\) is obtained by squishing a tangent vector in \(X_{i + 1}\): in this case there is a short exact sequence \[0 \to (k, +) \to \Pic(X_i) \to \Pic(X_{i + 1}) \to 0,\] and \(h^0_{i + 1} = h^0_i\), \(h^1_{i + 1} = h^1_i - 1\).

To prove the statements on dimensions of cohomology groups of the structure sheaf, use the exact sequences in Examples 0C1J and 0C1K. Since \(k\) is algebraically closed of characteristic prime to \(\ell\) we see that \((k, +)\) and \(k^*\) are \(\ell\)-divisible and with \(\ell\)-torsion \((k, +)[\ell] = 0\) and \(k^*[\ell] \cong \mathbf{F}_\ell\). Hence \[\dim_{\mathbf{F}_\ell} \Pic(X_{i + 1})[\ell] - \dim_{\mathbf{F}_\ell}\Pic(X_i)[\ell]\] is zero, except in case (2) where it is equal to \(-1\). At the end of this process we get the normalization \(X^\nu = X_n\) which is a disjoint union of smooth projective curves over \(k\). Hence we have

1. \(h^1_n = g_{geom}\) and

2. \(\dim_{\mathbf{F}_\ell} \Pic(X_n)[\ell] = 2g_{geom}\).

The last equality by Lemma 0C1Z. Since \(g = h^1_0\) we see that the number of steps of type (2) and (3) is at most \(h^1_0 - h^1_n = g - g_{geom}\). By our comptation of the differences in ranks we conclude that \[\dim_{\mathbf{F}_\ell} \Pic(X)[\ell] \leq g - g_{geom} + 2g_{geom} = g + g_{geom}\] and equality holds if and only if no steps of type (3) occur. This indeed means that all singularities of \(X\) are multicross by Lemma 0C1X. Conversely, if all the singularities are multicross, then Lemma 0C1X guarantees that we can find a sequence \(X^\nu = X_n \to \ldots \to X_0 = X\) as above such that no steps of type (3) occur in the sequence and we find equality holds in the lemma (just glue the local sequences for each point to find one that works for all singular points of \(x\); some details omitted).

Genus versus geometric genus

Let \(k\) be a field with algebraic closure \(\overline{k}\). Let \(X\) be a proper scheme of dimension \(\leq 1\) over \(k\). We define \(g_{geom}(X/k)\) to be the sum of the geometric genera of the irreducible components of \(X_{\overline{k}}\) which have dimension \(1\).

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme of dimension \(\leq 1\) over \(k\). Then \[g_{geom}(X/k) = \sum\nolimits_{C \subset X} g_{geom}(C/k)\] where the sum is over irreducible components \(C \subset X\) of dimension \(1\).

Proof

This is immediate from the definition and the fact that an irreducible component \(\overline{Z}\) of \(X_{\overline{k}}\) maps onto an irreducible component \(Z\) of \(X\) (Varieties, Lemma 04KX) of the same dimension (Morphisms, Lemma 02FY applied to the generic point of \(\overline{Z}\)).

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme of dimension \(\leq 1\) over \(k\). Then

1. We have \(g_{geom}(X/k) = g_{geom}(X_{red}/k)\).

2. If \(X' \to X\) is a birational proper morphism, then \(g_{geom}(X'/k) = g_{geom}(X/k)\).

3. If \(X^\nu \to X\) is the normalization morphism, then \(g_{geom}(X^\nu/k) = g_{geom}(X/k)\).

Proof

Part (1) is immediate from Lemma 0CE1. If \(X' \to X\) is proper birational, then it is finite and an isomorphism over a dense open (see Varieties, Lemmas 0AB7 and 0BFP). Hence \(X'_{\overline{k}} \to X_{\overline{k}}\) is an isomorphism over a dense open. Thus the irreducible components of \(X'_{\overline{k}}\) and \(X_{\overline{k}}\) are in bijective correspondence and the corresponding components have isomorphic function fields. In particular these components have isomorphic nonsingular projective models and hence have the same geometric genera. This proves (2). Part (3) follows from (1) and (2) and the fact that \(X^\nu \to X_{red}\) is birational (Morphisms, Lemma 0BXC).

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme of dimension \(\leq 1\) over \(k\). Let \(f : Y \to X\) be a finite morphism such that there exists a dense open \(U \subset X\) over which \(f\) is a closed immersion. Then \[\dim_k H^1(X, \mathcal{O}_X) \geq \dim_k H^1(Y, \mathcal{O}_Y)\]

Proof

Consider the exact sequence \[0 \to \mathcal{G} \to \mathcal{O}_X \to f_*\mathcal{O}_Y \to \mathcal{F} \to 0\] of coherent sheaves on \(X\). By assumption \(\mathcal{F}\) is supported in finitely many closed points and hence has vanishing higher cohomology (Varieties, Lemma 0AYT). On the other hand, we have \(H^2(X, \mathcal{G}) = 0\) by Cohomology, Proposition 02UZ. It follows formally that the induced map \(H^1(X, \mathcal{O}_X) \to H^1(X, f_*\mathcal{O}_Y)\) is surjective. Since \(H^1(X, f_*\mathcal{O}_Y) = H^1(Y, \mathcal{O}_Y)\) (Cohomology of Schemes, Lemma 089W) we conclude the lemma holds.

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme of dimension \(\leq 1\) over \(k\). If \(X' \to X\) is a birational proper morphism, then \[\dim_k H^1(X, \mathcal{O}_X) \geq \dim_k H^1(X', \mathcal{O}_{X'})\] If \(X\) is reduced, \(H^0(X, \mathcal{O}_X) \to H^0(X', \mathcal{O}_{X'})\) is surjective, and equality holds, then \(X' = X\).

Proof

If \(f : X' \to X\) is proper birational, then it is finite and an isomorphism over a dense open (see Varieties, Lemmas 0AB7 and 0BFP). Thus the inequality by Lemma 0CE3. Assume \(X\) is reduced. Then \(\mathcal{O}_X \to f_*\mathcal{O}_{X'}\) is injective and we obtain a short exact sequence \[0 \to \mathcal{O}_X \to f_*\mathcal{O}_{X'} \to \mathcal{F} \to 0\] Under the assumptions given in the second statement, we conclude from the long exact cohomology sequence that \(H^0(X, \mathcal{F}) = 0\). Then \(\mathcal{F} = 0\) because \(\mathcal{F}\) is generated by global sections (Varieties, Lemma 0AYT). and \(\mathcal{O}_X = f_*\mathcal{O}_{X'}\). Since \(f\) is affine this implies \(X = X'\).

Lemma

Let \(k\) be a field. Let \(C\) be a proper curve over \(k\). Set \(\kappa = H^0(C, \mathcal{O}_C)\). Then \[[\kappa : k]_s \dim_\kappa H^1(C, \mathcal{O}_C) \geq g_{geom}(C/k)\]

Proof

Varieties, Lemma 04L2 implies \(\kappa\) is a field and a finite extension of \(k\). By Fields, Lemma 09HJ we have \([\kappa : k]_s = |\Mor_k(\kappa, \overline{k})|\) and hence \(\Spec(\kappa \otimes_k \overline{k})\) has \([\kappa : k]_s\) points each with residue field \(\overline{k}\). Thus \[C_{\overline{k}} = \bigcup\nolimits_{t \in \Spec(\kappa \otimes_k \overline{k})} C_t\] (set theoretic union). Here \(C_t = C \times_{\Spec(\kappa), t} \Spec(\overline{k})\) where we view \(t\) as a \(k\)-algebra map \(t : \kappa \to \overline{k}\). The conclusion is that \(g_{geom}(C/k) = \sum_t g_{geom}(C_t/\overline{k})\) and the sum is over an index set of size \([\kappa : k]_s\). We have \[H^0(C_t, \mathcal{O}_{C_t}) = \overline{k} \quad\text{and}\quad \dim_{\overline{k}} H^1(C_t, \mathcal{O}_{C_t}) = \dim_\kappa H^1(C, \mathcal{O}_C)\] by cohomology and base change (Cohomology of Schemes, Lemma 02KH). Observe that the normalization \(C_t^\nu\) is the disjoint union of the nonsingular projective models of the irreducible components of \(C_t\) (Morphisms, Lemma 0CDV). Hence \(\dim_{\overline{k}} H^1(C_t^\nu, \mathcal{O}_{C_t^\nu})\) is equal to \(g_{geom}(C_t/\overline{k})\). By Lemma 0CE3 we have \[\dim_{\overline{k}} H^1(C_t, \mathcal{O}_{C_t}) \geq \dim_{\overline{k}} H^1(C_t^\nu, \mathcal{O}_{C_t^\nu})\] and this finishes the proof.

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme of dimension \(\leq 1\) over \(k\). Let \(\ell\) be a prime number invertible in \(k\). Then \[\dim_{\mathbf{F}_\ell} \Pic(X)[\ell] \leq \dim_k H^1(X, \mathcal{O}_X) + g_{geom}(X/k)\] where \(g_{geom}(X/k)\) is as defined above.

Proof

The map \(\Pic(X) \to \Pic(X_{\overline{k}})\) is injective by Varieties, Lemma 0CC5. By Cohomology of Schemes, Lemma 02KH \(\dim_k H^1(X, \mathcal{O}_X)\) equals \(\dim_{\overline{k}} H^1(X_{\overline{k}}, \mathcal{O}_{X_{\overline{k}}})\). Hence we may assume \(k\) is algebraically closed.

Let \(X_{red}\) be the reduction of \(X\). Then the surjection \(\mathcal{O}_X \to \mathcal{O}_{X_{red}}\) induces a surjection \(H^1(X, \mathcal{O}_X) \to H^1(X, \mathcal{O}_{X_{red}})\) because cohomology of quasi-coherent sheaves vanishes in degrees \(\geq 2\) by Cohomology, Proposition 02UZ. Since \(X_{red} \to X\) induces an isomorphism on irreducible components over \(\overline{k}\) and an isomorphism on \(\ell\)-torsion in Picard groups (Picard Schemes of Curves, Lemma 0CD5) we may replace \(X\) by \(X_{red}\). In this way we reduce to Proposition 0C20.

Nodal curves

We have already defined ordinary double points over algebraically closed fields, see Definition 0C1W. Namely, if \(x \in X\) is a closed point of a \(1\)-dimensional algebraic scheme over an algebraically closed field \(k\), then \(x\) is an ordinary double point if and only if \[\mathcal{O}_{X, x}^\wedge \cong k[[x, y]]/(xy)\] See discussion following (0C1U) in Section 0C1P.

Definition

Let \(k\) be a field. Let \(X\) be a \(1\)-dimensional locally algebraic \(k\)-scheme.

1. We say a closed point \(x \in X\) is a node, or an ordinary double point, or defines a nodal singularity if there exists an ordinary double point \(\overline{x} \in X_{\overline{k}}\) mapping to \(x\).

2. We say the singularities of \(X\) are at-worst-nodal if all closed points of \(X\) are either in the smooth locus of the structure morphism \(X \to \Spec(k)\) or are ordinary double points.

Often a \(1\)-dimensional algebraic scheme \(X\) is called a nodal curve if the singularities of \(X\) are at worst nodal. Sometimes a nodal curve is required to be proper. Since a nodal curve so defined need not be irreducible, this conflicts with our earlier definition of a curve as a variety of dimension \(1\).

Lemma

Let \((A, \mathfrak m)\) be a regular local ring of dimension \(2\). Let \(I \subset \mathfrak m\) be an ideal.

1. If \(A/I\) is reduced, then \(I = (0)\), \(I = \mathfrak m\), or \(I = (f)\) for some nonzero \(f \in \mathfrak m\).

2. If \(A/I\) has depth \(1\), then \(I = (f)\) for some nonzero \(f \in \mathfrak m\).

Proof

Assume \(I \not = 0\). Write \(I = (f_1, \ldots, f_r)\). As \(A\) is a UFD (More on Algebra, Lemma 0AG0) we can write \(f_i = fg_i\) where \(f\) is the gcd of \(f_1, \ldots, f_r\). Thus the gcd of \(g_1, \ldots, g_r\) is \(1\) which means that there is no height \(1\) prime ideal over \(g_1, \ldots, g_r\). Then either \((g_1, \ldots, g_r) = A\) which implies \(I = (f)\) or if not, then \(\dim(A) = 2\) implies that \(V(g_1, \ldots, g_r) = \{\mathfrak m\}\), i.e., \(\mathfrak m = \sqrt{(g_1, \ldots, g_r)}\).

Assume \(A/I\) reduced, i.e., \(I\) radical. If \(f\) is a unit, then since \(I\) is radical we see that \(I = \mathfrak m\). If \(f \in \mathfrak m\), then we see that \(f^n\) maps to zero in \(A/I\). Hence \(f \in I\) by reducedness and we conclude \(I = (f)\).

Assume \(A/I\) has depth \(1\). Then \(\mathfrak m\) is not an associated prime of \(A/I\). Since the class of \(f\) modulo \(I\) is annihilated by \(g_1, \ldots, g_r\), this implies that the class of \(f\) is zero in \(A/I\). Thus \(I = (f)\) as desired.

Let \(\kappa\) be a field and let \(V\) be a vector space over \(\kappa\). We will say \(q \in \text{Sym}^2_\kappa(V)\) is nondegenerate if the induced \(\kappa\)-linear map \(V^\vee \to V\) is an isomorphism. If \(q = \sum_{i \leq j} a_{ij} x_i x_j\) for some \(\kappa\)-basis \(x_1, \ldots, x_n\) of \(V\), then this means that the determinant of the matrix \[\left( \begin{matrix} 2a_{11} & a_{12} & \ldots \\ a_{12} & 2a_{22} & \ldots \\ \ldots & \ldots & \ldots \end{matrix} \right)\] is nonzero. This is equivalent to the condition that the partial derivatives of \(q\) with respect to the \(x_i\) cut out \(0\) scheme theoretically.

Lemma

Let \(k\) be a field. Let \((A, \mathfrak m, \kappa)\) be a Noetherian local \(k\)-algebra. The following are equivalent

1. \(\kappa/k\) is separable, \(A\) is reduced, \(\dim_\kappa(\mathfrak m/\mathfrak m^2) = 2\), and there exists a nondegenerate \(q \in \text{Sym}^2_\kappa(\mathfrak m/\mathfrak m^2)\) which maps to zero in \(\mathfrak m^2/\mathfrak m^3\),

2. \(\kappa/k\) is separable, \(\text{depth}(A) = 1\), \(\dim_\kappa(\mathfrak m/\mathfrak m^2) = 2\), and there exists a nondegenerate \(q \in \text{Sym}^2_\kappa(\mathfrak m/\mathfrak m^2)\) which maps to zero in \(\mathfrak m^2/\mathfrak m^3\),

3. \(\kappa/k\) is separable, \(A^\wedge \cong \kappa[[x, y]]/(ax^2 + bxy + cy^2)\) as a \(k\)-algebra where \(ax^2 + bxy + cy^2\) is a nondegenerate quadratic form over \(\kappa\).

Proof

Assume (3). Then \(A^\wedge\) is reduced because \(ax^2 + bxy + cy^2\) is either irreducible or a product of two nonassociated prime elements. Hence \(A \subset A^\wedge\) is reduced. It follows that (1) is true.

Assume (1). Then \(A\) cannot be Artinian, since it would not be reduced because \(\mathfrak m \not = (0)\). Hence \(\dim(A) \geq 1\), hence \(\text{depth}(A) \geq 1\) by Algebra, Lemma 031R. On the other hand \(\dim(A) = 2\) implies \(A\) is regular which contradicts the existence of \(q\) by Algebra, Lemma 00NO. Thus \(\dim(A) \leq 1\) and we conclude \(\text{depth}(A) = 1\) by Algebra, Lemma 00LK. It follows that (2) is true.

Assume (2). Since the depth of \(A\) is the same as the depth of \(A^\wedge\) (More on Algebra, Lemma 07NW) and since the other conditions are insensitive to completion, we may assume that \(A\) is complete. Choose \(\kappa \to A\) as in More on Algebra, Lemma 0C34. Since \(\dim_\kappa(\mathfrak m/\mathfrak m^2) = 2\) we can choose \(x_0, y_0 \in \mathfrak m\) which map to a basis. We obtain a continuous \(\kappa\)-algebra map \[\kappa[[x, y]] \longrightarrow A\] by the rules \(x \mapsto x_0\) and \(y \mapsto y_0\). Let \(q\) be the class of \(ax_0^2 + bx_0y_0 + cy_0^2\) in \(\text{Sym}^2_\kappa(\mathfrak m/\mathfrak m^2)\). Write \(Q(x, y) = ax^2 + bxy + cy^2\) viewed as a polynomial in two variables. Then we see that \[Q(x_0, y_0) = ax_0^2 + bx_0y_0 + cy_0^2 = \sum\nolimits_{i + j = 3} a_{ij} x_0^iy_0^j\] for some \(a_{ij}\) in \(A\). We want to prove that we can increase the order of vanishing by changing our choice of \(x_0\), \(y_0\). Suppose that \(x_1, y_1 \in \mathfrak m^2\). Then \[Q(x_0 + x_1, y_0 + y_1) = Q(x_0, y_0) + (2ax_0 + by_0)x_1 + (bx_0 + 2cy_0)y_1 \bmod \mathfrak m^4\] Nondegeneracy of \(Q\) means exactly that \(2ax_0 + by_0\) and \(bx_0 + 2cy_0\) are a \(\kappa\)-basis for \(\mathfrak m/\mathfrak m^2\), see discussion preceding the lemma. Hence we can certainly choose \(x_1, y_1 \in \mathfrak m^2\) such that \(Q(x_0 + x_1, y_0 + y_1) \in \mathfrak m^4\). Continuing in this fashion by induction we can find \(x_i, y_i \in \mathfrak m^{i + 1}\) such that \[Q(x_0 + x_1 + \ldots + x_n, y_0 + y_1 + \ldots + y_n) \in \mathfrak m^{n + 3}\] Since \(A\) is complete we can set \(x_\infty = \sum x_i\) and \(y_\infty = \sum y_i\) and we can consider the map \(\kappa[[x, y]] \longrightarrow A\) sending \(x\) to \(x_\infty\) and \(y\) to \(y_\infty\). This map induces a surjection \(\kappa[[x, y]]/(Q) \longrightarrow A\) by Algebra, Lemma 0315. By Lemma 0C48 the kernel of \(k[[x, y]] \to A\) is principal. But the kernel cannot contain a proper divisor of \(Q\) as such a divisor would have degree \(1\) in \(x, y\) and this would contradict \(\dim(\mathfrak m/\mathfrak m^2) = 2\). Hence \(Q\) generates the kernel as desired.

Lemma

Let \(k\) be a field. Let \((A, \mathfrak m, \kappa)\) be a Nagata local \(k\)-algebra. The following are equivalent

1. \(k \to A\) is as in Lemma 0C49,

2. \(\kappa/k\) is separable, \(A\) is reduced of dimension \(1\), the \(\delta\)-invariant of \(A\) is \(1\), and \(A\) has \(2\) geometric branches.

If this holds, then the integral closure \(A'\) of \(A\) in its total ring of fractions has either \(1\) or \(2\) maximal ideals \(\mathfrak m'\) and the extensions \(\kappa(\mathfrak m')/k\) are separable.

Proof

In both cases \(A\) and \(A^\wedge\) are reduced. In case (2) because the completion of a reduced local Nagata ring is reduced (More on Algebra, Lemma 07NZ). In both cases \(A\) and \(A^\wedge\) have dimension \(1\) (More on Algebra, Lemma 07NV). The \(\delta\)-invariant and the number of geometric branches of \(A\) and \(A^\wedge\) agree by Varieties, Lemma 0C3W and More on Algebra, Lemma 0C2D. Let \(A'\) be the integral closure of \(A\) in its total ring of fractions as in Varieties, Lemma 0C3S. By Varieties, Lemma 0C3V we see that \(A' \otimes_A A^\wedge\) plays the same role for \(A^\wedge\). Thus we may replace \(A\) by \(A^\wedge\) and assume \(A\) is complete.

Assume (1) holds. It suffices to show that \(A\) has two geometric branches and \(\delta\)-invariant \(1\). We may assume \(A = \kappa[[x, y]]/(ax^2 + bxy + cy^2)\) with \(q = ax^2 + bxy + cy^2\) nondegenerate. There are two cases.

Case I: \(q\) splits over \(\kappa\). In this case we may after changing coordinates assume that \(q = xy\). Then we see that \[A' = \kappa[[x, y]]/(x) \times \kappa[[x, y]]/(y)\]

Case II: \(q\) does not split. In this case \(c \not = 0\) and nondegenerate means \(b^2 - 4ac \not = 0\). Hence \(\kappa' = \kappa[t]/(a + bt + ct^2)\) is a degree \(2\) separable extension of \(\kappa\). Then \(t = y/x\) is integral over \(A\) and we conclude that \[A' = \kappa'[[x]]\] with \(y\) mapping to \(tx\) on the right hand side.

In both cases one verifies by hand that the \(\delta\)-invariant is \(1\) and the number of geometric branches is \(2\). In this way we see that (1) implies (2). Moreover we conclude that the final statement of the lemma holds.

Assume (2) holds. More on Algebra, Lemma 0C37 implies \(A'\) either has two maximal ideals or \(A'\) has one maximal ideal and \([\kappa(\mathfrak m') : \kappa]_s = 2\).

Case I: \(A'\) has two maximal ideals \(\mathfrak m'_1\), \(\mathfrak m'_2\) with residue fields \(\kappa_1\), \(\kappa_2\). Since the \(\delta\)-invariant is the length of \(A'/A\) and since there is a surjection \(A'/A \to (\kappa_1 \times \kappa_2)/\kappa\) we see that \(\kappa = \kappa_1 = \kappa_2\). Since \(A\) is complete (and henselian by Algebra, Lemma 04GM) and \(A'\) is finite over \(A\) we see that \(A' = A_1 \times A_2\) (by Algebra, Lemma 04GH). Since \(A'\) is a normal ring it follows that \(A_1\) and \(A_2\) are discrete valuation rings. Hence \(A_1\) and \(A_2\) are isomorphic to \(\kappa[[t]]\) (as \(k\)-algebras) by More on Algebra, Lemma 0C35. Since the \(\delta\)-invariant is \(1\) we conclude that \(A\) is the wedge of \(A_1\) and \(A_2\) (Varieties, Definition 0C41). It follows easily that \(A \cong \kappa[[x, y]]/(xy)\).

Case II: \(A'\) has a single maximal ideal \(\mathfrak m'\) with residue field \(\kappa'\) and \([\kappa' : \kappa]_s = 2\). Arguing exactly as in Case I we see that \([\kappa' : \kappa] = 2\) and \(\kappa'\) is separable over \(\kappa\). Since \(A'\) is normal we see that \(A'\) is isomorphic to \(\kappa'[[t]]\) (see reference above). Since \(A'/A\) has length \(1\) we conclude that \[A = \{f \in \kappa'[[t]] \mid f(0) \in \kappa\}\] Then a simple computation shows that \(A\) as in case (1).

Lemma

Let \(k\) be a field. Let \(A = k[[x_1, \ldots, x_n]]\). Let \(I = (f_1, \ldots, f_m) \subset A\) be an ideal. For any \(r \geq 0\) the ideal in \(A/I\) generated by the \(r \times r\)-minors of the matrix \((\partial f_j/\partial x_i)\) is independent of the choice of the generators of \(I\) or the regular system of parameters \(x_1, \ldots, x_n\) of \(A\).

Proof

The “correct” proof of this lemma is to prove that this ideal is the \((n - r)\)th Fitting ideal of a module of continuous differentials of \(A/I\) over \(k\). Here is a direct proof. If \(g_1, \ldots g_l\) is a second set of generators of \(I\), then we can write \(g_s = \sum a_{sj}f_j\) and we have the equality of matrices \[(\partial g_s/\partial x_i) = (a_{sj}) (\partial f_j/\partial x_i) + (\partial a_{sj}/\partial x_i f_j)\] The final term is zero in \(A/I\). By the Cauchy-Binet formula we see that the ideal of minors for the \(g_s\) is contained in the ideal for the \(f_j\). By symmetry these ideals are the same. If \(y_1, \ldots, y_n \in \mathfrak m_A\) is a second regular system of parameters, then the matrix \((\partial y_j/\partial x_i)\) is invertible and we can use the chain rule for differentiation. Some details omitted.

Lemma

Let \(k\) be a field. Let \(A = k[[x_1, \ldots, x_n]]\). Let \(I = (f_1, \ldots, f_m) \subset \mathfrak m_A\) be an ideal. The following are equivalent

1. \(k \to A/I\) is as in Lemma 0C49,

2. \(A/I\) is reduced and the \((n - 1) \times (n - 1)\) minors of the matrix \((\partial f_j/\partial x_i)\) generate \(I + \mathfrak m_A\),

3. \(\text{depth}(A/I) = 1\) and the \((n - 1) \times (n - 1)\) minors of the matrix \((\partial f_j/\partial x_i)\) generate \(I + \mathfrak m_A\).

Proof

By Lemma 0C4B we may change our system of coordinates and the choice of generators during the proof.

If (1) holds, then we may change coordinates such that \(x_1, \ldots, x_{n - 2}\) map to zero in \(A/I\) and \(A/I = k[[x_{n - 1}, x_n]]/(a x_{n - 1}^2 + b x_{n - 1}x_n + c x_n^2)\) for some nondegenerate quadric \(a x_{n - 1}^2 + b x_{n - 1}x_n + c x_n^2\). Then we can explicitly compute to show that both (2) and (3) are true.

Assume the \((n - 1) \times (n - 1)\) minors of the matrix \((\partial f_j/\partial x_i)\) generate \(I + \mathfrak m_A\). Suppose that for some \(i\) and \(j\) the partial derivative \(\partial f_j/\partial x_i\) is a unit in \(A\). Then we may use the system of parameters \(f_j, x_1, \ldots, x_{i - 1}, \hat x_i, x_{i + 1}, \ldots, x_n\) and the generators \(f_j, f_1, \ldots, f_{j - 1}, \hat f_j, f_{j + 1}, \ldots, f_m\) of \(I\). Then we get a regular system of parameters \(x_1, \ldots, x_n\) and generators \(x_1, f_2, \ldots, f_m\) of \(I\). Next, we look for an \(i \geq 2\) and \(j \geq 2\) such that \(\partial f_j/\partial x_i\) is a unit in \(A\). If such a pair exists, then we can make a replacement as above and assume that we have a regular system of parameters \(x_1, \ldots, x_n\) and generators \(x_1, x_2, f_3, \ldots, f_m\) of \(I\). Continuing, in finitely many steps we reach the situation where we have a regular system of parameters \(x_1, \ldots, x_n\) and generators \(x_1, \ldots, x_t, f_{t + 1}, \ldots, f_m\) of \(I\) such that \(\partial f_j/\partial x_i \in \mathfrak m_A\) for all \(i, j \geq t + 1\).

In this case the matrix of partial derivatives has the following block shape \[\left( \begin{matrix} I_{t \times t} & * \\ 0 & \mathfrak m_A \end{matrix} \right)\] Hence every \((n - 1) \times (n - 1)\)-minor is in \(\mathfrak m_A^{n - 1 - t}\). Note that \(I \not = \mathfrak m_A\) otherwise the ideal of minors would contain \(1\). It follows that \(n - 1 - t \leq 1\) because there is an element of \(\mathfrak m_A \setminus \mathfrak m_A^2 + I\) (otherwise \(I = \mathfrak m_A\) by Nakayama). Thus \(t \geq n - 2\). We have seen that \(t \not = n\) above and similarly if \(t = n - 1\), then there is an invertible \((n - 1) \times (n - 1)\)-minor which is disallowed as well. Hence \(t = n - 2\). Then \(A/I\) is a quotient of \(k[[x_{n - 1}, x_n]]\) and Lemma 0C48 implies in both cases (2) and (3) that \(I\) is generated by \(x_1, \ldots, x_{n - 2}, f\) for some \(f = f(x_{n - 1}, x_n)\). In this case the condition on the minors exactly says that the quadratic term in \(f\) is nondegenerate, i.e., \(A/I\) is as in Lemma 0C49.

Lemma

Let \(k\) be a field. Let \(X\) be a \(1\)-dimensional algebraic \(k\)-scheme. Let \(x \in X\) be a closed point. The following are equivalent

1. \(x\) is a node,

2. \(k \to \mathcal{O}_{X, x}\) is as in Lemma 0C49,

3. any \(\overline{x} \in X_{\overline{k}}\) mapping to \(x\) defines a nodal singularity,

4. \(\kappa(x)/k\) is separable, \(\mathcal{O}_{X, x}\) is reduced, and the first Fitting ideal of \(\Omega_{X/k}\) generates \(\mathfrak m_x\) in \(\mathcal{O}_{X, x}\),

5. \(\kappa(x)/k\) is separable, \(\text{depth}(\mathcal{O}_{X, x}) = 1\), and the first Fitting ideal of \(\Omega_{X/k}\) generates \(\mathfrak m_x\) in \(\mathcal{O}_{X, x}\),

6. \(\kappa(x)/k\) is separable and \(\mathcal{O}_{X, x}\) is reduced, has \(\delta\)-invariant \(1\), and has \(2\) geometric branches.

Proof

First assume that \(k\) is algebraically closed. In this case the equivalence of (1) and (3) is trivial. The equivalence of (1) and (3) with (2) holds because the only nondegenerate quadric in two variables is \(xy\) up to change in coordinates. The equivalence of (1) and (6) is Lemma 0C1V. After replacing \(X\) by an affine neighbourhood of \(x\), we may assume there is a closed immersion \(X \to \mathbf{A}^n_k\) mapping \(x\) to \(0\). Let \(f_1, \ldots, f_m \in k[x_1, \ldots, x_n]\) be generators for the ideal \(I\) of \(X\) in \(\mathbf{A}^n_k\). Then \(\Omega_{X/k}\) corresponds to the \(R = k[x_1, \ldots, x_n]/I\)-module \(\Omega_{R/k}\) which has a presentation \[R^{\oplus m} \xrightarrow{(\partial f_j/\partial x_i)} R^{\oplus n} \to \Omega_{R/k} \to 0\] (See Algebra, Sections 00RM and 00S0.) The first Fitting ideal of \(\Omega_{R/k}\) is thus the ideal generated by the \((n - 1) \times (n - 1)\)-minors of the matrix \((\partial f_j/\partial x_i)\). Hence (2), (4), (5) are equivalent by Lemma 0C4C applied to the completion of \(k[x_1, \ldots, x_n] \to R\) at the maximal ideal \((x_1, \ldots, x_n)\).

Now assume \(k\) is an arbitrary field. In cases (2), (4), (5), (6) the residue field \(\kappa(x)\) is separable over \(k\). Let us show this holds as well in cases (1) and (3). Namely, let \(Z \subset X\) be the closed subscheme of \(X\) defined by the first Fitting ideal of \(\Omega_{X/k}\). The formation of \(Z\) commutes with field extension (Divisors, Lemma 0C3I). If (1) or (3) is true, then there exists a point \(\overline{x}\) of \(X_{\overline{k}}\) such that \(\overline{x}\) is an isolated point of multiplicity \(1\) of \(Z_{\overline{k}}\) (as we have the equivalence of the conditions of the lemma over \(\overline{k}\)). In particular \(Z_{\overline{x}}\) is geometrically reduced at \(\overline{x}\) (because \(\overline{k}\) is algebraically closed). Hence \(Z\) is geometrically reduced at \(x\) (Varieties, Lemma 0384). In particular, \(Z\) is reduced at \(x\), hence \(Z = \Spec(\kappa(x))\) in a neighbourhood of \(x\) and \(\kappa(x)\) is geometrically reduced over \(k\). This means that \(\kappa(x)/k\) is separable (Algebra, Lemma 030W).

The argument of the previous paragraph shows that if (1) or (3) holds, then the first Fitting ideal of \(\Omega_{X/k}\) generates \(\mathfrak m_x\). Since \(\mathcal{O}_{X, x} \to \mathcal{O}_{X_{\overline{k}}, \overline{x}}\) is flat and since \(\mathcal{O}_{X_{\overline{k}}, \overline{x}}\) is reduced and has depth \(1\), we see that (4) and (5) hold (use Algebra, Lemmas 033F and 0337). Conversely, (4) implies (5) by Algebra, Lemma 031R. If (5) holds, then \(Z\) is geometrically reduced at \(x\) (because \(\kappa(x)/k\) separable and \(Z\) is \(x\) in a neighbourhood). Hence \(Z_{\overline{k}}\) is reduced at any point \(\overline{x}\) of \(X_{\overline{k}}\) lying over \(x\). In other words, the first fitting ideal of \(\Omega_{X_{\overline{k}}/\overline{k}}\) generates \(\mathfrak m_{\overline{x}}\) in \(\mathcal{O}_{X_{\overline{k}, \overline{x}}}\). Moreover, since \(\mathcal{O}_{X, x} \to \mathcal{O}_{X_{\overline{k}}, \overline{x}}\) is flat we see that \(\text{depth}(\mathcal{O}_{X_{\overline{k}}, \overline{x}}) = 1\) (see reference above). Hence (5) holds for \(\overline{x} \in X_{\overline{k}}\) and we conclude that (3) holds (because of the equivalence over algebraically closed fields). In this way we see that (1), (3), (4), (5) are equivalent.

The equivalence of (2) and (6) follows from Lemma 0C4A.

Finally, we prove the equivalence of (2) \(=\) (6) with (1) \(=\) (3) \(=\) (4) \(=\) (5). First we note that the geometric number of branches of \(X\) at \(x\) and the geometric number of branches of \(X_{\overline{k}}\) at \(\overline{x}\) are equal by Varieties, Lemma 0C40. We conclude from the information available to us at this point that in all cases this number is equal to \(2\). On the other hand, in case (1) it is clear that \(X\) is geometrically reduced at \(x\), and hence \[\delta\text{-invariant of }X\text{ at }x \leq \delta\text{-invariant of }X_{\overline{k}}\text{ at }\overline{x}\] by Varieties, Lemma 0C3X. Since in case (1) the right hand side is \(1\), this forces the \(\delta\)-invariant of \(X\) at \(x\) to be \(1\) (because if it were zero, then \(\mathcal{O}_{X, x}\) would be a discrete valuation ring by Varieties, Lemma 0C3U which is unibranch, a contradiction). Thus (5) holds. Conversely, if (2) \(=\) (5) is true, then assumptions (a), (b), (c) of Varieties, Lemma 0C3P hold for \(x \in X\) by Lemma 0C4A. Thus Varieties, Lemma 0C3Y applies and shows that we have equality in the above displayed inequality. We conclude that (5) holds for \(\overline{x} \in X_{\overline{k}}\) and we are back in case (1) by the equivalence of the conditions over an algebraically closed field.

Remark

Let \(k\) be a field. Let \((A, \mathfrak m, \kappa)\) be a Noetherian local \(k\)-algebra. Assume that either \((A, \mathfrak m, \kappa)\) is as in Lemma 0C49, or \(A\) is Nagata as in Lemma 0C4A, or \(A\) is complete and as in Lemma 0C4C. Then \(A\) defines canonically a degree \(2\) separable \(\kappa\)-algebra \(\kappa'\) as follows

1. let \(q = ax^2 + bxy + cy^2\) be a nondegenerate quadric as in Lemma 0C49 with coordinates \(x, y\) chosen such that \(a \not = 0\) and set \(\kappa' = \kappa[x]/(ax^2 + bx + c)\),

2. let \(A' \supset A\) be the integral closure of \(A\) in its total ring of fractions and set \(\kappa' = A'/\mathfrak m A'\), or

3. let \(\kappa'\) be the \(\kappa\)-algebra such that \(\text{Proj}(\bigoplus_{n \geq 0} \mathfrak m^n/\mathfrak m^{n + 1}) = \Spec(\kappa')\).

The equivalence of (1) and (2) was shown in the proof of Lemma 0C4A. We omit the equivalence of this with (3). If \(X\) is a locally Noetherian \(k\)-scheme and \(x \in X\) is a point such that \(\mathcal{O}_{X, x} = A\), then (3) shows that \(\Spec(\kappa') = X^\nu \times_X \Spec(\kappa)\) where \(\nu : X^\nu \to X\) is the normalization morphism.

Remark

Let \(k\) be a field. Let \((A, \mathfrak m, \kappa)\) be as in Remark 0CBT and let \(\kappa'/\kappa\) be the associated separable algebra of degree \(2\). Then the following are equivalent

1. \(\kappa' \cong \kappa \times \kappa\) as \(\kappa\)-algebra,

2. the form \(q\) of Lemma 0C49 can be chosen to be \(xy\),

3. \(A\) has two branches,

4. the extension \(A'/A\) of Lemma 0C4A has two maximal ideals, and

5. \(A^\wedge \cong \kappa[[x, y]]/(xy)\) as a \(k\)-algebra.

The equivalence between these conditions has been shown in the proof of Lemma 0C4A. If \(X\) is a locally Noetherian \(k\)-scheme and \(x \in X\) is a point such that \(\mathcal{O}_{X, x} = A\), then this means exactly that there are two points \(x_1, x_2\) of the normalization \(X^\nu\) lying over \(x\) and that \(\kappa(x) = \kappa(x_1) = \kappa(x_2)\).

Definition

Let \(k\) be a field. Let \(X\) be a \(1\)-dimensional algebraic \(k\)-scheme. Let \(x \in X\) be a closed point. We say \(x\) is a split node if \(x\) is a node, \(\kappa(x) = k\), and the equivalent assertions of Remark 0CBU hold for \(A = \mathcal{O}_{X, x}\).

We formulate the obligatory lemma stating what we already know about this concept.

Lemma

Let \(k\) be a field. Let \(X\) be a \(1\)-dimensional algebraic \(k\)-scheme. Let \(x \in X\) be a closed point. The following are equivalent

1. \(x\) is a split node,

2. \(x\) is a node and there are exactly two points \(x_1, x_2\) of the normalization \(X^\nu\) lying over \(x\) with \(k = \kappa(x_1) = \kappa(x_2)\),

3. \(\mathcal{O}_{X, x}^\wedge \cong k[[x, y]]/(xy)\) as a \(k\)-algebra, and

4. add more here.

Proof

This follows from the discussion in Remark 0CBU and Lemma 0C4D.

Lemma

Let \(K/k\) be an extension of fields. Let \(X\) be a locally algebraic \(k\)-scheme of dimension \(1\). Let \(y \in X_K\) be a point with image \(x \in X\). The following are equivalent

1. \(x\) is a closed point of \(X\) and a node, and

2. \(y\) is a closed point of \(Y\) and a node.

Proof

If \(x\) is a closed point of \(X\), then \(y\) is too (look at residue fields). But conversely, this need not be the case, i.e., it can happen that a closed point of \(Y\) maps to a nonclosed point of \(X\). However, in this case \(y\) cannot be a node. Namely, then \(X\) would be geometrically unibranch at \(x\) (because \(x\) would be a generic point of \(X\) and \(\mathcal{O}_{X, x}\) would be Artinian and any Artinian local ring is geometrically unibranch), hence \(Y\) is geometrically unibranch at \(y\) (Varieties, Lemma 0C55), which means that \(y\) cannot be a node by Lemma 0C4D. Thus we may and do assume that both \(x\) and \(y\) are closed points.

Choose algebraic closures \(\overline{k}\), \(\overline{K}\) and a map \(\overline{k} \to \overline{K}\) extending the given map \(k \to K\). Using the equivalence of (1) and (3) in Lemma 0C4D we reduce to the case where \(k\) and \(K\) are algebraically closed. In this case we can argue as in the proof of Lemma 0C4D that the geometric number of branches and \(\delta\)-invariants of \(X\) at \(x\) and \(Y\) at \(y\) are the same. Another argument can be given by choosing an isomorphism \(k[[x_1, \ldots, x_n]]/(g_1, \ldots, g_m) \to \mathcal{O}_{X, x}^\wedge\) of \(k\)-algebras as in Varieties, Lemma 0C52. By Varieties, Lemma 0C54 this gives an isomorphism \(K[[x_1, \ldots, x_n]]/(g_1, \ldots, g_m) \to \mathcal{O}_{Y, y}^\wedge\) of \(K\)-algebras. By definition we have to show that \[k[[x_1, \ldots, x_n]]/(g_1, \ldots, g_m) \cong k[[s, t]]/(st)\] if and only if \[K[[x_1, \ldots, x_n]]/(g_1, \ldots, g_m) \cong K[[s, t]]/(st)\] We encourage the reader to prove this for themselves. Since \(k\) and \(K\) are algebraically closed fields, this is the same as asking these rings to be as in Lemma 0C49. Via Lemma 0C4C this translates into a statement about the \((n - 1) \times (n - 1)\)-minors of the matrix \((\partial g_j/\partial x_i)\) which is clearly independent of the field used. We omit the details.

Lemma

Let \(k\) be a field. Let \(X\) be a locally algebraic \(k\)-scheme of dimension \(1\). Let \(Y \to X\) be an étale morphism. Let \(y \in Y\) be a point with image \(x \in X\). The following are equivalent

1. \(x\) is a closed point of \(X\) and a node, and

2. \(y\) is a closed point of \(Y\) and a node.

Proof

By Lemma 0C56 we may base change to the algebraic closure of \(k\). Then the residue fields of \(x\) and \(y\) are \(k\). Hence the map \(\mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{Y, y}^\wedge\) is an isomorphism (for example by Étale Morphisms, Lemma 039M or More on Algebra, Lemma 0AGX). Thus the lemma is clear.

Lemma

Let \(k'/k\) be a finite separable field extension. Let \(X\) be a locally algebraic \(k'\)-scheme of dimension \(1\). Let \(x \in X\) be a closed point. The following are equivalent

1. \(x\) is a node, and

2. \(x\) is a node when \(X\) viewed as a locally algebraic \(k\)-scheme.

Proof

Follows immediately from the characterization of nodes in Lemma 0C4D.

Lemma

Let \(k\) be a field. Let \(X\) be a locally algebraic \(k\)-scheme equidimensional of dimension \(1\). The following are equivalent

1. the singularities of \(X\) are at-worst-nodal, and

2. \(X\) is a local complete intersection over \(k\) and the closed subscheme \(Z \subset X\) cut out by the first fitting ideal of \(\Omega_{X/k}\) is unramified over \(k\).

Proof

We urge the reader to find their own proof of this lemma; what follows is just putting together earlier results and may hide what is really going on.

Assume (2). Since \(Z \to \Spec(k)\) is quasi-finite (Morphisms, Lemma 02V5) we see that the residue fields of points \(x \in Z\) are finite over \(k\) (as well as separable) by Morphisms, Lemma 01TG. Hence each \(x \in Z\) is a closed point of \(X\) by Morphisms, Lemma 01TE. The local ring \(\mathcal{O}_{X, x}\) is Cohen-Macaulay by Algebra, Lemma 00SB. Since \(\dim(\mathcal{O}_{X, x}) = 1\) by dimension theory (Varieties, Section 06LF), we conclude that \(\text{depth}(\mathcal{O}_{X, x})) = 1\). Thus \(x\) is a node by Lemma 0C4D. If \(x \in X\), \(x \not \in Z\), then \(X \to \Spec(k)\) is smooth at \(x\) by Divisors, Lemma 0C3K.

Assume (1). Under this assumption \(X\) is geometrically reduced at every closed point (see Varieties, Lemma 0384). Hence \(X \to \Spec(k)\) is smooth on a dense open by Varieties, Lemma 056V. Thus \(Z\) is closed and consists of closed points. By Divisors, Lemma 0C3K the morphism \(X \setminus Z \to \Spec(k)\) is smooth. Hence \(X \setminus Z\) is a local complete intersection by Morphisms, Lemma 01VD and the definition of a local complete intersection in Morphisms, Definition 01UC. By Lemma 0C4D for every point \(x \in Z\) the local ring \(\mathcal{O}_{Z, x}\) is equal to \(\kappa(x)\) and \(\kappa(x)\) is separable over \(k\). Thus \(Z \to \Spec(k)\) is unramified (Morphisms, Lemma 02G7). Finally, Lemma 0C4D via part (3) of Lemma 0C49, shows that \(\mathcal{O}_{X, x}\) is a complete intersection in the sense of Divided Power Algebra, Definition 09Q3. However, Divided Power Algebra, Lemma 09Q6 and Morphisms, Lemma 01UG show that this agrees with the notion used to define a local complete intersection scheme over a field and the proof is complete.

Lemma

Let \(k\) be a field. Let \(X\) be a locally algebraic \(k\)-scheme equidimensional of dimension \(1\) whose singularities are at-worst-nodal. Then \(X\) is Gorenstein and geometrically reduced.

Proof

The Gorenstein assertion follows from Lemma 0C4E and Duality for Schemes, Lemma 0BVA. Or you can use that it suffices to check after passing to the algebraic closure (Duality for Schemes, Lemma 0C03), then use that a Noetherian local ring is Gorenstein if and only if its completion is so (by Dualizing Complexes, Lemma 0BJL), and then prove that the local rings \(k[[t]]\) and \(k[[x, y]]/(xy)\) are Gorenstein by hand.

To see that \(X\) is geometrically reduced, it suffices to show that \(X_{\overline{k}}\) is reduced (Varieties, Lemmas 020I and 035X). But \(X_{\overline{k}}\) is a nodal curve over an algebraically closed field. Thus the complete local rings of \(X_{\overline{k}}\) are isomorphic to either \(\overline{k}[[t]]\) or \(\overline{k}[[x, y]]/(xy)\) which are reduced as desired.

Lemma

Let \(k\) be a field. Let \(X\) be a locally algebraic \(k\)-scheme equidimensional of dimension \(1\) whose singularities are at-worst-nodal. If \(Y \subset X\) is a reduced closed subscheme equidimensional of dimension \(1\), then

1. the singularities of \(Y\) are at-worst-nodal, and

2. if \(Z \subset X\) is the scheme theoretic closure of \(X \setminus Y\), then

   1. the scheme theoretic intersection \(Y \cap Z\) is the disjoint union of spectra of finite separable extensions of \(k\),

   2. each point of \(Y \cap Z\) is a node of \(X\), and

   3. \(Y \to \Spec(k)\) is smooth at every point of \(Y \cap Z\).

Proof

Since \(X\) and \(Y\) are reduced and equidimensional of dimension \(1\), we see that \(Y\) is the scheme theoretic union of a subset of the irreducible components of \(X\) (in a reduced ring \((0)\) is the intersection of the minimal primes). Let \(y \in Y\) be a closed point. If \(y\) is in the smooth locus of \(X \to \Spec(k)\), then \(y\) is on a unique irreducible component of \(X\) and we see that \(Y\) and \(X\) agree in an open neighbourhood of \(y\). Hence \(Y \to \Spec(k)\) is smooth at \(y\). If \(y\) is a node of \(X\) but still lies on a unique irreducible component of \(X\), then \(y\) is a node on \(Y\) by the same argument. Suppose that \(y\) lies on more than \(1\) irreducible component of \(X\). Since the number of geometric branches of \(X\) at \(y\) is \(2\) by Lemma 0C4D, there can be at most \(2\) irreducible components passing through \(y\) by Properties, Lemma 0E20. If \(Y\) contains both of these, then again \(Y = X\) in an open neighbourhood of \(y\) and \(y\) is a node of \(Y\). Finally, assume \(Y\) contains only one of the irreducible components. After replacing \(X\) by an open neighbourhood of \(x\) we may assume \(Y\) is one of the two irreducble components and \(Z\) is the other. By Properties, Lemma 0E20 again we see that \(X\) has two branches at \(y\), i.e., the local ring \(\mathcal{O}_{X, y}\) has two branches and that these branches come from \(\mathcal{O}_{Y, y}\) and \(\mathcal{O}_{Z, y}\). Write \(\mathcal{O}_{X, y}^\wedge \cong \kappa(y)[[u, v]]/(uv)\) as in Remark 0CBU. The field \(\kappa(y)\) is finite separable over \(k\) by Lemma 0C4D for example. Thus, after possibly switching the roles of \(u\) and \(v\), the completion of the map \(\mathcal{O}_{X, y} \to \mathcal{O}_{Y, Y}\) corresponds to \(\kappa(y)[[u, v]]/(uv) \to \kappa(y)[[u]]\) and the completion of the map \(\mathcal{O}_{X, y} \to \mathcal{O}_{Y, Y}\) corresponds to \(\kappa(y)[[u, v]]/(uv) \to \kappa(y)[[v]]\). The scheme theoretic intersection of \(Y \cap Z\) is cut out by the sum of their ideas which in the completion is \((u, v)\), i.e., the maximal ideal. Thus (2)(a) and (2)(b) are clear. Finally, (2)(c) holds: the completion of \(\mathcal{O}_{Y, y}\) is regular, hence \(\mathcal{O}_{Y, y}\) is regular (More on Algebra, Lemma 07NY) and \(\kappa(y)/k\) is separable, hence smoothness in an open neighbourhood by Algebra, Lemma 00TV.

Families of nodal curves

In the Stacks project curves are irreducible varieties of dimension \(1\), but in the literature a “semi-stable curve” or a “nodal curve” is usually not irreducible and often assumed to be proper, especially when used in a phrase such as “family of semistable curves” or “family of nodal curves”, or “nodal family”. Thus it is a bit difficult for us to choose a terminology which is consistent with the literature as well as internally consistent. Moreover, we really want to first study the notion introduced in the following lemma (which is local on the source).

Lemma

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

1. \(f\) is flat, locally of finite presentation, every nonempty fibre \(X_s\) is equidimensional of dimension \(1\), and \(X_s\) has at-worst-nodal singularities, and

2. \(f\) is syntomic of relative dimension \(1\) and the closed subscheme \(\text{Sing}(f) \subset X\) defined by the first Fitting ideal of \(\Omega_{X/S}\) is unramified over \(S\).

Proof

Recall that the formation of \(\text{Sing}(f)\) commutes with base change, see Divisors, Lemma 0C3I. Thus the lemma follows from Lemma 0C4E, Morphisms, Lemma 01UF, and Morphisms, Lemma 02G8. (We also use the trivial Morphisms, Lemmas 01UK and 01UL.)

Definition

Let \(f : X \to S\) be a morphism of schemes. We say \(f\) is at-worst-nodal of relative dimension \(1\) if \(f\) satisfies the equivalent conditions of Lemma 0C59.

Here are some reasons for the cumbersome terminology⁶. First, we want to make sure this notion is not confused with any of the other notions in the literature (see introduction to this section). Second, we can imagine several generalizations of this notion to morphisms of higher relative dimension (for example, one can ask for morphisms which are étale locally compositions of at-worst-nodal morphisms or one can ask for morphisms whose fibres are higher dimensional but have at worst ordinary double points).

Lemma

A smooth morphism of relative dimension \(1\) is at-worst-nodal of relative dimension \(1\).

Proof

Omitted.

Lemma

Let \(f : X \to S\) be at-worst-nodal of relative dimension \(1\). Then the same is true for any base change of \(f\).

Proof

This is true because the base change of a syntomic morphism is syntomic (Morphisms, Lemma 01UI), the base change of a morphism of relative dimension \(1\) has relative dimension \(1\) (Morphisms, Lemma 02NK), the formation of \(\text{Sing}(f)\) commutes with base change (Divisors, Lemma 0C3I), and the base change of an unramified morphism is unramified (Morphisms, Lemma 02GA).

The following lemma tells us that we can check whether a morphism is at-worst-nodal of relative dimension \(1\) on the fibres.

Lemma

Let \(f : X \to S\) be a morphism of schemes which is flat and locally of finite presentation. Then there is a maximal open subscheme \(U \subset X\) such that \(f|_U : U \to S\) is at-worst-nodal of relative dimension \(1\). Moreover, formation of \(U\) commutes with arbitrary base change.

Proof

By Morphisms, Lemma 02V3 we find that there is such an open where \(f\) is syntomic. Hence we may assume that \(f\) is a syntomic morphism. In particular \(f\) is a Cohen-Macaulay morphism (Duality for Schemes, Lemmas 0C15 and 0C06). Thus \(X\) is a disjoint union of open and closed subschemes on which \(f\) has given relative dimension, see Morphisms, Lemma 02NM. This decomposition is preserved by arbitrary base change, see Morphisms, Lemma 02NK. Discarding all but one piece we may assume \(f\) is syntomic of relative dimension \(1\). Let \(\text{Sing}(f) \subset X\) be the closed subscheem defined by the first fitting ideal of \(\Omega_{X/S}\). There is a maximal open subscheme \(W \subset \text{Sing}(f)\) such that \(W \to S\) is unramified and its formation commutes with base change (Morphisms, Lemma 0475). Since also formation of \(\text{Sing}(f)\) commutes with base change (Divisors, Lemma 0C3I), we see that \[U = (X \setminus \text{Sing}(f)) \cup W\] is the maximal open subscheme of \(X\) such that \(f|_U : U \to S\) is at-worst-nodal of relative dimension \(1\) and that formation of \(U\) commutes with base change.

Lemma

Let \(f : X \to S\) be at-worst-nodal of relative dimension \(1\). If \(Y \to X\) is an étale morphism, then the composition \(g : Y \to S\) is at-worst-nodal of relative dimension \(1\).

Proof

Observe that \(g\) is flat and locally of finite presentation as a composition of morphisms which are flat and locally of finite presentation (use Morphisms, Lemmas 02GR, 02GS, 01TR, and 01U7). Thus it suffices to prove the fibres have at-worst-nodal singularities. This follows from Lemma 0C57 (and the fact that the composition of an étale morphism and a smooth morphism is smooth by Morphisms, Lemmas 02GK and 01VA).

Lemma

Let \(S' \to S\) be an étale morphism of schemes. Let \(f : X \to S'\) be at-worst-nodal of relative dimension \(1\). Then the composition \(g : X \to S\) is at-worst-nodal of relative dimension \(1\).

Proof

Observe that \(g\) is flat and locally of finite presentation as a composition of morphisms which are flat and locally of finite presentation (use Morphisms, Lemmas 02GR, 02GS, 01TR, and 01U7). Thus it suffices to prove the fibres of \(g\) have at-worst-nodal singularities. This follows from Lemma 0CD6 and the analogous result for smooth points.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(\{U_i \to X\}\) be an étale covering. The following are equivalent

1. \(f\) is at-worst-nodal of relative dimension \(1\),

2. each \(U_i \to S\) is at-worst-nodal of relative dimension \(1\).

In other words, being at-worst-nodal of relative dimension \(1\) is étale local on the source.

Proof

One direction we have seen in Lemma 0C5C. For the other direction, observe that being locally of finite presentation, flat, or to have relative dimension \(1\) is étale local on the source (Descent, Lemmas 036N, 036K, and 04NL). Taking fibres we reduce to the case where \(S\) is the spectrum of a field. In this case the result follows from Lemma 0C57 (and the fact that being smooth is étale local on the source by Descent, Lemma 036U).

Lemma

Let \(f : X \to S\) be a morphism of schemes. Let \(\{U_i \to S\}\) be an fpqc covering. The following are equivalent

1. \(f\) is at-worst-nodal of relative dimension \(1\),

2. each \(X \times_S U_i \to U_i\) is at-worst-nodal of relative dimension \(1\).

In other words, being at-worst-nodal of relative dimension \(1\) is fpqc local on the target.

Proof

One direction we have seen in Lemma 0C5B. For the other direction, observe that being locally of finite presentation, flat, or to have relative dimension \(1\) is fpqc local on the target (Descent, Lemmas 02KY, 02L2, and Morphisms, Lemma 02FY). Taking fibres we reduce to the case where \(S\) is the spectrum of a field. In this case the result follows from Lemma 0C56 (and the fact that being smooth is fpqc local on the target by Descent, Lemma 02VL).

Lemma

Let \(S = \lim S_i\) be a limit of a directed system of schemes with affine transition morphisms. Let \(0 \in I\) and let \(f_0 : X_0 \to Y_0\) be a morphism of schemes over \(S_0\). Assume \(S_0\), \(X_0\), \(Y_0\) are quasi-compact and quasi-separated. Let \(f_i : X_i \to Y_i\) be the base change of \(f_0\) to \(S_i\) and let \(f : X \to Y\) be the base change of \(f_0\) to \(S\). If

1. \(f\) is at-worst-nodal of relative dimension \(1\), and

2. \(f_0\) is locally of finite presentation,

then there exists an \(i \geq 0\) such that \(f_i\) is at-worst-nodal of relative dimension \(1\).

Proof

By Limits, Lemma 0C3L there exists an \(i\) such that \(f_i\) is syntomic. Then \(X_i = \coprod_{d \geq 0} X_{i, d}\) is a disjoint union of open and closed subschemes such that \(X_{i, d} \to Y_i\) has relative dimension \(d\), see Morphisms, Lemma 02K1. Because of the behaviour of dimensions of fibres under base change given in Morphisms, Lemma 02FY we see that \(X \to X_i\) maps into \(X_{i, 1}\). Then there exists an \(i' \geq i\) such that \(X_{i'} \to X_i\) maps into \(X_{i, 1}\), see Limits, Lemma 05F4. Thus \(f_{i'} : X_{i'} \to Y_{i'}\) is syntomic of relative dimension \(1\) (by Morphisms, Lemma 02FY again). Consider the morphism \(\text{Sing}(f_{i'}) \to Y_{i'}\). We know that the base change to \(Y\) is an unramified morphism. Hence by Limits, Lemma 0C4W we see that after increasing \(i'\) the morphism \(\text{Sing}(f_{i'}) \to Y_{i'}\) becomes unramified. This finishes the proof.

Lemma

Let \(f : T \to S\) be a morphism of schemes. Let \(t \in T\) with image \(s \in S\). Assume

1. \(f\) is flat at \(t\),

2. \(\mathcal{O}_{S, s}\) is Noetherian,

3. \(f\) is locally of finite type,

4. \(t\) is a split node of the fibre \(T_s\).

Then there exists an \(h \in \mathfrak m_s^\wedge\) and an isomorphism \[\mathcal{O}_{T, t}^\wedge \cong \mathcal{O}_{S, s}^\wedge[[x, y]]/(xy - h)\] of \(\mathcal{O}_{S, s}^\wedge\)-algebras.

Proof

We replace \(S\) by \(\Spec(\mathcal{O}_{S, s})\) and \(T\) by the base change to \(\Spec(\mathcal{O}_{S, s})\). Then \(T\) is locally Noetherian and hence \(\mathcal{O}_{T, t}\) is Noetherian. Set \(A = \mathcal{O}_{S, s}^\wedge\), \(\mathfrak m = \mathfrak m_A\), and \(B = \mathcal{O}_{T, t}^\wedge\). By More on Algebra, Lemma 0C4G we see that \(A \to B\) is flat. Since \(\mathcal{O}_{T, t}/\mathfrak m_s \mathcal{O}_{T, t} = \mathcal{O}_{T_s, t}\) we see that \(B/\mathfrak m B = \mathcal{O}_{T_s, t}^\wedge\). By assumption (4) and Lemma 0CBW we conclude there exist \(\overline{u}, \overline{v} \in B/\mathfrak m B\) such that the map \[(A/\mathfrak m)[[x, y]] \longrightarrow B/\mathfrak m B,\quad x \longmapsto \overline{u}, x \longmapsto \overline{v}\] is surjective with kernel \((xy)\).

Assume we have \(n \geq 1\) and \(u, v \in B\) mapping to \(\overline{u}, \overline{v}\) such that \[u v = h + \delta\] for some \(h \in A\) and \(\delta \in \mathfrak m^nB\). We claim that there exist \(u', v' \in B\) with \(u - u', v - v' \in \mathfrak m^n B\) such that \[u' v' = h' + \delta'\] for some \(h' \in A\) and \(\delta' \in \mathfrak m^{n + 1}B\). To see this, write \(\delta = \sum f_i b_i\) with \(f_i \in \mathfrak m^n\) and \(b_i \in B\). Then write \(b_i = a_i + u b_{i, 1} + v b_{i, 2} + \delta_i\) with \(a_i \in A\), \(b_{i, 1}, b_{i, 2} \in B\) and \(\delta_i \in \mathfrak m B\). This is possible because the residue field of \(B\) agrees with the residue field of \(A\) and the images of \(u\) and \(v\) in \(B/\mathfrak m B\) generate the maximal ideal. Then we set \[u' = u - \sum b_{i, 2}f_i,\quad v' = v - \sum b_{i, 1}f_i\] and we obtain \[u'v' = h + \delta - \sum (b_{i, 1}u + b_{i, 2}v)f_i + \sum c_{ij}f_if_j = h + \sum a_if_i + \sum f_i \delta_i + \sum c_{ij}f_if_j\] for some \(c_{i, j} \in B\). Thus we get a formula as above with \(h' = h + \sum a_if_i\) and \(\delta' = \sum f_i \delta_i + \sum c_{ij}f_if_j\).

Arguing by induction and starting with any lifts \(u_1, v_1 \in B\) of \(\overline{u}, \overline{v}\) the result of the previous paragraph shows that we find a sequence of elements \(u_n, v_n \in B\) and \(h_n \in A\) such that \(u_n - u_{n + 1} \in \mathfrak m^n B\), \(v_n - v_{n + 1} \in \mathfrak m^n B\), \(h_n - h_{n + 1} \in \mathfrak m^n\), and such that \(u_n v_n - h_n \in \mathfrak m^n B\). Since \(A\) and \(B\) are complete we can set \(u_\infty = \lim u_n\), \(v_\infty = \lim v_n\), and \(h_\infty = \lim h_n\), and then we obtain \(u_\infty v_\infty = h_\infty\) in \(B\). Thus we have an \(A\)-algebra map \[A[[x, y]]/(xy - h_\infty) \longrightarrow B\] sending \(x\) to \(u_\infty\) and \(v\) to \(v_\infty\). This is a map of flat \(A\)-algebras which is an isomorphism after dividing by \(\mathfrak m\). It is surjective modulo \(\mathfrak m\) and hence surjective by completeness and Algebra, Lemma 0315. Then we can apply Algebra, Lemma 00ME to conclude it is an isomorphism.

Lemma

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

1. \(f\) is proper,

2. \(f\) is at-worst-nodal of relative dimension \(1\), and

3. the geometric fibres of \(f\) are connected.

Then (a) \(f_*\mathcal{O}_X = \mathcal{O}_S\) and this holds after any base change, (b) \(R^1f_*\mathcal{O}_X\) is a finite locally free \(\mathcal{O}_S\)-module whose formation commutes with any base change, and (c) \(R^qf_*\mathcal{O}_X = 0\) for \(q \geq 2\).

Proof

Part (a) follows from Derived Categories of Schemes, Lemma 0E0L. By Derived Categories of Schemes, Lemma 0E62 locally on \(S\) we can write \(Rf_*\mathcal{O}_X = \mathcal{O}_S \oplus P\) where \(P\) is perfect of tor amplitude in \([1, \infty)\). Recall that formation of \(Rf_*\mathcal{O}_X\) commutes with arbitrary base change (Derived Categories of Schemes, Lemma 0B91). Thus for \(s \in S\) we have \[H^i(P \otimes_{\mathcal{O}_S}^\mathbf{L} \kappa(s)) = H^i(X_s, \mathcal{O}_{X_s}) \text{ for }i \geq 1\] This is zero unless \(i = 1\) since \(X_s\) is a \(1\)-dimensional Noetherian scheme, see Cohomology, Proposition 02UZ. Then \(P = H^1(P)[-1]\) and \(H^1(P)\) is finite locally free for example by More on Algebra, Lemma 0BCD. Since everything is compatible with base change we conclude.

Étale local structure of nodal families

Consider the morphism of schemes \[\Spec(\mathbf{Z}[u, v, a]/(uv - a)) \longrightarrow \Spec(\mathbf{Z}[a])\] The next lemma shows that this morphism is a model for the étale local structure of a nodal family of curves.

Lemma

Let \(f : X \to S\) be a morphism of schemes. Assume that \(f\) is at-worst-nodal of relative dimension \(1\). Let \(x \in X\) be a point which is a singular point of the fibre \(X_s\). Then there exists a commutative diagram of schemes \[\xymatrix{ X \ar[d] & U \ar[rr] \ar[l] \ar[rd] & & W \ar[r] \ar[ld] & \Spec(\mathbf{Z}[u, v, a]/(uv - a)) \ar[d] \\ S & & V \ar[ll] \ar[rr] & & \Spec(\mathbf{Z}[a]) }\] with \(X \leftarrow U\), \(S \leftarrow V\), and \(U \to W\) étale morphisms, and with the right hand square cartesian, such that there exists a point \(u \in U\) mapping to \(x\) in \(X\).

Proof

We first use absolute Noetherian approximation to reduce to the case of schemes of finite type over \(\mathbf{Z}\). The question is local on \(X\) and \(S\). Hence we may assume that \(X\) and \(S\) are affine. Then we can write \(S = \Spec(R)\) and write \(R\) as a filtered colimit \(R = \colim R_i\) of finite type \(\mathbf{Z}\)-algebras. Using Limits, Lemma 01ZM we can find an \(i\) and a morphism \(f_i : X_i \to \Spec(R_i)\) whose base change to \(S\) is \(f\). After increasing \(i\) we may assume that \(f_i\) is at-worst-nodal of relative dimension \(1\), see Lemma 0C5F. The image \(x_i \in X_i\) of \(x\) will be a singular point of its fibre, for example because the formation of \(\text{Sing}(f)\) commutes with base change (Divisors, Lemma 0C3I). If we can prove the lemma for \(f_i : X_i \to S_i\) and \(x_i\), then the lemma follows for \(f : X \to S\) by base change. Thus we reduce to the case studied in the next paragraph.

Assume \(S\) is of finite type over \(\mathbf{Z}\). Let \(s \in S\) be the image of \(x\). Recall that \(\kappa(x)\) is a finite separable extension of \(\kappa(s)\), for example because \(\text{Sing}(f) \to S\) is unramified or because \(x\) is a node of the fibre \(X_s\) and we can apply Lemma 0C4D. Furthermore, let \(\kappa'/\kappa(x)\) be the degree \(2\) separable algebra associated to \(\mathcal{O}_{X_s, x}\) in Remark 0CBT. By More on Morphisms, Lemma 02LF we can choose an étale neighbourhood \((V, v) \to (S, s)\) such that the extension \(\kappa(v)/\kappa(s)\) realizes either the extension \(\kappa(x)/\kappa(s)\) in case \(\kappa' \cong \kappa(x) \times \kappa(x)\) or the extension \(\kappa'/\kappa(s)\) if \(\kappa'\) is a field. After replacing \(X\) by \(X \times_S V\) and \(S\) by \(V\) we reduce to the situation described in the next paragraph.

Assume \(S\) is of finite type over \(\mathbf{Z}\) and \(x \in X_s\) is a split node, see Definition 0CBV. By Lemma 0CBX we see that there exists an \(\mathcal{O}_{S, s}\)-algebra isomorphism \[\mathcal{O}_{X, x}^\wedge \cong \mathcal{O}_{S, s}^\wedge[[s, t]]/(st - h)\] for some \(h \in \mathfrak m_s^\wedge \subset \mathcal{O}_{S, s}^\wedge\). In other words, if we consider the homomorphism \[\sigma : \mathbf{Z}[a] \longrightarrow \mathcal{O}_{S, s}^\wedge\] sending \(a\) to \(h\), then there exists an \(\mathcal{O}_{S, s}\)-algebra isomorphism \[\mathcal{O}_{X, x}^\wedge \longrightarrow \mathcal{O}_{Y_\sigma, y_\sigma}^\wedge\] where \[Y_\sigma = \Spec(\mathbf{Z}[u, v, t]/(uv - a)) \times_{\Spec(\mathbf{Z}[a]), \sigma} \Spec(\mathcal{O}_{S, s}^\wedge)\] and \(y_\sigma\) is the point of \(Y_\sigma\) lying over the closed point of \(\Spec(\mathcal{O}_{S, s}^\wedge)\) and having coordinates \(u, v\) equal to zero. Since \(\mathcal{O}_{S, s}\) is a G-ring by More on Algebra, Proposition 07PX we may apply More on Morphisms, Lemma 0GDX to conclude.

Proof

We only sketch this proof; it was contributed by Mohan Swaminathan and it mimicks an argument of [ACG, Proposition X.2.1]. Contrary to the previous proof, this proof uses the actual equations.

In exactly the same manner as in the first proof we reduce to the case where \(S\) is of finite type over \(\mathbf{Z}\) and \(x \in X_s\) is a split node, see Definition 0CBV. By Lemma 0CBW we see that there exists an \(\kappa(s)\)-algebra isomorphism \[\mathcal{O}_{X_s, x}^\wedge \cong \kappa(s)[[u, v]]/(uv)\] Note that the tangent space of \(X_s\) at \(x\) has dimension \(2\). Thus by Varieties, Lemma 0CBL after Zariski shrinking we may assume there exists a closed immersion \(X \to Y\) of schemes over \(S\) with \(Y \to S\) smooth of relative dimension \(2\). Since \(X\) is syntomic of relative dimension \(1\) (by definition), we see that \(X\) is an effective Cartier divisor in \(Y\) (follows from Divisors, Lemma 067T). Thus after shrinking \(Y\) and \(X\) we may assume \(Y = \Spec(B)\) is affine and that there exists a nonzerodivisor \(g \in B\) such that \(X = \Spec(B/gB)\).

Write \(S = \Spec(R)\) and denote \(\mathfrak p \subset R\) be the prime corresponding to \(s\). Let \(\mathfrak q \subset B\) be the prime ideal corresponding to \(x\) (viewed as a point of \(Y\)). We have maps \[B_\mathfrak q / \mathfrak p B_\mathfrak q \longrightarrow B_\mathfrak q / g B_\mathfrak q + \mathfrak p B_\mathfrak q \longrightarrow \kappa(s)[[u, v]]/(uv)\] where the second arrow becomes an isomorphism after completion. After replacing \(B\) by a suitable princpal localization, we may assume that there exist \(U, V \in B\) which map to \(u\) and \(v\) up to \((u, v)^2\). Then the completion of \(B_\mathfrak q/\mathfrak p B_\mathfrak q\) is \(\kappa(s)[[U, V]]\). Furthermore, after shrinking \(Y\) and after multiplying \(g\) by a unit we may assume that \(g\) maps to \(UV\) plus higher order terms in \(\kappa(s)[[U, V]]\). After shrinking \(Y\) we may further assume

1. \(\mathfrak q = \mathfrak pB + (U, V)\)

2. \(\Omega_{B/R}\) is free on \(\text{d}U\) and \(\text{d}V\)

Write \(\text{d}(g) = g_2 \text{d}U + g_1 \text{d}V\) with \(g_1, g_2 \in B\). Note that the ideal \(J = (g_1, g_2) \subset B\) is independent of the choice of the “coordinates” \(U\) and \(V\) (for fixed \(g\)). Looking modulo \(\mathfrak p B_\mathfrak q + (U, V)^3B_\mathfrak q\) we see that \(g_1\) and \(g_2\) are congruent to \(U\) and \(V\) modulo \(\mathfrak p B_\mathfrak q + (U, V)^2B_\mathfrak q\). Thus we may redo the argument above with \(U\) replaced by \(g_1\) and \(V\) replaced by \(g_2\). Then we find that \(J = (U, V)\) which implies that \(g_1, g_2 \in (U, V) = J\) (of course these are different from the \(g_1\) and \(g_2\) with our previous choice of coordinates).

Note that \(\Spec(B/J) \to S = \Spec(R)\) is étale at \(x\). Thus after replacing \(S\) by an étale neighbourhood and shrinking \(Y\) again, we may assume that \(R \to B \to B/J\) is an isomorphism. Denote \(r \in R\) the element whose image in \(B/J\) is the image of \(-g\). Then we can write \[g + r \equiv a U + b V \bmod (U, V)^2\] for some \(a, b \in R\). Taking derivatives using that \(\text{d}g \in J\Omega_{B/R}\) we find \(a = b = 0\). Thus we can write \[g + r = \alpha U^2 + \beta UV + \gamma V^2\] for some \(\alpha, \beta, \gamma \in B\). Then \(\beta\) maps to \(1\) in \(\kappa(x)\) and \(\alpha\) and \(\gamma\) map to \(0\) in \(\kappa(x)\). Thus we may assume \(\beta\) and \(\alpha + \beta + \gamma\) are invertible in \(B\). Consider the cover \(Y'\) of \(Y\) given by \[Y' = \Spec(B[t]/(t(1 - t) - \alpha \gamma / \beta^2))\] with point \(x'\) the unique point over \(x\) with \(t = 1\). Note that \(Y' \to Y\) is étale at \(x'\). Consider the elements \[T_1 = (\alpha + t \beta)U + ((1 - t)\beta + \gamma)V \quad\text{and}\quad T_2 = \frac{(\alpha + (1 - t) \beta)U + (t\beta + \gamma)V}{\alpha + \beta + \gamma}\] of \(\Gamma(Y', \mathcal{O}_{Y'})\). Then we see that \[g = T_1 T_2 - r\] Since \(T_1\) and \(T_2\) are also coordinates in the sense above, we see that the morphism \(Y' \to \mathbf{A}^2_S\) given by \(T_1\) and \(T_2\) is étale in a neighbourhood of \(x'\) and that the pullback \(X' \subset Y'\) of \(X\) is the inverse image of the closed subscheme of \(\mathbf{A}^2_S = \Spec(R[x_1, x_2])\) given by the equation \(x_1x_2 - r = 0\) (again after possibly shrinking \(Y'\)). We leave it to the reader to put everything together to obtain the desired diagram.

More vanishing results

Continuation of Section 0B5C.

Lemma

In Situation 0B5D assume \(X\) is integral and has genus \(g\). Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. Let \(Z \subset X\) be a \(0\)-dimensional closed subscheme with ideal sheaf \(\mathcal{I} \subset \mathcal{O}_X\). If \(H^1(X, \mathcal{I}\mathcal{L})\) is nonzero, then \[\deg(\mathcal{L}) \leq 2g - 2 + \deg(Z)\] with strict inequality unless \(\mathcal{I}\mathcal{L} \cong \omega_X\).

Proof

Any curve, e.g. \(X\), is Cohen-Macaulay. If \(H^1(X, \mathcal{I}\mathcal{L})\) is nonzero, then there is a nonzero map \(\mathcal{I}\mathcal{L} \to \omega_X\), see Lemma 0BS3. Since \(\mathcal{I}\mathcal{L}\) is torsion free, this map is injective. Since a field is Gorenstein and \(X\) is reduced, we find that the Gorenstein locus \(U \subset X\) of \(X\) is nonempty, see Duality for Schemes, Lemma 0BFQ. This lemma also tells us that \(\omega_X|_U\) is invertible. In this way we see we have a short exact sequence \[0 \to \mathcal{I}\mathcal{L} \to \omega_X \to \mathcal{Q} \to 0\] where the support of \(\mathcal{Q}\) is zero dimensional. Hence we have \[\begin{align*} 0 & \leq \dim \Gamma(X, \mathcal{Q})\\ & = \chi(\mathcal{Q}) \\ & = \chi(\omega_X) - \chi(\mathcal{I}\mathcal{L}) \\ & = \chi(\omega_X) - \deg(\mathcal{L}) - \chi(\mathcal{I}) \\ & = 2g - 2 - \deg(\mathcal{L}) + \deg(Z) \end{align*}\] by Lemmas 0BS5 and 0BS6, by (0BY8), and by Varieties, Lemmas 0AYT and 0AYV. We have also used that \(\deg(Z) = \dim_k \Gamma(Z, \mathcal{O}_Z) = \chi(\mathcal{O}_Z)\) and the short exact sequence \(0 \to \mathcal{I} \to \mathcal{O}_X \to \mathcal{O}_Z \to 0\). The lemma follows.

Lemma

In Situation 0B5D assume \(X\) is integral and has genus \(g\). Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. Let \(Z \subset X\) be a \(0\)-dimensional closed subscheme with ideal sheaf \(\mathcal{I} \subset \mathcal{O}_X\). If \(\deg(\mathcal{L}) > 2g - 2 + \deg(Z)\), then \(H^1(X, \mathcal{I}\mathcal{L}) = 0\) and one of the following possibilities occurs

1. \(H^0(X, \mathcal{I}\mathcal{L}) \not = 0\), or

2. \(g = 0\) and \(\deg(\mathcal{L}) = \deg(Z) - 1\).

In case (2) if \(Z = \emptyset\), then \(X \cong \mathbf{P}^1_k\) and \(\mathcal{L}\) corresponds to \(\mathcal{O}_{\mathbf{P}^1}(-1)\).

Proof

The vanishing of \(H^1(X, \mathcal{I}\mathcal{L})\) follows from Lemma 0E3A. If \(H^0(X, \mathcal{I}\mathcal{L}) = 0\), then \(\chi(\mathcal{I}\mathcal{L}) = 0\). From the short exact sequence \(0 \to \mathcal{I}\mathcal{L} \to \mathcal{L} \to \mathcal{O}_Z \to 0\) we conclude \(\deg(\mathcal{L}) = g - 1 + \deg(Z)\). Thus \(g - 1 + \deg(Z) > 2g - 2 + \deg(Z)\) which implies \(g = 0\) hence (2) holds. If \(Z = \emptyset\) in case (2), then \(\mathcal{L}^{-1}\) is an invertible sheaf of degree \(1\). This implies there is an isomorphism \(X \to \mathbf{P}^1_k\) and \(\mathcal{L}^{-1}\) is the pullback of \(\mathcal{O}_{\mathbf{P}^1}(1)\) by Lemma 0C6T.

Lemma

In Situation 0B5D assume \(X\) is integral and has genus \(g\). Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. If \(\deg(\mathcal{L}) \geq 2g\), then \(\mathcal{L}\) is globally generated.

Proof

Let \(Z \subset X\) be the closed subscheme cut out by the global sections of \(\mathcal{L}\). By Lemma 0E3B we see that \(Z \not = X\). Let \(\mathcal{I} \subset \mathcal{O}_X\) be the ideal sheaf cutting out \(Z\). Consider the short exact sequence \[0 \to \mathcal{I}\mathcal{L} \to \mathcal{L} \to \mathcal{O}_Z \to 0\] If \(Z \not = \emptyset\), then \(H^1(X, \mathcal{I}\mathcal{L})\) is nonzero as follows from the long exact sequence of cohomology. By Lemma 0BS3 this gives a nonzero and hence injective map \[\mathcal{I}\mathcal{L} \longrightarrow \omega_X\] In particular, we find an injective map \(H^0(X, \mathcal{L}) = H^0(X, \mathcal{I}\mathcal{L}) \to H^0(X, \omega_X)\). This is impossible as \[\dim_k H^0(X, \mathcal{L}) = \dim_k H^1(X, \mathcal{L}) + \deg(\mathcal{L}) + 1 - g \geq g + 1\] and \(\dim H^0(X, \omega_X) = g\) by (0BY8).

Lemma

In Situation 0B5D assume \(X\) is integral and has genus \(g\). Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. Let \(Z \subset X\) be a nonempty \(0\)-dimensional closed subscheme. If \(\deg(\mathcal{L}) \geq 2g - 1 + \deg(Z)\), then \(\mathcal{L}\) is globally generated and \(H^0(X, \mathcal{L}) \to H^0(X, \mathcal{L}|_Z)\) is surjective.

Proof

Global generation by Lemma 0E3C. If \(\mathcal{I} \subset \mathcal{O}_X\) is the ideal sheaf of \(Z\), then \(H^1(X, \mathcal{I}\mathcal{L}) = 0\) by Lemma 0E3A. Hence surjectivity.

Lemma

In Situation 0B5D, assume \(X\) is geometrically integral over \(k\) and has genus \(g\). Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. If \(\deg(\mathcal{L}) \geq 2g + 1\), then \(\mathcal{L}\) is very ample.

Proof

By Lemma 0E3C, \(\mathcal{L}\) is globally generated, and so it determines a morphism \(f : X \to \mathbf{P}^n_k\) where \(n = h^0(X,\mathcal{L}) - 1\). To show that \(\mathcal{L}\) is very ample means to show that \(f\) is a closed immersion. It suffices to check that the base change of \(f\) to an algebraic closure \(\overline{k}\) of \(k\) is a closed immersion (Descent, Lemma 02L6). So we may assume that \(k\) is algebraically closed; \(X\) remains integral, by assumption. Lemma 0E3D gives that for every \(0\)-dimensional closed subscheme \(Z\subset X\) of degree 2, the restriction map \(H^0(X, \mathcal{L}) \to H^0(X, \mathcal{L}|_Z)\) is surjective. By Varieties, Lemma 0E8T, \(\mathcal{L}\) is very ample.

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) which is reduced, connected, and of dimension \(1\). Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. Let \(Z \subset X\) be a \(0\)-dimensional closed subscheme with ideal sheaf \(\mathcal{I} \subset \mathcal{O}_X\). If \(H^1(X, \mathcal{I}\mathcal{L}) \not = 0\), then there exists a reduced connected closed subscheme \(Y \subset X\) of dimension \(1\) such that \[\deg(\mathcal{L}|_Y) \leq -2\chi(Y, \mathcal{O}_Y) + \deg(Z \cap Y)\] where \(Z \cap Y\) is the scheme theoretic intersection.

Proof

If \(H^1(X, \mathcal{I}\mathcal{L})\) is nonzero, then there is a nonzero map \(\varphi : \mathcal{I}\mathcal{L} \to \omega_X\), see Lemma 0BS3. Let \(Y \subset X\) be the union of the irreducible components \(C\) of \(X\) such that \(\varphi\) is nonzero in the generic point of \(C\). Then \(Y\) is a reduced closed subscheme. Let \(\mathcal{J} \subset \mathcal{O}_X\) be the ideal sheaf of \(Y\). Since \(\mathcal{J}\mathcal{I}\mathcal{L}\) has no embedded associated points (as a submodule of \(\mathcal{L}\)) and as \(\varphi\) is zero in the generic points of the support of \(\mathcal{J}\) (by choice of \(Y\) and as \(X\) is reduced), we find that \(\varphi\) factors as \[\mathcal{I}\mathcal{L} \to \mathcal{I}\mathcal{L}/\mathcal{J}\mathcal{I}\mathcal{L} \to \omega_X\] We can view \(\mathcal{I}\mathcal{L}/\mathcal{J}\mathcal{I}\mathcal{L}\) as the pushforward of a coherent sheaf on \(Y\) which by abuse of notation we indicate with the same symbol. Since \(\omega_Y = \SheafHom(\mathcal{O}_Y, \omega_X)\) by Lemma 0E33 we find a map \[\mathcal{I}\mathcal{L}/ \mathcal{J}\mathcal{I}\mathcal{L} \to \omega_Y\] of \(\mathcal{O}_Y\)-modules which is injective in the generic points of \(Y\). Let \(\mathcal{I}' \subset \mathcal{O}_Y\) be the ideal sheaf of \(Z \cap Y\). There is a map \(\mathcal{I}\mathcal{L}/\mathcal{J}\mathcal{I}\mathcal{L} \to \mathcal{I}'\mathcal{L}|_Y\) whose kernel is supported in closed points. Since \(\omega_Y\) is a Cohen-Macaulay module, the map above factors through an injective map \(\mathcal{I}'\mathcal{L}|_Y \to \omega_Y\). We see that we get an exact sequence \[0 \to \mathcal{I}'\mathcal{L}|_Y \to \omega_Y \to \mathcal{Q} \to 0\] of coherent sheaves on \(Y\) where \(\mathcal{Q}\) is supported in dimension \(0\) (this uses that \(\omega_Y\) is an invertible module in the generic points of \(Y\)). We conclude that \[0 \leq \dim \Gamma(Y, \mathcal{Q}) = \chi(\mathcal{Q}) = \chi(\omega_Y) - \chi(\mathcal{I}'\mathcal{L}) = -2\chi(\mathcal{O}_Y) - \deg(\mathcal{L}|_Y) + \deg(Z \cap Y)\] by Lemma 0BS5 and Varieties, Lemma 0AYT. If \(Y\) is connected, then this proves the lemma. If not, then we repeat the last part of the argument for one of the connected components of \(Y\).

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) which is reduced, connected, and of dimension \(1\). Let \(\mathcal{L}\) be an invertible \(\mathcal{O}_X\)-module. Assume that for every reduced connected closed subscheme \(Y \subset X\) of dimension \(1\) we have \[\deg(\mathcal{L}|_Y) \geq 2\dim_k H^1(Y, \mathcal{O}_Y)\] Then \(\mathcal{L}\) is globally generated.

Proof

By induction on the number of irreducible components of \(X\). If \(X\) is irreducible, then the lemma holds by Lemma 0E3C applied to \(X\) viewed as a scheme over the field \(k' = H^0(X, \mathcal{O}_X)\). Assume \(X\) is not irreducible. Before we continue, if \(k\) is finite, then we replace \(k\) by a purely transcendental extension \(K\). This is allowed by Varieties, Lemmas 0B57, 0B59, 035Z, and 038F, Cohomology of Schemes, Lemma 02KH, Lemma 0E32 and the elementary fact that \(K\) is geometrically integral over \(k\).

Assume that \(\mathcal{L}\) is not globally generated to get a contradiction. Then we may choose a coherent ideal sheaf \(\mathcal{I} \subset \mathcal{O}_X\) such that \(H^0(X, \mathcal{I}\mathcal{L}) = H^0(X, \mathcal{L})\) and such that \(\mathcal{O}_X/\mathcal{I}\) is nonzero with support of dimension \(0\). For example, take \(\mathcal{I}\) the ideal sheaf of any closed point in the common vanishing locus of the global sections of \(\mathcal{L}\). We consider the short exact sequence \[0 \to \mathcal{I}\mathcal{L} \to \mathcal{L} \to \mathcal{L}/\mathcal{I}\mathcal{L} \to 0\] Since the support of \(\mathcal{L}/\mathcal{I}\mathcal{L}\) has dimension \(0\) we see that \(\mathcal{L}/\mathcal{I}\mathcal{L}\) is generated by global sections (Varieties, Lemma 0AYT). From the short exact sequence, and the fact that \(H^0(X, \mathcal{I}\mathcal{L}) = H^0(X, \mathcal{L})\) we get an injection \(H^0(X, \mathcal{L}/\mathcal{I}\mathcal{L}) \to H^1(X, \mathcal{I}\mathcal{L})\).

Recall that the \(k\)-vector space \(H^1(X, \mathcal{I}\mathcal{L})\) is dual to \(\Hom(\mathcal{I}\mathcal{L}, \omega_X)\). Choose \(\varphi : \mathcal{I}\mathcal{L} \to \omega_X\). By Lemma 0E3E we have \(H^1(X, \mathcal{L}) = 0\). Hence \[\dim_k H^0(X, \mathcal{I}\mathcal{L}) = \dim_k H^0(X, \mathcal{L}) = \deg(\mathcal{L}) + \chi(\mathcal{O}_X) > \dim_k H^1(X, \mathcal{O}_X) = \dim_k H^0(X, \omega_X)\] We conclude that \(\varphi\) is not injective on global sections, in particular \(\varphi\) is not injective. For every generic point \(\eta \in X\) of an irreducible component of \(X\) denote \(V_\eta \subset \Hom(\mathcal{I}\mathcal{L}, \omega_X)\) the \(k\)-subvector space consisting of those \(\varphi\) which are zero at \(\eta\). Since every associated point of \(\mathcal{I}\mathcal{L}\) is a generic point of \(X\), the above shows that \(\Hom(\mathcal{I}\mathcal{L}, \omega_X) = \bigcup V_\eta\). As \(X\) has finitely many generic points and \(k\) is infinite, we conclude \(\Hom(\mathcal{I}\mathcal{L}, \omega_X) = V_\eta\) for some \(\eta\). Let \(\eta \in C \subset X\) be the corresponding irreducible component. Let \(Y \subset X\) be the union of the other irreducible components of \(X\). Then \(Y\) is a nonempty reduced closed subscheme not equal to \(X\). Let \(\mathcal{J} \subset \mathcal{O}_X\) be the ideal sheaf of \(Y\). Please keep in mind that the support of \(\mathcal{J}\) is \(C\).

Let \(\varphi : \mathcal{I}\mathcal{L} \to \omega_X\) be arbitrary. Since \(\mathcal{J}\mathcal{I}\mathcal{L}\) has no embedded associated points (as a submodule of \(\mathcal{L}\)) and as \(\varphi\) is zero in the generic point \(\eta\) of the support of \(\mathcal{J}\), we find that \(\varphi\) factors as \[\mathcal{I}\mathcal{L} \to \mathcal{I}\mathcal{L}/\mathcal{J}\mathcal{I}\mathcal{L} \to \omega_X\] We can view \(\mathcal{I}\mathcal{L}/\mathcal{J}\mathcal{I}\mathcal{L}\) as the pushforward of a coherent sheaf on \(Y\) which by abuse of notation we indicate with the same symbol. Since \(\omega_Y = \SheafHom(\mathcal{O}_Y, \omega_X)\) by Lemma 0E33 we find a factorization \[\mathcal{I}\mathcal{L} \to \mathcal{I}\mathcal{L}/ \mathcal{J}\mathcal{I}\mathcal{L} \xrightarrow{\varphi'} \omega_Y \to \omega_X\] of \(\varphi\). Let \(\mathcal{I}' \subset \mathcal{O}_Y\) be the image of \(\mathcal{I} \subset \mathcal{O}_X\). There is a surjective map \(\mathcal{I}\mathcal{L}/\mathcal{J}\mathcal{I}\mathcal{L} \to \mathcal{I}'\mathcal{L}|_Y\) whose kernel is supported in closed points. Since \(\omega_Y\) is a Cohen-Macaulay module on \(Y\), the map \(\varphi'\) factors through a map \(\varphi'' : \mathcal{I}'\mathcal{L}|_Y \to \omega_Y\). Thus we have commutative diagrams \[\vcenter{ \xymatrix{ 0 \ar[r] & \mathcal{I}\mathcal{L} \ar[r] \ar[d] & \mathcal{L} \ar[r] \ar[d] & \mathcal{L}/\mathcal{I}\mathcal{L} \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \mathcal{I}'\mathcal{L}|_Y \ar[r] & \mathcal{L}|_Y \ar[r] & \mathcal{L}|_Y/\mathcal{I}'\mathcal{L}|_Y \ar[r] & 0 } } \quad\text{and}\quad \vcenter{ \xymatrix{ \mathcal{I}\mathcal{L} \ar[r]_\varphi \ar[d] & \omega_X \\ \mathcal{I}'\mathcal{L}|_Y \ar[r]^{\varphi''} & \omega_Y \ar[u] } }\] Now we can finish the proof as follows: Since for every \(\varphi\) we have a \(\varphi''\) and since \(\omega_X \in \textit{Coh}(\mathcal{O}_X)\) represents the functor \(\mathcal{F} \mapsto \Hom_k(H^1(X, \mathcal{F}), k)\), we find that \(H^1(X, \mathcal{I}\mathcal{L}) \to H^1(Y, \mathcal{I}'\mathcal{L}|_Y)\) is injective. Since the boundary \(H^0(X, \mathcal{L}/\mathcal{I}\mathcal{L}) \to H^1(X, \mathcal{I}\mathcal{L})\) is injective, we conclude the composition \[H^0(X, \mathcal{L}/\mathcal{I}\mathcal{L}) \to H^0(X, \mathcal{L}|_Y/\mathcal{I}'\mathcal{L}|_Y) \to H^1(X, \mathcal{I}'\mathcal{L}|_Y)\] is injective. Since \(\mathcal{L}/\mathcal{I}\mathcal{L} \to \mathcal{L}|_Y/\mathcal{I}'\mathcal{L}|_Y\) is a surjective map of coherent modules whose supports have dimension \(0\), we see that the first map \(H^0(X, \mathcal{L}/\mathcal{I}\mathcal{L}) \to H^0(X, \mathcal{L}|_Y/\mathcal{I}'\mathcal{L}|_Y)\) is surjective (and hence bijective). But by induction we have that \(\mathcal{L}|_Y\) is globally generated (if \(Y\) is disconnected this still works of course) and hence the boundary map \[H^0(X, \mathcal{L}|_Y/\mathcal{I}'\mathcal{L}|_Y) \to H^1(X, \mathcal{I}'\mathcal{L}|_Y)\] cannot be injective. This contradiction finishes the proof.

Contracting rational tails

In this section we discuss the simplest possible case of contracting a scheme to improve positivity properties of its canonical sheaf.

Example

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) having dimension \(1\) and \(H^0(X, \mathcal{O}_X) = k\). Assume the singularities of \(X\) are at-worst-nodal. A rational tail will be an irreducible component \(C \subset X\) (viewed as an integral closed subscheme) with the following properties

1. \(X' \not = \emptyset\) where \(X' \subset X\) is the scheme theoretic closure of \(X \setminus C\),

2. the scheme theoretic intersection \(C \cap X'\) is a single reduced point \(x\),

3. \(H^0(C, \mathcal{O}_C)\) maps isomorphically to the residue field of \(x\), and

4. \(C\) has genus zero.

Since there are at least two irreducible components of \(X\) passing through \(x\), we conclude that \(x\) is a node. Set \(k' = H^0(C, \mathcal{O}_C) = \kappa(x)\). Then \(k'/k\) is a finite separable extension of fields (Lemma 0C4D). There is a canonical morphism \[c : X \longrightarrow X'\] inducing the identity on \(X'\) and mapping \(C\) to \(x \in X'\) via the canonical morphism \(C \to \Spec(k') = x\). This follows from Morphisms, Lemma 0C4J since \(X\) is the scheme theoretic union of \(C\) and \(X'\) (as \(X\) is reduced). Moreover, we claim that \[c_*\mathcal{O}_X = \mathcal{O}_{X'} \quad\text{and}\quad R^1c_*\mathcal{O}_X = 0\] To see this, denote \(i_C : C \to X\), \(i_{X'} : X' \to X\) and \(i_x : x \to X\) the embeddings and use the exact sequence \[0 \to \mathcal{O}_X \to i_{C, *}\mathcal{O}_C \oplus i_{X', *}\mathcal{O}_{X'} \to i_{x, *}\kappa(x) \to 0\] of Morphisms, Lemma 0C4J. Looking at the long exact sequence of higher direct images, it follows that it suffices to show \(H^0(C, \mathcal{O}_C) = k'\) and \(H^1(C, \mathcal{O}_C) = 0\) which follows from the assumptions. Observe that \(X'\) is also a proper scheme over \(k\), of dimension \(1\) whose singularities are at-worst-nodal (Lemma 0E38) has \(H^0(X', \mathcal{O}_{X'}) = k\), and \(X'\) has the same genus as \(X\). We will say \(c : X \to X'\) is the contraction of a rational tail.

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) having dimension \(1\) and \(H^0(X, \mathcal{O}_X) = k\). Assume the singularities of \(X\) are at-worst-nodal. Let \(C \subset X\) be a rational tail (Example 0E3H). Then \(\deg(\omega_X|_C) < 0\).

Proof

Let \(X' \subset X\) be as in the example. Then we have a short exact sequence \[0 \to \omega_C \to \omega_X|_C \to \mathcal{O}_{C \cap X'} \to 0\] See Lemmas 0E34, 0E37, and 0E38. With \(k'\) as in the example we see that \(\deg(\omega_C) = -2[k' : k]\) as \(C \cong \mathbf{P}^1_{k'}\) by Proposition 0C6U and \(\deg(C \cap X') = [k' : k]\). Hence \(\deg(\omega_X|_C) = -[k' : k]\) which is negative.

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) having dimension \(1\) and \(H^0(X, \mathcal{O}_X) = k\). Assume the singularities of \(X\) are at-worst-nodal. Let \(C \subset X\) be a rational tail (Example 0E3H). For any field extension \(K/k\) the base change \(C_K \subset X_K\) is a finite disjoint union of rational tails.

Proof

Let \(x \in C\) and \(k' = \kappa(x)\) be as in the example. Observe that \(C \cong \mathbf{P}^1_{k'}\) by Proposition 0C6U. Since \(k'/k\) is finite separable, we see that \(k' \otimes_k K = K'_1 \times \ldots \times K'_n\) is a finite product of finite separable extensions \(K'_i/K\). Set \(C_i = \mathbf{P}^1_{K'_i}\) and denote \(x_i \in C_i\) the inverse image of \(x\). Then \(C_K = \coprod C_i\) and \(X'_K \cap C_i = x_i\) as desired.

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) having dimension \(1\) and \(H^0(X, \mathcal{O}_X) = k\). Assume the singularities of \(X\) are at-worst-nodal. If \(X\) does not have a rational tail (Example 0E3H), then for every reduced connected closed subscheme \(Y \subset X\), \(Y \not = X\) of dimension \(1\) we have \(\deg(\omega_X|_Y) \geq \dim_k H^1(Y, \mathcal{O}_Y)\).

Proof

Let \(Y \subset X\) be as in the statement. Then \(k' = H^0(Y, \mathcal{O}_Y)\) is a field and a finite extension of \(k\) and \([k' : k]\) divides all numerical invariants below associated to \(Y\) and coherent sheaves on \(Y\), see Varieties, Lemma 0C6P. Let \(Z \subset X\) be as in Lemma 0E34. We will use the results of this lemma and of Lemmas 0E37 and 0E38 without further mention. Then we get a short exact sequence \[0 \to \omega_Y \to \omega_X|_Y \to \mathcal{O}_{Y \cap Z} \to 0\] See Lemma 0E34. We conclude that \[\deg(\omega_X|_Y) = \deg(Y \cap Z) + \deg(\omega_Y) = \deg(Y \cap Z) - 2\chi(Y, \mathcal{O}_Y)\] Hence, if the lemma is false, then \[2[k' : k] > \deg(Y \cap Z) + \dim_k H^1(Y, \mathcal{O}_Y)\] Since \(Y \cap Z\) is nonempty and by the divisiblity mentioned above, this can happen only if \(Y \cap Z\) is a single \(k'\)-rational point of the smooth locus of \(Y\) and \(H^1(Y, \mathcal{O}_Y) = 0\). If \(Y\) is irreducible, then this implies \(Y\) is a rational tail. If \(Y\) is reducible, then since \(\deg(\omega_X|_Y) = -[k' : k]\) we find there is some irreducible component \(C\) of \(Y\) such that \(\deg(\omega_X|_C) < 0\), see Varieties, Lemma 0AYW. Then the analysis above applied to \(C\) gives that \(C\) is a rational tail.

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) having dimension \(1\) and \(H^0(X, \mathcal{O}_X) = k\). Assume the singularities of \(X\) are at-worst-nodal. Assume \(X\) does not have a rational tail (Example 0E3H). If

1. the genus of \(X\) is \(0\), then \(X\) is isomorphic to an irreducible plane conic and \(\omega_X^{\otimes -1}\) is very ample,

2. the genus of \(X\) is \(1\), then \(\omega_X \cong \mathcal{O}_X\),

3. the genus of \(X\) is \(\geq 2\), then \(\omega_X^{\otimes m}\) is globally generated for \(m \geq 2\).

Proof

By Lemma 0E37 we find that \(X\) is Gorenstein, i.e., \(\omega_X\) is an invertible \(\mathcal{O}_X\)-module.

If the genus of \(X\) is zero, then \(\deg(\omega_X) < 0\), hence if \(X\) has more than one irreducible component, we get a contradiction with Lemma 0E3J. In the irreducible case we see that \(X\) is isomorphic to an irreducible plane conic and \(\omega_X^{\otimes -1}\) is very ample by Lemma 0C6N.

If the genus of \(X\) is \(1\), then \(\omega_X\) has a global section and \(\deg(\omega_X|_C) = 0\) for all irreducible components. Namely, \(\deg(\omega_X|_C) \geq 0\) for all irreducible components \(C\) by Lemma 0E3J, the sum of these numbers is \(0\) by Lemma 0C19, and we can apply Varieties, Lemma 0AYW. Then \(\omega_X \cong \mathcal{O}_X\) by Varieties, Lemma 0E22.

Assume the genus \(g\) of \(X\) is greater than or equal to \(2\). If \(X\) is irreducible, then we are done by Lemma 0E3C. Assume \(X\) reducible. By Lemma 0E3J the inequalities of Lemma 0E3F hold for every \(Y \subset X\) as in the statement, except for \(Y = X\). Analyzing the proof of Lemma 0E3F we see that (in the reducible case) the only inequality used for \(Y = X\) are \[\deg(\omega_X^{\otimes m}) > -2 \chi(\mathcal{O}_X) \quad\text{and}\quad \deg(\omega_X^{\otimes m}) + \chi(\mathcal{O}_X) > \dim_k H^1(X, \mathcal{O}_X)\] Since these both hold under the assumption \(g \geq 2\) and \(m \geq 2\) we win.

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) of dimension \(1\) with \(H^0(X, \mathcal{O}_X) = k\). Assume the singularities of \(X\) are at-worst-nodal. Consider a sequence \[X = X_0 \to X_1 \to \ldots \to X_n = X'\] of contractions of rational tails (Example 0E3H) until none are left. Then

1. if the genus of \(X\) is \(0\), then \(X'\) is an irreducible plane conic,

2. if the genus of \(X\) is \(1\), then \(\omega_{X'} \cong \mathcal{O}_X\),

3. if the genus of \(X\) is \(> 1\), then \(\omega_{X'}^{\otimes m}\) is globally generated for \(m \geq 2\).

If the genus of \(X\) is \(\geq 1\), then the morphism \(X \to X'\) is independent of choices and formation of this morphism commutes with base field extensions.

Proof

We proceed by contracting rational tails until there are none left. Then we see that (1), (2), (3) hold by Lemma 0E3K.

Uniqueness. To see that \(f : X \to X'\) is independent of the choices made, it suffices to show: any rational tail \(C \subset X\) is mapped to a point by \(X \to X'\); some details omitted. If not, then we can find a section \(s \in \Gamma(X', \omega_{X'}^{\otimes 2})\) which does not vanish in the generic point of the irreducible component \(f(C)\). Since in each of the contractions \(X_i \to X_{i + 1}\) we have a section \(X_{i + 1} \to X_i\), there is a section \(X' \to X\) of \(f\). Then we have an exact sequence \[0 \to \omega_{X'} \to \omega_X \to \omega_X|_{X''} \to 0\] where \(X'' \subset X\) is the union of the irreducible components contracted by \(f\). See Lemma 0E34. Thus we get a map \(\omega_{X'}^{\otimes 2} \to \omega_X^{\otimes 2}\) and we can take the image of \(s\) to get a section of \(\omega_X^{\otimes 2}\) not vanishing in the generic point of \(C\). This is a contradiction with the fact that the restriction of \(\omega_X\) to a rational tail has negative degree (Lemma 0E63).

The statement on base field extensions follows from Lemma 0E3I. Some details omitted.

Contracting rational bridges

In this section we discuss the next simplest possible case (after the case discussed in Section 0E3G) of contracting a scheme to improve positivity properties of its canonical sheaf.

Example

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) having dimension \(1\) and \(H^0(X, \mathcal{O}_X) = k\). Assume the singularities of \(X\) are at-worst-nodal. A rational bridge will be an irreducible component \(C \subset X\) (viewed as an integral closed subscheme) with the following properties

1. \(X' \not = \emptyset\) where \(X' \subset X\) is the scheme theoretic closure of \(X \setminus C\),

2. the scheme theoretic intersection \(C \cap X'\) has degree \(2\) over \(H^0(C, \mathcal{O}_C)\), and

3. \(C\) has genus zero.

Set \(k' = H^0(C, \mathcal{O}_C)\) and \(k'' = H^0(C \cap X', \mathcal{O}_{C \cap X'})\). Then \(k'\) is a field (Varieties, Lemma 0BUG) and \(\dim_{k'}(k'') = 2\). Since there are at least two irreducible components of \(X\) passing through each point of \(C \cap X'\), we conclude these points are nodes of \(X\) and smooth points on both \(C\) and \(X'\) (Lemma 0E38). Hence \(k'/k\) is a finite separable extension of fields and \(k''/k'\) is either a degree \(2\) separable extension of fields or \(k'' = k' \times k'\) (Lemma 0C4D). By Section 0E35 there exists a pushout \[\xymatrix{ C \cap X' \ar[r] \ar[d] & X' \ar[d]^a \\ \Spec(k') \ar[r] & Y }\] with many good properties (all of which we will use below without further mention). Let \(y \in Y\) be the image of \(\Spec(k') \to Y\). Then \[\mathcal{O}_{Y, y}^\wedge \cong k'[[s, t]]/(st) \quad\text{or}\quad \mathcal{O}_{Y, y}^\wedge \cong \{f \in k''[[s]] : f(0) \in k'\}\] depending on whether \(C \cap X'\) has \(2\) or \(1\) points. This follows from Lemma 0E36 and the fact that \(\mathcal{O}_{X', p} \cong \kappa(p)[[t]]\) for \(p \in C \cap X'\) by More on Algebra, Lemma 0C35. Thus we see that \(y \in Y\) is a node, see Lemmas 0C4D and 0C4A and in particular the discussion of Case II in the proof of (2) \(\Rightarrow\) (1) in Lemma 0C4A. Thus the singularities of \(Y\) are at-worst-nodal.

We can extend the commutative diagram above to a diagram \[\xymatrix{ C \cap X' \ar[r] \ar[d] & X' \ar[d]^a \ar[r] & X \ar[ld]^c & C \ar[ld] \ar[l] \\ \Spec(k') \ar[r] & Y & \Spec(k') \ar[l] }\] where the two lower horizontal arrows are the same. Namely, \(X\) is the scheme theoretic union of \(X'\) and \(C\) (thus a pushout by Morphisms, Lemma 0C4J) and the morphisms \(C \to Y\) and \(X' \to Y\) agree on \(C \cap X'\). Finally, we claim that \[c_*\mathcal{O}_X = \mathcal{O}_Y \quad\text{and}\quad R^1c_*\mathcal{O}_X = 0\] To see this use the exact sequence \[0 \to \mathcal{O}_X \to \mathcal{O}_C \oplus \mathcal{O}_{X'} \to \mathcal{O}_{C \cap X'} \to 0\] of Morphisms, Lemma 0C4J. The long exact sequence of higher direct images is \[0 \to c_*\mathcal{O}_X \to c_*\mathcal{O}_C \oplus c_*\mathcal{O}_{X'} \to c_*\mathcal{O}_{C \cap X'} \to R^1c_*\mathcal{O}_X \to R^1c_*\mathcal{O}_C \oplus R^1c_*\mathcal{O}_{X'}\] Since \(c|_{X'} = a\) is affine we see that \(R^1c_*\mathcal{O}_{X'} = 0\). Since \(c|_C\) factors as \(C \to \Spec(k') \to X\) and since \(C\) has genus zero, we find that \(R^1c_*\mathcal{O}_C = 0\). Since \(\mathcal{O}_{X'} \to \mathcal{O}_{C \cap X'}\) is surjective and since \(c|_{X'}\) is affine, we see that \(c_*\mathcal{O}_{X'} \to c_*\mathcal{O}_{C \cap X'}\) is surjective. This proves that \(R^1c_*\mathcal{O}_X = 0\). Finally, we have \(\mathcal{O}_Y = c_*\mathcal{O}_X\) by the exact sequence and the description of the structure sheaf of the pushout in More on Morphisms, Proposition 0E25.

All of this means that \(Y\) is also a proper scheme over \(k\) having dimension \(1\) and \(H^0(Y, \mathcal{O}_Y) = k\) whose singularities are at-worst-nodal (Lemma 0E38) and that \(Y\) has the same genus as \(X\). We will say \(c : X \to Y\) is the contraction of a rational bridge.

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) having dimension \(1\) and \(H^0(X, \mathcal{O}_X) = k\). Assume the singularities of \(X\) are at-worst-nodal. Let \(C \subset X\) be a rational bridge (Example 0E3M). Then \(\deg(\omega_X|_C) = 0\).

Proof

Let \(X' \subset X\) be as in the example. Then we have a short exact sequence \[0 \to \omega_C \to \omega_X|_C \to \mathcal{O}_{C \cap X'} \to 0\] See Lemmas 0E34, 0E37, and 0E38. With \(k''/k'/k\) as in the example we see that \(\deg(\omega_C) = -2[k' : k]\) as \(C\) has genus \(0\) (Lemma 0BS6) and \(\deg(C \cap X') = [k'' : k] = 2[k' : k]\). Hence \(\deg(\omega_X|_C) = 0\).

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) having dimension \(1\) and \(H^0(X, \mathcal{O}_X) = k\). Assume the singularities of \(X\) are at-worst-nodal. Let \(C \subset X\) be a rational bridge (Example 0E3M). For any field extension \(K/k\) the base change \(C_K \subset X_K\) is a finite disjoint union of rational bridges.

Proof

Let \(k''/k'/k\) be as in the example. Since \(k'/k\) is finite separable, we see that \(k' \otimes_k K = K'_1 \times \ldots \times K'_n\) is a finite product of finite separable extensions \(K'_i/K\). The corresponding product decomposition \(k'' \otimes_k K = \prod K''_i\) gives degree \(2\) separable algebra extensions \(K''_i/K'_i\). Set \(C_i = C_{K'_i}\). Then \(C_K = \coprod C_i\) and therefore each \(C_i\) has genus \(0\) (viewed as a curve over \(K'_i\)), because \(H^1(C_K, \mathcal{O}_{C_K}) = 0\) by flat base change. Finally, we have \(X'_K \cap C_i = \Spec(K''_i)\) has degree \(2\) over \(K'_i\) as desired.

Lemma

Let \(c : X \to Y\) be the contraction of a rational bridge (Example 0E3M). Then \(c^*\omega_Y \cong \omega_X\).

Proof

You can prove this by direct computation, but we prefer to use the characterization of \(\omega_X\) as the coherent \(\mathcal{O}_X\)-module which represents the functor \(\textit{Coh}(\mathcal{O}_X) \to \textit{Sets}\), \(\mathcal{F} \mapsto \Hom_k(H^1(X, \mathcal{F}), k) = H^1(X, \mathcal{F})^\vee\), see Lemma 0BS3 or Duality for Schemes, Lemma 0AWP.

To be precise, denote \(\mathcal{C}_Y\) the category whose objects are invertible \(\mathcal{O}_Y\)-modules and whose maps are \(\mathcal{O}_Y\)-module homomorphisms. Denote \(\mathcal{C}_X\) the category whose objects are invertible \(\mathcal{O}_X\)-modules \(\mathcal{L}\) with \(\mathcal{L}|_C \cong \mathcal{O}_C\) and whose maps are \(\mathcal{O}_Y\)-module homomorphisms. We claim that the functor \[c^* : \mathcal{C}_Y \to \mathcal{C}_X\] is an equivalence of categories. Namely, by More on Morphisms, Lemma 0E24 it is essentially surjective. Then the projection formula (Cohomology, Lemma 01E8) shows \(c_*c^*\mathcal{N} = \mathcal{N}\) and hence \(c^*\) is an equivalence with quasi-inverse given by \(c_*\).

We claim \(\omega_X\) is an object of \(\mathcal{C}_X\). Namely, we have a short exact sequence \[0 \to \omega_C \to \omega_X|_C \to \mathcal{O}_{C \cap X'} \to 0\] See Lemma 0E34. Taking degrees we find \(\deg(\omega_X|_C) = 0\) (small detail omitted). Thus \(\omega_X|_C\) is trivial by Lemma 0C6M and \(\omega_X\) is an object of \(\mathcal{C}_X\).

Since \(R^1c_*\mathcal{O}_X = 0\) the projection formula shows that \(R^1c_*c^*\mathcal{N} = 0\) for \(\mathcal{N} \in \Ob(\mathcal{C}_Y)\). Therefore the Leray spectral sequence (Cohomology, Lemma 01F4) the diagram \[\xymatrix{ \mathcal{C}_Y \ar[rr]_{c^*} \ar[dr]_{H^1(Y, -)^\vee} & & \mathcal{C}_X \ar[ld]^{H^1(X, -)^\vee} \\ & \textit{Sets} }\] of categories and functors is commutative. Since \(\omega_Y \in \Ob(\mathcal{C}_Y)\) represents the south-east arrow and \(\omega_X \in \Ob(\mathcal{C}_X)\) represents the south-east arrow we conclude by the Yoneda lemma (Categories, Lemma 001P).

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) having dimension \(1\) and \(H^0(X, \mathcal{O}_X) = k\). Assume

1. the singularities of \(X\) are at-worst-nodal,

2. \(X\) does not have a rational tail (Example 0E3H),

3. \(X\) does not have a rational bridge (Example 0E3M),

4. the genus \(g\) of \(X\) is \(\geq 2\).

Then \(\omega_X\) is ample.

Proof

It suffices to show that \(\deg(\omega_X|_C) > 0\) for every irreducible component \(C\) of \(X\), see Varieties, Lemma 0B5Y. If \(X = C\) is irreducible, this follows from \(g \geq 2\) and Lemma 0C19. Otherwise, set \(k' = H^0(C, \mathcal{O}_C)\). This is a field and a finite extension of \(k\) and \([k' : k]\) divides all numerical invariants below associated to \(C\) and coherent sheaves on \(C\), see Varieties, Lemma 0C6P. Let \(X' \subset X\) be the closure of \(X \setminus C\) as in Lemma 0E34. We will use the results of this lemma and of Lemmas 0E37 and 0E38 without further mention. Then we get a short exact sequence \[0 \to \omega_C \to \omega_X|_C \to \mathcal{O}_{C \cap X'} \to 0\] See Lemma 0E34. We conclude that \[\deg(\omega_X|_C) = \deg(C \cap X') + \deg(\omega_C) = \deg(C \cap X') - 2\chi(C, \mathcal{O}_C)\] Hence, if the lemma is false, then \[2[k' : k] \geq \deg(C \cap X') + 2\dim_k H^1(C, \mathcal{O}_C)\] Since \(C \cap X'\) is nonempty and by the divisiblity mentioned above, this can happen only if either

1. \(C \cap X'\) is a single \(k'\)-rational point of \(C\) and \(H^1(C, \mathcal{O}_C) = 0\), and

2. \(C \cap X'\) has degree \(2\) over \(k'\) and \(H^1(C, \mathcal{O}_C) = 0\).

The first possibility means \(C\) is a rational tail and the second that \(C\) is a rational bridge. Since both are excluded the proof is complete.

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) of dimension \(1\) with \(H^0(X, \mathcal{O}_X) = k\) having genus \(g \geq 2\). Assume the singularities of \(X\) are at-worst-nodal and that \(X\) has no rational tails. Consider a sequence \[X = X_0 \to X_1 \to \ldots \to X_n = X'\] of contractions of rational bridges (Example 0E3M) until none are left. Then \(\omega_{X'}\) ample. The morphism \(X \to X'\) is independent of choices and formation of this morphism commutes with base field extensions.

Proof

We proceed by contracting rational bridges until there are none left. Then \(\omega_{X'}\) is ample by Lemma 0E3P.

Denote \(f : X \to X'\) the composition. By Lemma 0E3N and induction we see that \(f^*\omega_{X'} = \omega_X\). We have \(f_*\mathcal{O}_X = \mathcal{O}_{X'}\) because this is true for contraction of a rational bridge. Thus the projection formula says that \(f_*f^*\mathcal{L} = \mathcal{L}\) for all invertible \(\mathcal{O}_{X'}\)-modules \(\mathcal{L}\). Hence \[\Gamma(X', \omega_{X'}^{\otimes m}) = \Gamma(X, \omega_X^{\otimes m})\] for all \(m\). Since \(X'\) is the Proj of the direct sum of these by Morphisms, Lemma 0C6J we conclude that the morphism \(X \to X'\) is completely canonical.

Let \(K/k\) be an extension of fields, then \(\omega_{X_K}\) is the pullback of \(\omega_X\) (Lemma 0E32) and we have \(\Gamma(X, \omega_X^{\otimes m}) \otimes_k K\) is equal to \(\Gamma(X_K, \omega_{X_K}^{\otimes m})\) by Cohomology of Schemes, Lemma 02KH. Thus formation of \(f : X \to X'\) commutes with base change by \(K/k\) by the arguments given above. Some details omitted.

Contracting to a stable curve

In this section we combine the contraction morphisms found in Sections 0E3G and 0E7M. Namely, suppose that \(k\) is a field and let \(X\) be a proper scheme over \(k\) of dimension \(1\) with \(H^0(X, \mathcal{O}_X) = k\) having genus \(g \geq 2\). Assume the singularities of \(X\) are at-worst-nodal. Composing the morphism of Lemma 0E3L with the morphism of Lemma 0E3Q we get a morphism \[c : X \longrightarrow Y\] such that \(Y\) also is a proper scheme over \(k\) of dimension \(1\) whose singularities are at worst nodal, with \(k = H^0(Y, \mathcal{O}_Y)\) and having genus \(g\), such that \(\mathcal{O}_Y = c_*\mathcal{O}_X\) and \(R^1c_*\mathcal{O}_X = 0\), and such that \(\omega_Y\) is ample on \(Y\). Lemma 0E7Q shows these conditions in fact characterize this morphism.

Lemma

Let \(k\) be a field. Let \(c : X \to Y\) be a morphism of proper schemes over \(k\). Assume

1. \(\mathcal{O}_Y = c_*\mathcal{O}_X\) and \(R^1c_*\mathcal{O}_X = 0\),

2. \(X\) and \(Y\) are reduced, Gorenstein, and have dimension \(1\),

3. \(\exists\ m \in \mathbf{Z}\) with \(H^1(X, \omega_X^{\otimes m}) = 0\) and \(\omega_X^{\otimes m}\) generated by global sections.

Then \(c^*\omega_Y \cong \omega_X\).

Proof

The fibres of \(c\) are geometrically connected by More on Morphisms, Theorem 03H0. In particular \(c\) is surjective. There are finitely many closed points \(y = y_1, \ldots, y_r\) of \(Y\) where \(X_y\) has dimension \(1\) and over \(Y \setminus \{y_1, \ldots, y_r\}\) the morphism \(c\) is an isomorphism. Some details omitted; hint: outside of \(\{y_1, \ldots, y_r\}\) the morphism \(c\) is finite, see Cohomology of Schemes, Lemma 02OG.

Let us carefully construct a map \(b : c^*\omega_Y \to \omega_X\). Denote \(f : X \to \Spec(k)\) and \(g : Y \to \Spec(k)\) the structure morphisms. We have \(f^!k = \omega_X[1]\) and \(g^!k = \omega_Y[1]\), see Lemma 0BS2 and its proof. Then \(f^! = c^! \circ g^!\) and hence \(c^!\omega_Y = \omega_X\). Thus there is a functorial isomorphism \[\Hom_{D(\mathcal{O}_X)}(\mathcal{F}, \omega_X) \longrightarrow \Hom_{D(\mathcal{O}_Y)}(Rc_*\mathcal{F}, \omega_Y)\] for coherent \(\mathcal{O}_X\)-modules \(\mathcal{F}\) by definition of \(c^!\)⁷. This isomorphism is induced by a trace map \(t : Rc_*\omega_X \to \omega_Y\) (the counit of the adjunction). By the projection formula (Cohomology, Lemma 01E8) the canonical map \(a : \omega_Y \to Rc_*c^*\omega_Y\) is an isomorphism. Combining the above we see there is a canonical map \(b : c^*\omega_Y \to \omega_X\) such that \[t \circ Rc_*(b) = a^{-1}\] In particular, if we restrict \(b\) to \(c^{-1}(Y \setminus \{y_1, \ldots, y_r\})\) then it is an isomorphism (because it is a map between invertible modules whose composition with another gives the isomorphism \(a^{-1}\)).

Choose \(m \in \mathbf{Z}\) as in (3) consider the map \[b^{\otimes m} : \Gamma(Y, \omega_Y^{\otimes m}) \longrightarrow \Gamma(X, \omega_X^{\otimes m})\] This map is injective because \(Y\) is reduced and by the last property of \(b\) mentioned in its construction. By Riemann-Roch (Lemma 0BS6) we have \(\chi(X, \omega_X^{\otimes m}) =\chi(Y, \omega_Y^{\otimes m})\). Thus \[\dim_k \Gamma(Y, \omega_Y^{\otimes m}) \geq \dim_k \Gamma(X, \omega_X^{\otimes m}) = \chi(X, \omega_X^{\otimes m})\] and we conclude \(b^{\otimes m}\) induces an isomorphism on global sections. So \(b^{\otimes m} : c^*\omega_Y^{\otimes m} \to \omega_X^{\otimes m}\) is surjective as generators of \(\omega_X^{\otimes m}\) are in the image. Hence \(b^{\otimes m}\) is an isomorphism. Thus \(b\) is an isomorphism.

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) of dimension \(1\) with \(H^0(X, \mathcal{O}_X) = k\) having genus \(g \geq 2\). Assume the singularities of \(X\) are at-worst-nodal. There is a unique morphism (up to unique isomorphism) \[c : X \longrightarrow Y\] of schemes over \(k\) having the following properties:

1. \(Y\) is proper over \(k\), \(\dim(Y) = 1\), the singularities of \(Y\) are at-worst-nodal,

2. \(\mathcal{O}_Y = c_*\mathcal{O}_X\) and \(R^1c_*\mathcal{O}_X = 0\), and

3. \(\omega_Y\) is ample on \(Y\).

Proof

Existence: A morphism with all the properties listed exists by combining Lemmas 0E3L and 0E3Q as discussed in the introduction to this section. Moreover, we see that it can be written as a composition \[X \to X_1 \to X_2 \ldots \to X_n \to X_{n + 1} \to \ldots \to X_{n + n'}\] where the first \(n\) morphisms are contractions of rational tails and the last \(n'\) morphisms are contractions of rational bridges. Note that property (2) holds for each contraction of a rational tail (Example 0E3H) and contraction of a rational bridge (Example 0E3M). It is easy to see that this property is inherited by compositions of morphisms.

Uniqueness: Let \(c : X \to Y\) be a morphism satisfying conditions (1), (2), and (3). We will show that there is a unique isomorphism \(X_{n + n'} \to Y\) compatible with the morphisms \(X \to X_{n + n'}\) and \(c\).

Before we start the proof we make some observations about \(c\). We first observe that the fibres of \(c\) are geometrically connected by More on Morphisms, Theorem 03H0. In particular \(c\) is surjective. For a closed point \(y \in Y\) the fibre \(X_y\) satisfies \[H^1(X_y, \mathcal{O}_{X_y}) = 0 \quad\text{and}\quad H^0(X_y, \mathcal{O}_{X_y}) = \kappa(y)\] The first equality by More on Morphisms, Lemma 0E7F and the second by More on Morphisms, Lemma 0E7I. Thus either \(X_y = x\) where \(x\) is the unique point of \(X\) mapping to \(y\) and has the same residue field as \(y\), or \(X_y\) is a \(1\)-dimensional proper scheme over \(\kappa(y)\). Observe that in the second case \(X_y\) is Cohen-Macaulay (Lemma 0BY5). However, since \(X\) is reduced, we see that \(X_y\) must be reduced at all of its generic points (details omitted), and hence \(X_y\) is reduced by Properties, Lemma 0344. It follows that the singularities of \(X_y\) are at-worst-nodal (Lemma 0E38). Note that the genus of \(X_y\) is zero (see above). Finally, there are only a finite number of points \(y\) where the fibre \(X_y\) has dimension \(1\), say \(\{y_1, \ldots, y_r\}\), and \(c^{-1}(Y \setminus \{y_1, \ldots, y_r\})\) maps isomorphically to \(Y \setminus \{y_1, \ldots, y_r\}\) by \(c\). Some details omitted; hint: outside of \(\{y_1, \ldots, y_r\}\) the morphism \(c\) is finite, see Cohomology of Schemes, Lemma 02OG.

Let \(C \subset X\) be a rational tail. We claim that \(c\) maps \(C\) to a point. Assume that this is not the case to get a contradiction. Then the image of \(C\) is an irreducible component \(D \subset Y\). Recall that \(H^0(C, \mathcal{O}_C) = k'\) is a finite separable extension of \(k\) and that \(C\) has a \(k'\)-rational point \(x\) which is also the unique intersection of \(C\) with the “rest” of \(X\). We conclude from the general discussion above that \(C \setminus \{x\} \subset c^{-1}(Y \setminus \{y_1, \ldots, y_r\})\) maps isomorphically to an open \(V\) of \(D\). Let \(y = c(x) \in D\). Observe that \(y\) is the only point of \(D\) meeting the “rest” of \(Y\). If \(y \not \in \{y_1, \ldots, y_r\}\), then \(C \cong D\) and it is clear that \(D\) is a rational tail of \(Y\) which is a contradiction with the ampleness of \(\omega_Y\) (Lemma 0E63). Thus \(y \in \{y_1, \ldots, y_r\}\) and \(\dim(X_y) = 1\). Then \(x \in X_y \cap C\) and \(x\) is a smooth point of \(X_y\) and \(C\) (Lemma 0E38). If \(y \in D\) is a singular point of \(D\), then \(y\) is a node and then \(Y = D\) (because there cannot be another component of \(Y\) passing through \(y\) by Lemma 0E38). Then \(X = X_y \cup C\) which means \(g = 0\) because it is equal to the genus of \(X_y\) by the discussion in Example 0E3H; a contradiction. If \(y \in D\) is a smooth point of \(D\), then \(C \to D\) is an isomorphism (because the nonsingular projective model is unique and \(C\) and \(D\) are birational, see Section 0BXX). Then \(D\) is a rational tail of \(Y\) which is a contradiction with ampleness of \(\omega_Y\).

Assume \(n \geq 1\). If \(C \subset X\) is the rational tail contracted by \(X \to X_1\), then we see that \(C\) is mapped to a point of \(Y\) by the previous paragraph. Hence \(c : X \to Y\) factors through \(X \to X_1\) (because \(X\) is the pushout of \(C\) and \(X_1\), see discussion in Example 0E3H). After replacing \(X\) by \(X_1\) we have decreased \(n\). By induction we may assume \(n = 0\), i.e., \(X\) does not have a rational tail.

Assume \(n = 0\), i.e., \(X\) does not have any rational tails. Then \(\omega_X^{\otimes 2}\) and \(\omega_X^{\otimes 3}\) are globally generated by Lemma 0E3K. It follows that \(H^1(X, \omega_X^{\otimes 3}) = 0\) by Lemma 0B62. By Lemma 0E7P applied with \(m = 3\) we find that \(c^*\omega_Y \cong \omega_X\). We also have that \(\omega_X = (X \to X_{n'})^*\omega_{X_{n'}}\) by Lemma 0E3N and induction. Applying the projection formula for both \(c\) and \(X \to X_{n'}\) we conclude that \[\Gamma(X_{n'}, \omega_{X_{n'}}^{\otimes m}) = \Gamma(X, \omega_X^{\otimes m}) = \Gamma(Y, \omega_Y^{\otimes m})\] for all \(m\). Since \(X_{n'}\) and \(Y\) are the Proj of the direct sum of these by Morphisms, Lemma 0C6J we conclude that there is a canonical isomorphism \(X_{n'} = Y\) as desired. We omit the verification that this is the unique isomorphism making the diagram commute.

Lemma

Let \(k\) be a field. Let \(X\) be a proper scheme over \(k\) of dimension \(1\) with \(H^0(X, \mathcal{O}_X) = k\) having genus \(g \geq 2\). Assume the singularities of \(X\) are at-worst-nodal and \(\omega_X\) is ample. Then \(\omega_X^{\otimes 3}\) is very ample and \(H^1(X, \omega_X^{\otimes 3}) = 0\).

Proof

Combining Varieties, Lemma 0B5Y and Lemmas 0E63 and 0E64 we see that \(X\) contains no rational tails or bridges. Then we see that \(\omega_X^{\otimes 3}\) is globally generated by Lemma 0E3L. Choose a \(k\)-basis \(s_0, \ldots, s_n\) of \(H^0(X, \omega_X^{\otimes 3})\). We get a morphism \[\varphi_{\omega_X^{\otimes 3}, (s_0, \ldots, s_n)} : X \longrightarrow \mathbf{P}^n_k\] See Constructions, Section 01ND. The lemma asserts that this morphism is a closed immersion. To check this we may replace \(k\) by its algebraic closure, see Descent, Lemma 02L6. Thus we may assume \(k\) is algebraically closed.

Assume \(k\) is algebraically closed. We will use Varieties, Lemma 0E8T to prove the lemma. Let \(Z \subset X\) be a closed subscheme of degree \(2\) over \(Z\) with ideal sheaf \(\mathcal{I} \subset \mathcal{O}_X\). We have to show that \[H^0(X, \mathcal{L}) \to H^0(Z, \mathcal{L}|_Z)\] is surjective. Thus it suffices to show that \(H^1(X, \mathcal{I}\mathcal{L}) = 0\). To do this we will use Lemma 0E3E. Thus it suffices to show that \[3\deg(\omega_X|_Y) > -2\chi(Y, \mathcal{O}_Y) + \deg(Z \cap Y)\] for every reduced connected closed subscheme \(Y \subset X\). Since \(k\) is algebraically closed and \(Y\) connected and reduced we have \(H^0(Y, \mathcal{O}_Y) = k\) (Varieties, Lemma 0BUG). Hence \(\chi(Y, \mathcal{O}_Y) = 1 - \dim H^1(Y, \mathcal{O}_Y)\). Thus we have to show \[3\deg(\omega_X|_Y) > -2 + 2\dim H^1(Y, \mathcal{O}_Y) + \deg(Z \cap Y)\] which is true by Lemma 0E3J except possibly if \(Y = X\) or if \(\deg(\omega_X|_Y) = 0\). Since \(\omega_X\) is ample the second possibility does not occur (see first lemma cited in this proof). Finally, if \(Y = X\) we can use Riemann-Roch (Lemma 0BS6) and the fact that \(g \geq 2\) to see that the inquality holds. The same argument with \(Z = \emptyset\) shows that \(H^1(X, \omega_X^{\otimes 3}) = 0\).

Vector fields

In this section we study the space of vector fields on a curve. Vector fields correspond to infinitesimal automorphisms, see More on Morphisms, Section 04BU, hence play an important role in moduli theory.

Let \(k\) be an algebraically closed field. Let \(X\) be a finite type scheme over \(k\). Let \(x \in X\) be a closed point. We will say an element \(D \in \text{Der}_k(\mathcal{O}_X, \mathcal{O}_X)\) fixes \(x\) if \(D(\mathcal{I}) \subset \mathcal{I}\) where \(\mathcal{I} \subset \mathcal{O}_X\) is the ideal sheaf of \(x\).

Lemma

Let \(k\) be an algebraically closed field. Let \(X\) be a smooth, proper, connected curve over \(k\). Let \(g\) be the genus of \(X\).

1. If \(g \geq 2\), then \(\text{Der}_k(\mathcal{O}_X, \mathcal{O}_X)\) is zero,

2. if \(g = 1\) and \(D \in \text{Der}_k(\mathcal{O}_X, \mathcal{O}_X)\) is nonzero, then \(D\) does not fix any closed point of \(X\), and

3. if \(g = 0\) and \(D \in \text{Der}_k(\mathcal{O}_X, \mathcal{O}_X)\) is nonzero, then \(D\) fixes at most \(2\) closed points of \(X\).

Proof

Recall that we have a universal \(k\)-derivation \(d : \mathcal{O}_X \to \Omega_{X/k}\) and hence \(D = \theta \circ d\) for some \(\mathcal{O}_X\)-linear map \(\theta : \Omega_{X/k} \to \mathcal{O}_X\). Recall that \(\Omega_{X/k} \cong \omega_X\), see Lemma 0BS2. By Riemann-Roch we have \(\deg(\omega_X) = 2g - 2\) (Lemma 0BS6). Thus we see that \(\theta\) is forced to be zero if \(g > 1\) by Varieties, Lemma 0B40. This proves part (1). If \(g = 1\), then a nonzero \(\theta\) does not vanish anywhere and if \(g = 0\), then a nonzero \(\theta\) vanishes in a divisor of degree \(2\). Thus parts (2) and (3) follow if we show that vanishing of \(\theta\) at a closed point \(x \in X\) is equivalent to the statement that \(D\) fixes \(x\) (as defined above). Let \(z \in \mathcal{O}_{X, x}\) be a uniformizer. Then \(dz\) is a basis element for \(\Omega_{X, x}\), see Lemma 0C1E. Since \(D(z) = \theta(dz)\) we conclude.

Lemma

Let \(k\) be an algebraically closed field. Let \(X\) be an at-worst-nodal, proper, connected \(1\)-dimensional scheme over \(k\). Let \(\nu : X^\nu \to X\) be the normalization. Let \(S \subset X^\nu\) be the set of points where \(\nu\) is not an isomorphism. Then \[\text{Der}_k(\mathcal{O}_X, \mathcal{O}_X) = \{D' \in \text{Der}_k(\mathcal{O}_{X^\nu}, \mathcal{O}_{X^\nu}) \mid D' \text{ fixes every }x^\nu \in S\}\]

Proof

Let \(x \in X\) be a node. Let \(x', x'' \in X^\nu\) be the inverse images of \(x\). (Every node is a split node since \(k\) is algebriacally closed, see Definition 0CBV and Lemma 0CBW.) Let \(u \in \mathcal{O}_{X^\nu, x'}\) and \(v \in \mathcal{O}_{X^\nu, x''}\) be uniformizers. Observe that we have an exact sequence \[0 \to \mathcal{O}_{X, x} \to \mathcal{O}_{X^\nu, x'} \times \mathcal{O}_{X^\nu, x''} \to k \to 0\] This follows from Lemma 0C1X. Thus we can view \(u\) and \(v\) as elements of \(\mathcal{O}_{X, x}\) with \(uv = 0\).

Let \(D \in \text{Der}_k(\mathcal{O}_X, \mathcal{O}_X)\). Then \(0 = D(uv) = vD(u) + uD(v)\). Since \((u)\) is annihilator of \(v\) in \(\mathcal{O}_{X, x}\) and vice versa, we see that \(D(u) \in (u)\) and \(D(v) \in (v)\). As \(\mathcal{O}_{X^\nu, x'} = k + (u)\) we conclude that we can extend \(D\) to \(\mathcal{O}_{X^\nu, x'}\) and moreover the extension fixes \(x'\). This produces a \(D'\) in the right hand side of the equality. Conversely, given a \(D'\) fixing \(x'\) and \(x''\) we find that \(D'\) preserves the subring \(\mathcal{O}_{X, x} \subset \mathcal{O}_{X^\nu, x'} \times \mathcal{O}_{X^\nu, x''}\) and this is how we go from right to left in the equality.

Lemma

Let \(k\) be an algebraically closed field. Let \(X\) be an at-worst-nodal, proper, connected \(1\)-dimensional scheme over \(k\). Assume the genus of \(X\) is at least \(2\) and that \(X\) has no rational tails or bridges. Then \(\text{Der}_k(\mathcal{O}_X, \mathcal{O}_X) = 0\).

Proof

Let \(D \in \text{Der}_k(\mathcal{O}_X, \mathcal{O}_X)\). Let \(X^\nu\) be the normalization of \(X\). Let \(D' \in \text{Der}_k(\mathcal{O}_{X^\nu}, \mathcal{O}_{X^\nu})\) be the element corresponding to \(D\) via Lemma 0E68. Let \(C \subset X^\nu\) be an irreducible component. If the genus of \(C\) is \(> 1\), then \(D'|_{\mathcal{O}_C} = 0\) by Lemma 0E67 part (1). If the genus of \(C\) is \(1\), then there is at least one closed point \(c\) of \(C\) which maps to a node on \(X\) (since otherwise \(X \cong C\) would have genus \(1\)). By the correspondence this means that \(D'|_{\mathcal{O}_C}\) fixes \(c\) hence is zero by Lemma 0E67 part (2). Finally, if the genus of \(C\) is zero, then there are at least \(3\) pairwise distinct closed points \(c_1, c_2, c_3 \in C\) mapping to nodes in \(X\), since otherwise either \(X\) is \(C\) with two points glued (two points of \(C\) mapping to the same node), or \(C\) is a rational bridge (two points mapping to different nodes of \(X\)), or \(C\) is a rational tail (one point mapping to a node of \(X\)). These three possibilities are not permitted since \(C\) has genus \(\geq 2\) and has no rational bridges, or rational tails. Whence \(D'|_{\mathcal{O}_C}\) fixes \(c_1, c_2, c_3\) hence is zero by Lemma 0E67 part (3).