Lemma 53.6.1. 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 $1$.

## 53.6 Some vanishing results

This section contains some very weak vanishing results. Please see Section 53.21 for a few more and more interesting results.

**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 31.4.6) and is globally generated (Varieties, Lemma 33.33.3). 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 31.4.6). Thus $X$ has $(S_1)$ by Divisors, Lemma 31.4.3. Hence $X$ is Cohen-Macaulay.
$\square$

In this section we work in the following situation.

Situation 53.6.2. 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 53.6.1 the scheme $X$ is Cohen-Macaulay and equidimensional of dimension $1$. The dualizing module $\omega _ X$ discussed in Lemmas 53.4.1 and 53.4.2 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 53.6.3. In Situation 53.6.2. Given an exact sequence

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 30.9.10.) Since $H^0(X, \mathcal{O}_ X) = k$ is dual to $H^1(X, \omega _ X)$ (see Section 53.5) we see that $\dim H^1(X, \omega _ X) = 1$. The sheaf $\omega _ X$ represents the functor $\mathcal{F} \mapsto \mathop{\mathrm{Hom}}\nolimits _ k(H^1(X, \mathcal{F}), k)$ on the category of coherent $\mathcal{O}_ X$-modules (Duality for Schemes, Lemma 48.22.5). 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.
$\square$

Lemma 53.6.4. In Situation 53.6.2. 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 53.5 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.
$\square$

Lemma 53.6.5. In Situation 53.6.2. Given an exact sequence

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 53.6.3 and our assumption on $\mathcal{F}$. Consider the short exact sequence

If the quotient on the right is nonzero, then we obtain a contradiction because then $H^0(X, \mathcal{F})$ is bigger than $H^0(X, \mathcal{F}')$. $\square$

Here is an example global generation statement.

Lemma 53.6.6. In Situation 53.6.2 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 53.6.5 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 48.22.5). 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 $\mathcal{F}' \to \mathcal{Q}$.
$\square$

Lemma 53.6.7. In Situation 53.6.2. 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 33.22.2 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

By Lemma 53.6.5 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 $X_{\overline{k}} \to \mathop{\mathrm{Spec}}(\overline{k})$ is an example of Situation 53.6.2. By flat base change (Cohomology of Schemes, Lemma 30.5.2) we see that $H^0(X_{\overline{k}}, \mathcal{O}) = \overline{k}$. The scheme $X_{\overline{k}}$ is proper over $\overline{k}$ (Morphisms, Lemma 29.41.5) and equidimensional of dimension $1$ (Morphisms, Lemma 29.28.3). The pullback of $\omega _ X$ to $X_{\overline{k}}$ is the dualizing module of $X_{\overline{k}}$ by Lemma 53.4.4. The pullback of $\mathcal{L}$ to $X_{\overline{k}}$ is very ample (Morphisms, Lemma 29.38.8). The degree of the pullback of $\mathcal{L}$ to $X_{\overline{k}}$ is equal to the degree of $\mathcal{L}$ on $X$ (Varieties, Lemma 33.44.2). Finally, we see that $\omega _ X \otimes \mathcal{L}$ is globally generated if and only if its base change is so (Varieties, Lemma 33.22.1). In this way we see that the result follows from the result in the case of an algebraically closed ground field. $\square$

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