Riemann-Roch equation for proper integral curves (not necessarily non-singular!)

Question

In Qing Liu - Algebraic Geometry and Arithmetic Curves, chapter 7.3, Theorem 3.26 the author states the Theorem of Riemann-Roch as follows:

Let $f: X \rightarrow \text{Spec }k$ be a projective curve over a field $k$. Then for any Cartier divisor $D \in \text{Div}(X)$, we have $$\dim_kH^0(X,\mathcal{O}_x(D)) - \dim_k H^0(X,\omega_f \otimes \mathcal{O}_x(-D)) = \deg D +1 -p_a.$$

Here $\omega_f$ denotes the 1-dualizing sheaf for $f$. On the next page, in Remark 3.33, he points out that if $X$ is a local complete intersection and integral curve, then one obtains $$\dim_k H^0(X,\mathcal{O}_x(D)) = \deg D +1 -p_a(X)$$ if the degree of $D$ is sufficiently large.

That equation is the one I'm looking for, except that I want to drop the assumption that $X$ is a local complete intersection curve.

Thus the question:

Let $X$ be a proper integral curve over the field $k$. Does the equation $$\dim_k H^0(X,\mathcal{O}_x(D)) = \deg D +1 -p_a(X)$$ hold for any Cartier divisor $D$ on $X$ with $\deg D \geq 2p_a(X)-1$?

Thoughts: The desired equation holds in the case of a l.c.i. curve since then the dualizing sheaf is invertible and then we can work with the degree of $\omega_f \otimes \mathcal{O}_x(-D)$, i.e. for $\deg D$ large enough, $\omega_f \otimes \mathcal{O}_x(-D)$ has negative degree and thus no global sections. Maybe there is another way to show that this is the case if $\deg D$ increases.

Another approach is to think about the first cohomology term $\dim_k H^1(X,\mathcal{O}_X(D))$ instead of $\dim_k H^0(X,\omega_f \otimes \mathcal{O}_x(-D))$ (i.e. avoid using the dualizing sheaf at all) and to show it vanishes for $\deg D$ large enough. One possible approach: Assume $\pi: C \rightarrow \mathbb{P}^1$ to be a proper integral morphism and let $\mathbb{P}^1 = U_0 \cup U_\infty$ be an open affine cover where $x$ is regular on $U_0$ and $1/x$ regular on $U_\infty$. Let $V_0 = \pi^{-1}(U_0)$, $V_\infty = \pi^{-1}(U_\infty)$, then $C = V_0 \cup V_\infty$ is an open affine cover of $C$ where $V_0$ is the open subset where the function $x$ is regular and analogously $V_\infty$ the open subset where the function $1/x$ is regular. Following the definitions of Cech cohomology, $H^1(X,\mathcal{O}_X(D)) = 0$ if and only if $$\mathcal{O}_X(D)(V_0 \cap V_\infty) = \text{im}(d^0:C^0(\mathcal{O}_X(D))\rightarrow C^1(\mathcal{O}_X(D)))$$ where the latter is $$\left\{ (a_1 - a_0)_{V_0 \cap V_\infty} \mid (a_0,a_1) \in C^0(\mathcal{O}_X(D)) = \mathcal{O}_X(D)(V_0) \times \mathcal{O}_X(D)(V_\infty) \right\}.$$ Then all this provides the equivalence $H^1(X,\mathcal{O}_X(D)) = 0$ if and only if $$\forall \gamma \in \mathcal{O}_X(D)(V_0\cap V_\infty) \exists (a_0,a_1) \in \mathcal{O}_X(D)(V_0) \times \mathcal{O}_X(D)(V_\infty): \gamma = (a_1-a_0)_{V_0 \cap V_\infty}.$$ Can you show that the latter is true?

I found no reference for this equation with this few assumptions, but if you know one, please let me know. Further I'd be grateful for any kind of help.

Edit: It would be sufficient if there is a statement as follows:

Let $f: X \rightarrow \text{Spec }k$ be a projective curve over a field $k$. Let $\mathcal{L}$ be an invertible sheaf on $X$ and $\mathcal{F}$ an arbitrary quasi-coherent sheaf on $X$. Then if $\deg \mathcal{L}$ is positive enough, it would imply $$\text{Hom}_{\mathcal{O}_X}(\mathcal{L},\mathcal{F}) = 0.$$

Hence if anybody knows a reference for this (or even a proof) please let me know or post it!

Thank you very much.

Answer 1

I will assume the canonical map $k \to H^0(X, \mathcal{O}_X)$ is an isomorphism (I am not exactly sure how "curve" is defined in Liu's book). I will denote $g = \dim_k H^1(X, \mathcal{O}_X)$ and I think this agrees with what you call $p_a$ under the assumptions made sofar. As you explain in the post we have to show that if there is a nonzero map $$ \mathcal{O}_X(D) \to \omega_X $$ where $\omega_X$ is the dualizing module of $X$, then $\deg(D) < 2g - 1$. Since $X$ is a variety, the dualizing module is torsion free (this is a general fact). Hence if there is a nonzero map, then it is injective and hence the induced map $$ H^0(X, \mathcal{O}_X(D)) \to H^0(X, \omega_X) $$ is injective too. We have $$ \dim_k H^0(X, \mathcal{O}_X(D)) \geq \chi(\mathcal{O}_X(D)) = \deg(D) + 1 - g $$ and we have $\dim_k H^0(X, \omega_X) = g$. From injectivity we get $$ \deg(D) + 1 - g \leq g \Leftrightarrow \deg(D) \leq 2g - 1 $$ This proves what you want except in case that equality holds. In this case equality holds everywhere which means that $\dim_k H^0(X, \mathcal{O}_X(D) = \chi(\mathcal{O}_X(D))$ and hence $H^1(X, \mathcal{O}_X(D)) = 0$. Thus the result is true in this case as well.

Answer 2

I don't know how to prove that $h^0(\mathcal{O}_X(D)) = \deg D + 1 - p_a(X)$ holds as soon as $\deg D > 2p_a - 2$. But it is not too hard to show that there exists a positive integer $N$ such that the above equality holds as soon as $\deg D > N$.

Indeed there exists a divisor $D_0$ such that $H^1(X, \mathcal{O}_X(D_0)) = 0$ (take a sufficiently big multiple of an ample divisor on $X$). Let $N = \deg D_0 + p_a(X) - 1$. Let $\deg D > n$. Then $\deg (D - D_0) > p_a(X) - 1$ and$$h^0(X, \mathcal{O}_X(D - D_0)) \ge \deg (D - D_0) + 1 - p_a(X) > 0.$$So $D$ is linearly equivalent to $D_0 + E$ with $E$ effective. So $H^1(X, \mathcal{O}_X(D))$ is isomorphic to $H^1(X, \mathcal{O}_X(D_0 + E))$ and the exact sequence$$0 \to \mathcal{O}_X(D_0) \to \mathcal{O}_X(D_0 + E) \to \mathcal{O}_X(D_0 + E)|_E \to 0$$implies that $H^1(X, \mathcal{O}_X(D_0)) \to H^1(X, \mathcal{O}_X(D_0 + E))$ is surjective because the support of $\mathcal{O}_X(D_0 + E)|_E$ has dimension $0$.

To study $H^1(X, \mathcal{O}_X(D))$, one may first reduce to the case (by linear equivalence) reduce to the case where $D$ has support in the regular locus of $X$, and compare the $H^1$ on $X$ to the $H^1$ on the normalization of $X$. But I have to say that I can't conclude...

Let $f: X \rightarrow \text{Spec }k$ be a projective curve over a field $k$. Let $\mathcal{L}$ be an invertible sheaf on $X$ and $\mathcal{F}$ an arbitrary quasi-coherent sheaf on $X$. Then if $\deg \mathcal{L}$ is negative enough, it would imply $$\text{Hom}_{\mathcal{O}_X}(\mathcal{L},\mathcal{F}) = 0.$$

What you want is $\deg \mathcal{L}$ positive enough. Well, I mean you need some conditions on $\mathcal{F}$. Otherwise take $\mathcal{F}$ to be e.g. skyscraper sheaf at a point. If $\mathcal{F}$ is the skyscraper sheaf at a point, then the thing you wrote down is never nonzero.