Locally free resolution of coherent sheaves on nonsingular curves

algebraic-curvesalgebraic-geometrycoherent-sheaves

This question is from Exercise II 6.11 of Hartshorne.

Let $X$ be a nonsingular curve over an algebraically closed field $k$. For any coherent sheaf $\mathcal{F}$ on $X$, show that there exist locally free sheaves $\mathcal{E}_1$ and $\mathcal{E}_0$ of finite ranks and an exact sequence $0\to \mathcal{E}_1 \to \mathcal{E}_0 \to \mathcal{F}\to 0$.

I can prove that if $\mathcal{E}_0$ exists, then we can take $\mathcal{E}_1$ be the kernel of $\mathcal{E}_0\to \mathcal{F}$. So it suffices to prove that $\mathcal{F}$ can be written as a quotient of a locally free sheaf of finite rank.

The following is my attempt:

Firstly, we can take an open subscheme $U=\operatorname{Spec}(A)$ such that $\mathcal{F}|_U\simeq \widetilde{M}$, where $M$ is a finitely generated $A$-module with generators $m_1,\cdots,m_n$. I want to extend $m_i$ to the global section of $\mathcal{F}$. Since $X$ is a curve, $X-U$ consists of finite points. For $p\in X-U$, we can take $U'=\operatorname{Spec}(B)$ containing $p$ and $f\in B$ such that $p=V(f)$. Furthermore, we can assume that $m\in \mathcal{F}(D(f))\simeq \mathcal{F}(U')_f$, then $f^N m\in \mathcal{F}(U')$ for some $N$. In this way I can extend $m$ to $U'$ by multiplying $f^N$, but what is $f^N m$ in $\mathcal{F}(U)$?

Best Answer

By proposition II.6.7, a complete nonsingular curve over an algebraically closed field is projective. By corollary II.5.18, any coherent sheaf on a scheme projective over a noetherian ring can be written as a quotient of a finite direct sum of twists of the structure sheaf. So you don't have to do this yourself and you can just cite previous results.