Sorry for my bad English.
In Hartshorne "Algebraic Geometry " , III. Example 6.5.1
i.e. there is locally free resolution
$\dots \to \mathscr{L}_1\to \mathscr{L}_0\to \mathscr{F}\to 0$.
But I can't understand why $\mathscr{L}_1, \mathscr{L}_2$… are also
locally free sheaves of finite rank.
Please help me, thanks.
Best Answer
II, 5.18 states that for $\overline{X}$ projective over $A$, any coherent sheaf $\mathscr{F}$ can be written as an epimorphic image of some finite direct sum of the twists $\mathcal{O}(n)$ of the structure sheaf. Thus, as a quotient of a locally free sheaf $\mathscr{L}_0$ of finite rank. As the kernel of $\mathscr{L}_0\rightarrow \mathscr{F}$ is itself coherent, same applies, and one finds $\mathscr{L}_1\rightarrow \mathscr{L}_0 \rightarrow \mathscr{F}\rightarrow 0$ exact with $\mathscr{L}_1$ locally free of finite rank. And so on.
In the quasi-projective case, one can choose an open immersion $i:X \hookrightarrow \overline{X}$ into a projective $A$-scheme, and then by Exercise II.5.15 of Hartshorne, extend a coherent sheaf $\mathscr{F}$ on $X$ to a sheaf $\mathscr{F'}$ on $\overline{X}$ which is still coherent. By the previous, there is a resolution $\mathscr{L}_{\bullet}\rightarrow \mathscr{F}'\rightarrow 0$ by a finite rank locally free sheaves on $\overline{X}$. The restriction functor to the open subscheme $X$ is exact and preserves "being locally free of finite rank", hence ${\mathscr{L}_{\bullet}}_{\restriction_{X}} \rightarrow \mathscr{F'}_{\restriction_{X}}(=\mathscr{F})\rightarrow 0$ is a resolution by finite rank locally free sheaves on $X$ (in fact, finite direct sums of line bundles).