# Is locally free sheaf of finite rank coherent?

Let $\mathcal{F}$ be a locally free sheaf of finite rank of scheme $X$, is $\mathcal{F}$ coherent?

By the definition of locally free sheaf, there exists an open cover {$U_i$} of $X$ such that $\mathcal{F}|_{U_i}$ is isomorphic to the sheaf $\widetilde{\mathcal{O}(U_i)^n}$. But we don't know each $U_i$ affine or not!

So, it that true or not?

How about $X$ being locally noetherian? It $X$ is, we can find $V_{ij} \subset U_i$ s.t. $V_{ji} = Spec(A_{ji})$. And $\mathcal{F}|_{V_{ij}} = \mathcal{F}|_{U_i}|_{V_{ij}}$...?

For example, $X(\Delta)$ is a toric varity with $\Delta$ consists of strongly convex polyhedral cones.

Thank you very much!!

-
Being coherent involves a finiteness condition, whereas being locally free does not. –  Zhen Lin Jun 30 '12 at 11:15
If we only consider locally free sheaf of finite rank? –  Peter Hu Jun 30 '12 at 11:17

The exact condition for locally free sheaves on a ringed space $(X,\mathcal O_X)$ to be coherent is exactly that $\mathcal O_X$ be coherent.
a) The condition is clearly necessary since $\mathcal O_X$ is locally free.
b) It is sufficient because if the structure shaf is coherent, then coherence is a local property and because a direct sum of coherent sheaves is coherent: apply to $\mathcal F \mid U_i \cong (\mathcal O\mid U_i)^{\oplus r}$
And when is $\mathcal O_X$ coherent?
However, for locally noetherian schemes, it is the case that $\mathcal O_X$ is coherent, so for these schemes, yes, locally free sheaves are coherent.