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If $X$ is a locally noetherian scheme and $F$ is a coherent sheaf, I want to show the following equivalence:

$F$ is locally free iff its stalk is a free $O_{X,p}$-module for every $p$ in $X$.

=> follows from the definition of locally free.

<= is difficult for me: I don't see how to combine finite-type condition on $F$ with locally noetherian property.

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In fact it suffices that $X$ is an arbitrary ringed space and that $F$ is of finite presentation; of course, Georges' proof does not work in this generality. The basic idea of the general proof is that $\mathcal{H}om(F,\mathcal{O}_X^n)_x \cong Hom(F_x,\mathcal{O}_{X,x}^n)$ (by the presentation condition), so that an isomorphism $F_x \to \mathcal{O}_{X,x}^n$ lifts to an isomorphism $F|_U \to \mathcal{O}_X|_U^n$ on some open neighbourhood $U$ of $x$. –  Ingo Blechschmidt May 7 at 13:54

1 Answer 1

up vote 8 down vote accepted

As the name indicates, "locally free" is a local concept.
So we can assume that $X=Spec(A)$, the affine scheme associated to the noetherian ring $A$, and $F=\tilde M$, the coherent sheaf associated to the finitely generated $A$-module $M$.

The sheaf $F=\tilde M$ is locally free if and only if the module $M$ is projective.
And a finitely generated module $M$ over a noetherian ring $A$ is projective if and only if all its localizations $M_{\mathfrak p} \; ( \mathfrak p \in Spec(A)$ are $A_{\mathfrak p}$-free modules.
Since at a point $p\in X$ corresponding to the prime $\mathfrak p \in Spec(A)$ we have $\mathcal O_{X,x}=A_{\mathfrak p}$ and $ F_p=M_{\mathfrak p}$, we see that indeed freeness of all stalks of $F$ implies local freeness of the given coherent sheaf $F$.

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