# Sections of locally free sheaves

I tried to prove the following assertion, which I think is implicit in a text I have read, but I'm not sure about that:

Let $X$ be an integral scheme and $\mathcal F$ a locally free sheaf of finite rank on $X$.

Let $s$ be a section of $\mathcal F(X)$ which is zero in $\mathcal F(U)$, where $U$ is an open nonempty subset of $X$.

Is it true that then also $s$ itself is zero in $\mathcal F(X)$?

To check that $s$ is zero in $F(X)$ it is enough to do so on affine open subsets. Let $V\subseteq X$ be affine. Then $U\cap V$ is an open subset of $V$ which is dense in $V$, and on which $s$ is zero. Can you show that $s|_V$ is zero?
• @Fozad: I did not invent the hypotheses: I am using those in the statement of the question. It may well be the case, though, that the support of $\mathcal F$ is the complement of $U$, so something is needed. – Mariano Suárez-Álvarez Nov 18 '11 at 18:22
• Well, I think if one has the assumptions, then the following argument would do: On affines Spec(A) our sheaf is just $A^n$, and by assumption our element $s \in A^n$ vanishes in the generic point, i.e. in $Quot(A)^n$, but then it is zero. – Cyril Nov 18 '11 at 19:27
• @Cyril, it is $A^n$ on sufficiently small affine open sets, not in all of them (at least; but showing that it is $A^n$ on all affine open sets is at least harder than Serre's conjecture on projective modules over polynomial rings, which is now a difficult theorem of Suslin and Quillen) – Mariano Suárez-Álvarez Nov 19 '11 at 0:09