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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 $\mathcal F(X)$?

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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?

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    But I do find a cover of $X$ by such affine opens, where it is $A^n$, and this suffices I think.2011-11-19