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Let $X$ be a sufficiently nice scheme and $\mathcal F$ a locally free sheaf of finite rank, i.e., a vector bundle on $X$.

Let $x$ be point of $X$ and $k(x)$ it's residue field.

Let $f:\mathcal F \to \mathcal F$ be a morphism of sheaves; then $f$ induces a map

$$f_x: \mathcal F_x \rightarrow F_x ,$$

and a map

$$\bar{f_x}: \mathcal F_x \otimes k(x) \rightarrow \mathcal F_x \otimes k(x) .$$


If I know that $f_x$ is zero, then can I conclude that $f$ is zero in a sufficiently small neighborhood of $x$? And, still more important for me, if $\bar{f}_x$ is zero, can I conclude that $f$ is zero in a sufficiently small neighorhood of $x$?

Perhaps one may also assume $\mathcal F$ just as coherent, I'm not sure.

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up vote 2 down vote accepted

2) The answer to the second question, the one more important to you, is no. For example take $X=\mathbb A^1_k=Spec(k[z])$ and $\mathcal F = \mathcal O_X$.
The morphism $z: \mathcal O_X \to \mathcal O_X$ is zero at $x=O=(z)$, that is $\bar z_x=0: k\to k $, but nevertheless $z|U\neq0$ for every neighbourhood $U$ of $x$.

1) On the other hand, suppose $f_x:\mathcal F_x \to \mathcal F_x$ is zero where $f:\mathcal F\to \mathcal F$ is an endomorphism of a locally free sheaf.
Then since locally $\mathcal F=\mathcal O^r$, $f$ is given by a square matrix of regular functions $f_{ij}$ and $f_x=0$ means that all these functions $f_{ij}$ satisfy $(f_{ij})_x=0$. Hence all these $f_{ij}$ are zero on some neighbourhood of $x$ and $f$ is zero on that neighbourhood too.
So the answer to your first question is yes.

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@Cyril: If $f_x=0$, $f$ vanishes in an open neighborhood of $x$ which has no reason to be $X$ even when $X$ is irreducible (thought this is true if $X$ is moreover reduced). – user18119 Nov 20 '11 at 9:29
@QiL: Sorry for the unprecise comment: of course I meant: if $f_x=0$ for a point on an integral scheme, then one can conclude $f=0$ globally. I think this is right. – Cyril Nov 20 '11 at 19:23
@Cyril: yes this is right. – user18119 Nov 20 '11 at 19:41

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