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Let $f: X \rightarrow Y$ be a flat morphism of locally noetherian schemes and $\varphi: \mathcal U \rightarrow \mathcal V$ a map of vector bundles (locally free sheaves of finite rank) on $X$.

I want to test when $\varphi$ is the zero map. What I have is the following information:

For each point $y \in Y$ the induced map

$\varphi_{|X_y}:\mathcal U_{|X_y} \rightarrow \mathcal V_{|X_y}$

is zero. Here, $X_y$ denotes the scheme-theoretic fiber of $y$, and the notation $(.)_{|X_y}$ means that one takes pullback along the natural map $X_y\rightarrow X$.

Is this information sufficient to conclude that $\varphi =0$ ? If not, are there additional assumptions on the schemes or on $f$ that would imply the vanishing of $\varphi$ ?

Note that I do not want to assume that the cokernel of $\varphi$ is locally free. If necessary we may assume $X$ as reduced.

Maybe to consider the special case $Y=X$ and $f=id$ can also be enlightening.

Any helpful comment on this would be highly appreciated.

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    $\begingroup$ Consider the case that $X = Y$, $f = \mathrm{id}_X$, $\cal U = \cal V = \cal O$, and $X$ is not reduced. $\endgroup$
    – Angelo
    Commented Aug 9, 2013 at 18:50
  • $\begingroup$ Dear Angelo, thank you for the comment. It would be nice if you could slightly elaborate it. I also think that the assumption of $X$ reduced is not bad for me, so the question remains if the claim holds for $X$ reduced (I edited the question at this point). $\endgroup$
    – Falter
    Commented Aug 9, 2013 at 19:01
  • $\begingroup$ I think Angelo's comment dealt with the slightly different situation when you assume that the map is zero on the fibers of the two vector bundles. Then indeed $X=Y=Spec\, k[x]/(x^2)$, $\phi=x: O_X\to O_X$ is zero on the fiber i.e. after diving by $x$. $\endgroup$
    – Piotr Achinger
    Commented Aug 10, 2013 at 0:13
  • $\begingroup$ Yes, but this is indeed the question. So Angelo's example shows that the claim is false for schemes $X$ which are not reduced. The question remains if it holds for reduced $X$. $\endgroup$
    – Falter
    Commented Aug 10, 2013 at 6:41
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    $\begingroup$ Assume $X=Y$. Then locally the map of vector bundles is given by a matrix with entries regular functions and the question becomes: can a nonzero regular function be zero at every point? if this happens, then $X$ has an irreducible component all of whose points are non-reduced. $\endgroup$
    – rita
    Commented Aug 10, 2013 at 9:11

1 Answer 1

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The question is local. Assume $X$ and $Y$ are affine, so $X = Spec A$, $Y = Spec B$, and $A$ is a $B$-algebra. We have a morphism $\varphi:M \to N$ of $A$-modules such that $\varphi \otimes B/{\mathfrak{m}} = 0$ for any maximal ideal ${\mathfrak{m}} \subset B$. It follows that $\varphi(M) \subset {\mathfrak{m}}N$ for all $m$, hence $\varphi(M) \subset (\cap {\mathfrak{m}})N$. The intersection of all maximal ideals is the Jacobson radical, so if $B$ is finitely generated it is the nilradical. If you assume that $B$ has no nilpotents ($Y$ is reduced) then it follows that $\varphi(M) \subset 0$, so $\varphi = 0$.

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    $\begingroup$ You seem to use $\cap_{\mathfrak{m}} (\mathfrak{m} N) = (\cap_{\mathfrak{m}} \mathfrak{m}) N$, which fails in general. But it holds when $N$ is finitely generated free, which may be assumed here. $\endgroup$
    – Martin Brandenburg
    Commented Aug 10, 2013 at 8:32
  • $\begingroup$ I see the problem pointed out by Martin, too. But I am not sure if it is resolved by his comment: $N$ is only finitely generated free over $A$, but not over $B$. Note that the $\mathcal m$ are maximal ideals in $B$. $\endgroup$
    – Falter
    Commented Aug 10, 2013 at 18:35
  • $\begingroup$ Let me also remark that from my intuition the obstacle to the claim in question should come from the non-reducedness of $X$ and not from $Y$. $\endgroup$
    – Falter
    Commented Aug 10, 2013 at 18:36

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