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$X$: projective scheme over a scheme $S$.

$E, F$: $\mathscr{O}_X$-modules, flat/$S$

$\phi$: $E \rightarrow F$ : morphism s.t. $\phi_t$: $E_t \rightarrow F_t$ is zero morphism for all $t \in S$

Then, is $\phi$ zero morphism ?

I'd be glad if you could tell me something! (Please give me some comments about the comment below!)

Edit: especially I am interested in the case $X = Y \times S$ ,where $Y$: projective surface / $\mathbb{C}$, $S: \mathbb{C}$-scheme

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No. For instance, take $$ X = S = \mathrm{Spec}(\Bbbk[\epsilon]/\epsilon^2), \qquad E = F = \Bbbk[\epsilon]/\epsilon^2, \qquad \phi = \epsilon. $$ Then for the unique point $t \in S$ the morphism $\phi_t$ is zero, while $\phi$ is not.

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  • $\begingroup$ Thank you for the answer! If$Y, S$: $\mathbb{C}$-schemes and $Y$: projective over $\mathbb{C}$ of dim $>0$ and $X := Y \times S$,then is the question true ? $\endgroup$ – Walter field Dec 24 '19 at 15:51
  • $\begingroup$ @Walter field: I think you can cross the above example by P1. Let everything be over $\mathbb C$, $D:=Spec(\mathbb{C}[\epsilon]/(\epsilon^2)$, $X:=D\times\mathbb{P}^1$, $\pi:X\to D$, $E:=I$ where $I=\pi^{*}(\epsilon)$, $F:=\mathcal O_X$, and $E\to F$ the inclusion. $\endgroup$ – Qfwfq Dec 24 '19 at 16:25
  • $\begingroup$ Thank you for the comment! But, how do we prove $E$ is flat over $D$ ? $\endgroup$ – Walter field Dec 25 '19 at 2:00

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