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Let $f\colon X\rightarrow Y$ be a proper morphism of schemes and $\mathscr{F}$ a coherent $\mathscr{O}_X$-module. Given a point $y\in Y$, we have the following diagram.

$$\require{AMScd} \begin{CD} X_y @>{i}>> X\\ @VV{g}V @VV{f}V \\ \operatorname{Spec} k(y) @>{j}>> Y \end{CD}$$

It seems that if $\mathscr{F}_y:=i^*\mathscr{F}=0$, then by Nakayama's Lemma, there exists an affine open set $U\subset Y$ such that $\mathscr{F}|_{f^{-1}(U)}=0$?

For any $x\in X_y$, we have $(i^{*}\mathscr{F})_{x}=\mathscr{F}_x\otimes_{\mathscr{O}_{X,x}}\mathscr{O}_{X_y,x}=\mathscr{F}_x\otimes_{\mathscr{O}_{Y,y}}k(y)$. But $\mathscr{F}_x$ may not be a finitely generated $\mathscr{O}_{Y,y}$-module, so I can't use Nakayama's Lemma.

Or more specifically, to check that $f^*f_*\mathscr{L}\rightarrow \mathscr{L}$ is surjective, why is it enough to check this on any fiber $X_y$? This question arises in the proof that ampleness is an open condition in families(See [Kollár-Mori, Proposition 1.41]) and [Hartshorne, Proposition V.2.2].(Please read the image in the hyperlink.)

KReiser
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notime
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    Hi, a few tips: $\operatorname{Spec} k(y)$ for $\operatorname{Spec} k(y)$ produces better spacing than $\mathrm{Spec} k(y)$ $\mathrm{Spec} k(y)$; next, images of text are neither searchable nor accessible, and the images you've linked are rather large. Please consider placing the relevant portions inside your post, typeset with MathJax. – KReiser Oct 24 '24 at 03:57
  • This is related to https://math.stackexchange.com/q/1263724/571974 – notime Oct 24 '24 at 06:14

1 Answers1

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I'll answer the underlying question. Nakayama's lemma guarantees that if $\mathcal{F}$ is a quasicoherent sheaf of finite type over a scheme $X$, then $\mathcal{F}=0$ iff $\mathcal{F}_x\otimes_{\mathcal{O}_{X,x}} k(x)=0$ for all $x$. Letting $i_x:\operatorname{Spec} k(x)\to X$ be the natural inclusion, we can rewrite $\mathcal{F}_x\otimes_{\mathcal{O}_{X,x}} k(x)$ as $i_x^*\mathcal{F}$, so $\mathcal{F}=0$ iff $i_x^*\mathcal{F}=0$ for all $x\in X$. But if we have a map $f:X\to Y$ and we let $y=f(x)$, then $i_x:\operatorname{Spec} k(x)\to X$ factors as $$\operatorname{Spec} k(x)\stackrel{j_x}{\to} X_y \stackrel{i_{X_y}}{\to} X,$$ and clearly if $i_{X_y}^*\mathcal{F}=0$ for all $y\in Y$, then we get $i_x^*\mathcal{F}=j_x^*i_{X_y}^*\mathcal{F}=0$ for all $x$. So it suffices to prove that $i_{X_y}^*\mathcal{F}=0$ for all $y$ in order to show that $\mathcal{F}=0$.

Your question about an affine open neighborhood also has an affirmative answer. If $i_{X_y}^*\mathcal{F}=0$, then we have that $i_x^*\mathcal{F}=\mathcal{F}_x\otimes_{\mathcal{O}_{X,x}} k(x)=0$ for all $x\in X_y$, hence $\mathcal{F}_x=0$ for all $x\in X_y$, and so the support of $\mathcal{F}$ is contained in a closed subset $S\subset X$ which does not intersect $X_y$. As a proper morphism is closed, $f(S)\subset Y$ is a closed subset avoiding $y$, hence an affine open neighborhood of $y$ contained in $Y\setminus f(S)$ can be taken to play the role of $U$.

(NB, if you are working over a variety or any other situation where the support of a sheaf of finite type is closed and closed points are dense in every nonempty closed subset, you can do a little better than this - it suffices to check these conditions in the first paragraph over closed points in that case. For instance, this is useful in Hartshorne when he occasionally makes a reduction to only looking at closed points without necessarily saying so much.)

KReiser
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