# Stalk of a pushforward sheaf in algebraic geometry

Excuse me if this is a naive question. Let $f : X \to Y$ be a morphism of varieties over a field $k$ and $\mathcal{F}$ a quasi-coherent sheaf on $X$. I know that for general sheaves on spaces not much can be said about the stalk $(f_*\mathcal{F})_y$ at $y \in Y$ (let's just talk about closed points), but does anything nice happen in this situation? If $f$ is proper then the completion of the stalk is described by the formal function theorem, but I'm interested in the honest stalk.

Suppose, for instance, that $X = \text{Spec } A$ is affine (so $\mathcal{F} = \widetilde{M}$ for some $A$-module $M$) and that the scheme-theoretic fiber of $f$ over the closed point $y \in Y$ is reduced. Write this fiber as a union of irreducible components $Z_1 \cup \cdots \cup Z_r$ corresponding to prime ideals $\mathfrak{p}_1,\cdots,\mathfrak{p}_r \subset A$, so there is an associated semi-local ring $S^{-1}A$ with $S = A \setminus (\mathfrak{p}_1 \cup \cdots \cup \mathfrak{p}_r)$. Hopefully here $(f_*\mathcal{F})_y$ is just $S^{-1}M$ regarded as an $\mathcal{O}_{Y,y}$-module via the natural map $\mathcal{O}_{Y,y} \to S^{-1}A$.

Does this make sense? Is something like this true when $X$ is not affine and/or the fiber is nonreduced?

Edit: As Georges's answer shows, this cannot possibly be true in general. I wonder if there is still hope when $f$ is proper? My example in the comments below (the squaring map $\mathbb{A}^1 \to \mathbb{A}^1$ with $\mathcal{F} = \mathcal{O}$ and $y = 1$) is actually consistent with my guess above: $(f_*\mathcal{F})_y$ is the localization of $k[t]$ at $k[t^2] \setminus (t^2-1)k[t^2]$. It is not hard to see that this coincides with the localization at $k[t] \setminus ((t-1) \cup (t+1))$, i.e. the semilocal ring at $\{ \pm 1 \}$.

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Is the answer $(f_*\mathcal{F})_y=\Gamma(X_y, \mathcal{F}|_{X_y})$ complete enough for what you are looking for? –  Matt Jul 21 '12 at 20:56
@Matt: I don't think this is right. Consider the squaring map $f : \mathbb{A}^1 \to \mathbb{A}^1$ with $\mathcal{F} = \mathcal{O}_{\mathbb{A}^1}$ and $y = 1$. Then $(f_*\mathcal{F})_y$ is the localization of $k[t]$ at $(t-1) \cap k[t^2]$, but $\Gamma(X_y,\mathcal{F}|_y) = k[t]_{(t-1)} \times k[t]_{(t+1)}$. These appear to be different. –  Justin Campbell Jul 21 '12 at 21:59
Edit: Should be $\mathcal{F}|_{X_y}$, not $\mathcal{F}|_y$. –  Justin Campbell Jul 21 '12 at 22:07
Note, however, that the two sides of your equation have the same completion at $(t-1) \cap k[t^2] = (t^2-1)k[t^2]$, in accordance with the formal function theorem. –  Justin Campbell Jul 21 '12 at 22:17
Oops. I was being dumb. Of course what I wrote can't be true or else several major theorems in algebraic geometry would become trivial. What you wrote as an answer was actually what I was thinking when I wrote that. –  Matt Jul 21 '12 at 23:31

## 2 Answers

I think the best we can do is the following. The question is local on the base, so we may as well assume $Y = \text{Spec } A$ and $X = \text{Spec } B$ are affine and $\mathcal{F} = \widetilde{M}$ where $M$ is a $B$-module. Then if $y = \mathfrak{p} \in \text{Spec } A$ then $(f_*\mathcal{F})_y$ is the $A_{\mathfrak{p}}$-module $M_{\mathfrak{p}}$, where we think of $M$ now as an $A$-module.

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You need that $f$ is qcqs ,otherwise $f_*$ won't preserve quasi-coherence. –  Martin Brandenburg Oct 5 '13 at 9:47

No, I'm afraid what you say does not work, even in the simplest situation:

a) Let $k$ be a field , $X=\mathbb A^1_k=Spec (k[t])$ , $Y=Spec (k)=\lbrace y\rbrace$ and let $f:X\to Y$ be the obvious morphism.
Then for $\mathcal F=\mathcal O_X$ we have $(f_*\mathcal F)_y=k[t]$.

b) On the other hand, still in your notation, $X=Z_1$, $\mathfrak p_1=(0)$, $M=k[t]$, $S=k[t]\setminus \lbrace 0\rbrace$ and thus $S^{-1}M=k(t)$

c) Conclusion: $$(f_*\mathcal F)_y=k[t]\neq S^{-1}M=k(t)$$

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Thanks Georges, I was just realizing that my guess was wrong... you can see it in my example in the comments as well. –  Justin Campbell Jul 21 '12 at 22:13
I'm glad we came to the same conclusion, Justin. –  Georges Elencwajg Jul 21 '12 at 22:16
Actually, my example is consistent with my guess: I have edited the question accordingly. –  Justin Campbell Jul 22 '12 at 18:36