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I have the following problem:

Let $X$ be a scheme and $x$ a closed point on it. If $F$ is a sheaf on X with nontrivial stalk $F_x$ at $x$, then one has a canonical surjective morphism $F\rightarrow G$, where $G$ is the structure sheaf of the point $x$. I dont understand how to describe $G$ and where this morphism comes from and why it should be surjective. Perhaps one can explain that in detail for me.

Thank you very much

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Dear Descartes, What makes you think such a canonical map exists? (It doesn't.) Also, what do you mean by the structure sheaf of $x$? Are you thinking of $x$ as the underlying topological space of Spec $\kappa(x)$? Regards, – Matt E Aug 9 '11 at 15:17
well, I mean x as a closed subscheme – Descartes Aug 9 '11 at 16:05
Dear Descartes, Yes, but which scheme structure? The reduced one? Regards, – Matt E Aug 9 '11 at 16:38
@Descartes: If this coming from a textbook or some lecture notes, can you give a citation? It may be more helpful to see it in context. If I had to guess, I would say that $G$ is the direct image sheaf under the obvious morphism $\{ x \} \hookrightarrow X$. (This is also known as the skyscraper sheaf.) – Zhen Lin Aug 9 '11 at 16:46
well, the proof where it occurs is rather involved; but it comes down to the following: you have a sheaf F, take a closed point in its support and then you have this morphism from F to k(x); I have called G what the author calls k(x); but he doesn't say what he means with that – Descartes Aug 9 '11 at 17:00
up vote 3 down vote accepted

The map exists (under some hypotheses --- e.g. $\mathcal F$ is a coherent sheaf of $\mathcal O_X$-modules), but is not canonical.

We can form the stalk $\mathcal F_x$, which is a module over $\mathcal O_{X,x}$. We can then form the fibre $\mathcal F_x /\mathfrak m_x \mathcal F_x$, which is a vector space over $\kappa(x)$. If the original sheaf $\mathcal F$ was coherent, and $x$ is in its support, then the fibre will be non-zero. Any non-zero vector space over $\kappa(x)$ admits a surjective map to $\kappa(x)$. (If a vector space $V$ is non-zero then the same is true of its dual.)

Choosing such a non-zero map $\mathcal F_x/\mathfrak m_x \mathcal F_x \to \kappa(x)$, we obtain the map $\mathcal F \to \mathcal G$ that you are asking about.

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Perfect, that's what I wanted. Sorry for the error with "canonical", but in the text it looked as if it was. – Descartes Aug 10 '11 at 7:31

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