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I am interested in the graph $\Gamma_f$ of a morphism

$f: X\rightarrow Y$ between two sufficiently nice schemes $X,Y$.

One knows that it is a closed subset of $X\times Y$ (when the schemes are nice, say varieties over a field).

I would like to know the following: if you endow it with the reduced structure, what are the stalks of it's structure sheaf in a point $(x,f(x))$ ?

Thanks you very much!

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You're really posting a lot of questions here and on MO. Are you really thinking your questions over carefully before you ask them here? Do you really have time to digest all the helpful answers you've been getting? –  Dan Petersen Aug 22 '11 at 13:30

1 Answer 1

Let $f:X\to Y$ be a morphism of $S$-schemes. The graph morphism $\gamma_f:X\to X\times_S Y$ is the pull-back of the diagonal morphism $\delta: Y\to Y\times_S Y$ along $f\times id_Y: X\times_S Y \to Y\times_S Y$. This implies that if $\delta$ is a closed embedding (i.e. $Y$ is separated over $S$) so is $\gamma_f$. So $\gamma_f$ induces an isomorphism between $X$ and its image $\Gamma_f \subset X\times Y$.

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Why is it an isomorphism on it's image? A closed immersion only yields an epimorphism on the structure sheaves. –  Descartes Aug 25 '11 at 9:33
You are confusing the target (or codomain) $X\times Y$ and the schematic image $\Gamma_f$ of $\gamma_f$. By definition $\Gamma_f$ is the closed subscheme defined by the sheaf of ideal $\mathcal{I} = ker(\mathcal{O}_{X\times Y} \to (\gamma_f)_*\mathcal{O}_{X} )$. Since the morphism is surjective we get $\mathcal{O}_{X\times Y}/\mathcal{I} = \mathcal{O}_{\Gamma_f} = (\gamma_f)_*\mathcal{O}_{X}$. –  YBL Sep 4 '11 at 11:28

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