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The corollary says - An arbitrary morphism $f:X\longrightarrow Y$ is separated if and only if the image of the diagonal morphism is a closed subset of $X\times_{Y} X$.

I am studying the proof of this proposition. One way is obvious. To prove the other way, we need to prove that $\Delta$ is a homeomorphism onto $\Delta(X)$ and the morphism of sheaves $O_{X\times_{Y} X}\longrightarrow\Delta_*{O_X}$ is surjective. Homeomorphism again is easy to prove. But to prove the surjectivity, Hartshorne says it is a local question. Does that mean we check it at the stalk level? I don't understand this part of the proof.

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An arbitrary scheme is a collection of affine schemes patched together, and an affine scheme is of the form $ \operatorname{Spec}(A) $ for some ring $A$.

To see if a map of sheaves is surjective, it is enough to check it affine-locally. Hence we can pick a neighborhood $U$ of $P$ small enough so that $\Delta(U) \subset V$ for a neighborhood $V$ of $\Delta(P)$ in $X\times X$.

Such affine neighborhoods are affine schemes, so they are of the form $$ U= \operatorname{Spec}(A) \quad\text{and}\quad V=\operatorname{Spec}(B) $$ for some rings $A$ and $B$.

Here, locally, the map we want to prove to be surjective coincides with the homomorphism of rings $$ B\otimes_A B \to B $$ and the fact that restriction of the morphism $\Delta$ to $U$ $$ \Delta : \operatorname{Spec}(B) = U\to U\times U = \operatorname{Spec}(B\otimes_A B) $$ is a closed embedding exactly means (by definition) that the above map of rings is surjective.

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This may be very trivial, but why is it true that the map we want is locally the ring homomorphism? The map we want to prove is surjective is the map of sheaves right. And if at each open subset, the map is surjective then the map of sheaves is. Are you using that? –  poorna Dec 9 '13 at 17:59
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The point is that if $Spec(A) \subset X$ is an affine open, then the structure sheaf $\mathcal{O}_X$ of $X$ restricted to $U$ is just $A$. This follows directly from the definitions –  Abramo Dec 9 '13 at 18:04
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Ah okay! Thank you so much! Your answer is very helpful! –  poorna Dec 9 '13 at 18:09
    
Cool, I'm so happy when I can help other people! I'm also trying to learn AG, and it is a very hard task. But it's also fun! In my case Hartshorne was making it even harder, so I switched to Ravi Vakil's notes, which I think are much better for a newbie. You can find them here: math.stanford.edu/~vakil/216blog –  Abramo Dec 9 '13 at 18:14
    
Yes I find AG hard and sometimes overwhelming because I am new to it. I have looked at the notes, but haven't really used it. I should perhaps switch to that too! –  poorna Dec 9 '13 at 18:22

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