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Let $P$ be the skyscraper sheaf $k(x)$ of a closed point $x$ on a smooth projective variety $X/k$. Why is $\operatorname{Hom}_k(P,P)$ a field? I think this holds iff $k(x) = k$, but not for strict extensions.

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Wait: I assume all Hom's are finite dimensional. And that $P = \kappa(x)$. –  user5262 Nov 27 '11 at 17:15
    
But we have $\mathrm{Hom}_k(\kappa(x),\kappa(x)) = \mathrm{Mat}_{[\kappa(x):k]}(k)$, which is not a field for $[\kappa(x):k] > 1$? –  user5262 Nov 27 '11 at 17:41
    
By finite over $k$ I mean that the Hom is a finite dimensional $k$-vector space. –  user5262 Nov 27 '11 at 17:48
    
I'm asking if it is a finite field extension of $k$. –  user5262 Nov 27 '11 at 18:10
    
Yes, but why is the Hom-set a field in the 2nd question? –  user5262 Nov 27 '11 at 18:29
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1 Answer

up vote 4 down vote accepted

As usual, things will be clearer in a more general context:
Consider a scheme $X$, an $\mathcal O_X$-Module $\mathcal F $, a point $x\in X$, an $\mathcal O_{X,x}$-module $M$ and the associated skyscraper sheaf $\mathcal G^x (M)$, which is also an $\mathcal O_X$-Module. We then have the adjunction formula:
$$Hom_{\mathcal O_X}(\mathcal F, \mathcal G^x (M)) =Hom_{\mathcal O_{X,x}}(\mathcal F_x,M) $$
In the special case where $\mathcal F=\mathcal G^x (M)$, this yields $$ Hom_{\mathcal O_X}(\mathcal G^x (M), \mathcal G^x (M)) =Hom_{\mathcal O_{X,x}}(M,M)$$
[remember that $(\mathcal G^x (M))_x=M$]
And in your even more special case where $M=\kappa (x)$ ( and where your notation is $P=\mathcal G^x (M) $ ) :
$$Hom_{\mathcal O_X}(P,P)=Hom_{\mathcal O_{X,x}}(\kappa (x),\kappa (x)) $$
If you take into account that $\kappa (x)$ is killed by ${\mathfrak m}_x$ you get $$ Hom_{\mathcal O_X}(P,P)=Hom_{\kappa (x)}(\kappa (x),\kappa (x)) $$
which is indeed isomorphic to the field $\kappa (x)$.
(Smoothness and projectiveness of $X$ are irrelevant and so is the closedness of $x$)

Summary: $Hom_{\mathcal O_X}(P,P)=\kappa (x)$

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