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.
Hom(P,P) field => finite over k
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1 Answers
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: $Hom_{\mathcal O_X}(P,P)=\kappa (x)$