I was wondering why for a closed subscheme $Z \subset X$ we have $\text{Hom}_{\mathcal{O}_X}(\mathcal{O}_Z, \mathcal{O}_Z) \cong H^0(\mathcal{O}_Z)$? How does one construct such an isomorphism?
$\text{Hom}_{\mathcal{O}_X}(\mathcal{O}_Z, \mathcal{O}_Z) \cong H^0(\mathcal{O}_Z)$?
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algebraic-geometry
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4I would say that a morphism $\phi : O_Z \to O_Z$ is determined by the image of $1 \in O_Z$ so this is an element of $H^0(O_Z)$. Conversely any section $s$ gives a morphism of $O_X$ modules. – 2017-01-29
1 Answers
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One should note that technically $\operatorname{Hom}_{\mathcal O_X}(\mathcal O_Z, \mathcal O_Z)$ means $\operatorname{Hom}_{\mathcal O_X}(i_*\mathcal O_Z, i_*\mathcal O_Z)$ where $i: Z \hookrightarrow X$ is the closed immersion.
The adjunction between $i^*$ and $i_*$ yields a natural isomorphism $$\operatorname{Hom}_{\mathcal O_X}(i_*\mathcal O_Z, i_*\mathcal O_Z) = \operatorname{Hom}_{\mathcal O_Z}(i^*i_*\mathcal O_Z,\mathcal O_Z)$$ and we conclude by the fact that the natural map $$i^*i_*\mathcal O_Z \to \mathcal O_Z$$ is an isomorphism. This can be checked affine locally and there it is just the isomorphism $$R/I \otimes_R R/I \to R/I.$$
In particular we have a natural isomorphism
$$\operatorname{Hom}_{\mathcal O_X}(i_*\mathcal O_Z, i_*\mathcal O_Z) = \operatorname{Hom}_{\mathcal O_Z}(i^*i_*\mathcal O_Z,\mathcal O_Z) = \operatorname{Hom}_{\mathcal O_Z}(\mathcal O_Z,\mathcal O_Z) = H^0(Z,\mathcal O_Z).$$