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Hartshorne asserts in 8.9.1 that $\mathcal I$, the kernel of the diagonal morphism $X \to X \times_Y X$, has a natural $\mathcal O_{\Delta X}$-module structure.

My problem is that $\mathcal I$ is an ideal sheaf on the whole ${X \times_Y X}$, whereas $\mathcal O_{\Delta X}$ is the restriction of the sheaf $\mathcal O_{X \times_Y X}$ to $\Delta X$. So there are only two possibilities: Either we restrict $\mathcal I$ to $\Delta X$, where it is zero, or we extend $\mathcal O_{\Delta X}$ to $\mathcal O_{X \times_Y X}$ by zeros. In either way we have a trivial action, which I highly doubt to be the natural $\mathcal O_{\Delta X}$-module structure mentioned by Hartshorne.

Rodrigo
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  • See my answer here: http://math.stackexchange.com/questions/229986/induced-module-structure-of-the-sheaf-of-ideals-with-application-to-the-sheaf-of/231008#231008 – Keenan Kidwell Jun 17 '13 at 11:50

1 Answers1

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The point is that the discussion is about $\mathcal{I} / \mathcal{I}^{2}$. The module structure comes from the right factor in the isomorphism

$\mathcal{I} / \mathcal{I} ^{2} \simeq \mathcal{I} \otimes _{\mathcal{O}_{X}} \mathcal{O} _{X} / \mathcal{I} \simeq \mathcal{I} \otimes _{\mathcal{O}_{X}} \mathcal{O} _{\Delta(X)}$,

since $\mathcal{O} _{\Delta(X)} \simeq \mathcal{O} _{X} / \mathcal{I}$.

(Notice I have ignored the subtle fact that $\Delta(X) \subseteq X$ is possibly a non-closed subscheme in the non-separated case, because it seems it is not the source of your confusion.)

Piotr Pstrągowski
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