I'm having some silly confusion about the definition so I would really appreciate it if someone can help me out with this definition.

Let $i: X \hookrightarrow Y$ be a closed immersion cut out by $\mathscr{I} \subseteq \mathcal{O}_Y$. Then, Vakil defines the conormal sheaf of this closed immersion to be $\mathscr{I}/\mathscr{I}^2$ as viewed as a quasi-coherent sheaf on X. The italicized part is essentially my confusion.

How are we viewing this as a quasi-coherent sheaf on $X$? The most obvious way to do this is to take $i^*(\mathscr{I}/\mathscr{I}^2)$ but I don't think this is right. In particular, this question seems to take it to be $i^* \mathscr{I} = i^{-1}(\mathscr{I}/\mathscr{I}^2)$. I find this unsettling though.

By definition, $\Delta: X \to X\times_YX$ is a locally closed immersion and we define $\Omega_{X/Y}$ to be the conormal sheaf of this embedding. This would be fine, although Hartshorne defines $\Omega_{X/Y} = \Delta^*(\mathscr{I}/\mathscr{I}^2)$ instead of what I would have expected from above, $\Delta^*(\mathscr{I})$. What is going on here?

To summarize, is the conormal sheaf of $i: X \hookrightarrow Y$ defined to be $i^*(\mathscr{I})$, $i^*(\mathscr{I}/\mathscr{I}^2)$, or both?

Thank you very much for any help.


1 Answer 1


These are all the same sheaf.

Basically, this is the isomorphism $I/I^2 \cong I \otimes A / I$ for an ideal $I \subset A$. This tells you that if $\iota :Z \to X$ is a closed immersion cut out by an ideal $\mathcal{I}$ then the following sheaves are the same,

  1. $\iota^{*} (\mathcal{I} / \mathcal{I}^2) $
  2. $\iota^{-1} (\mathcal{I} / \mathcal{I}^2) $
  3. $\iota^{*} \mathcal{I}$

This is because $\iota^* \mathcal{F} = \iota^{-1} \mathcal{F} \otimes \mathcal{O}_Z$ but $\mathcal{I} / \mathcal{I}^2$ is already a $\mathcal{O}_Z = \mathcal{O}_X / \mathcal{I}$-module and $\mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{O}_Z = \mathcal{I} / \mathcal{I}^2$.

Furthermore, $\iota_*$ induced an equivalence of categories from $\mathcal{O}_Z$-modules to $\mathcal{O}_X$-modules $\mathcal{F}$ such that $\mathcal{I} \cdot \mathcal{F} = 0$ (in particular supported along $Z$). If this is confusing, think about what the difference between an $A$-module and an $A / I$-module is. Therefore, we can view $\mathcal{O}_Z$ and $\iota_* \mathcal{O}_Z$ as essentially the same sheaf as well as view, $$ \iota^* \mathcal{I} = \iota^* (\mathcal{I} / \mathcal{I}^2) = \iota^{-1} (\mathcal{I} / \mathcal{I}^2) $$ as essentially the same sheaf as $\mathcal{I} / \mathcal{I}^2$. This is why Ravi said viewed as a sheaf on $X$ rather than pulled back to $X$ because once you quotient its already a $\iota_* \mathcal{O}_Z$-module and therefore by the equivalence of categories already essentially the same as $\iota^{-1} (\mathcal{I} / \mathcal{I}^2)$ on $Z$.

  • $\begingroup$ @Daniel edited thanks $\endgroup$
    – Ben C
    Dec 17, 2021 at 1:18

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