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In the definition of the conormal sheaf, we are given a locally closed immersion $X \to Y$, which factors through some closed subscheme $Z$ so that we have $X \to Z \to Y$, where the first map is a closed immersion and the second an open immersion.

We then take the ideal sheaf $\mathcal{I}$ corresponding to the closed immersion $f:X\to Z$, and define the conormal to be $\mathcal{I}/\mathcal{I^2}$, viewed as an $\mathcal{O}_X$-module (is it correct to say that the conormal is $f^*(\mathcal{I}/\mathcal{I^2})$?).

I've tried to show that is independent of the choice of $Z$, but haven't really managed to work it out successfully (perhaps factoring through the scheme-theoretic closure might work?). How does one show that the conormal is well-defined?

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2 Answers 2

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Oops, this is rather silly. If $\Delta (X)\subset U,V\subset Y$ so that $X\to U\subset Y,X\to V\subset Y$ are both the same locally closed immersion $\Delta$, and also noting that if we ever have a morphism $f:X\to Y$ with $U\subset X$ maps into $V\supset f(U)$, then $f^*$ commutes with restriction, we get $\Delta^*(\mathcal{I}/{\mathcal{I}^2})|_X=\Delta_X^*(\mathcal{I}/\mathcal{I}^2|_V)=\Delta_X^*(\mathcal{I}/\mathcal{I}^2|_U)$. In particular, there is nothing special about $\mathcal{I}/\mathcal{I^2}$, If I'm not mistaken.

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  • $\begingroup$ You wrote $X\rightarrow U\subset Y$. Later you wrote $f: X\rightarrow Y$ with $U\subset X$. Is $X$ contained in $U$ or contains $U$? $\endgroup$
    – aaa acb
    Jun 24, 2023 at 12:20
  • $\begingroup$ I find [noting that if we ever have a morphism $f:X\to Y$ with $U\subset X$ maps into $V\supset f(U)$] confusing. $\endgroup$
    – aaa acb
    Jun 24, 2023 at 12:23
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Factor the locally closed immersion $i: X\rightarrow Y$ as below, such that $i_U: X\rightarrow U$ is a closed immersion and $U$ is an open subscheme of $X$.

factori1

Let $\mathscr{I}_U$ be the ideal sheaf for $X$ in $U$. The conormal sheaf is defined as:

$$ \mathscr{N}_{X/Y}^{\vee} = i_U^*\mathscr{I}_U. $$

To show the definition is independent of the choice of $U$, pick another open subscheme $V$ of $Y$ having $X$ as a closed subscheme. Then $X$ is a closed subscheme of $U\cap V$, and we have the following commutative diagram, where $i_{U\cap V}$ is a closed immersion with ideal sheaf $\mathscr{I}_{U\cap V}=\mathscr{I}_U|_{U\cap V}$.

ucapv

Thus $$ i_U^*\mathscr{I}_U = i_{U\cap V}^*\left(\mathscr{I}_{U}|_{U\cap V}\right) = i_{U\cap V}^*\mathscr{I}_{U\cap V}. $$ Similarly $i_V^*\mathscr{I}_V = i_{U\cap V}^*\mathscr{I}_{U\cap V}$.

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