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?
Best Answer
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.