Confused about why the conormal exact sequence is what it is on a scheme

Question

Consider a composition of morphisms of schemes, $$Z \stackrel{j}{\longrightarrow} X \stackrel{f}{\longrightarrow} Y,$$ where $j: Z \rightarrow X$ is a closed immersion with sheaf of ideals $\mathscr{I}$. The conormal exact sequence for the sheaves of differentials should read, $$ j^{*}(\mathscr{I} / \mathscr{I}^{2}) \longrightarrow j^{*} \Omega_{X / Y} \longrightarrow \Omega_{Z/Y} \longrightarrow 0. $$ I am trying to prove this via the corresponding conormal sequence for modules. So suppose the composition of morphisms of schemes above are of affine schemes. Say $X = \text{spec}B$, $Y = \text{spec}A$, and $Z = \text{spec}C$, with $C \simeq B / I$ for some ideal $I$ of $B$. Then the conormal exact sequence of modules reads, $$ I / I^{2} \longrightarrow \Omega_{B/A} \otimes_{B} C \longrightarrow \Omega_{C/A} \longrightarrow 0, $$ where this is an exact sequence of $C$-modules. Now we just have to take the corresponding sequence of quasicoherent $\mathcal{O}_{Z}$-modules. We are viewing $I/I^{2}$ as a $C$-module via the isomorphism, $$ I \otimes_{B} C \simeq I \otimes_{B} B / I \simeq I / I^{2}. $$ But then $I/I^{2}$ is just the $B$-module $I$ with the change of ring $- \otimes_{B} C$ applied. In that case, the exact sequence of $\mathcal{O}_{Z}$-modules should just be $$ j^{*}\mathscr{I} \longrightarrow j^{*} \Omega_{X / Y} \longrightarrow \Omega_{Z/Y} \longrightarrow 0. $$ So where is my flaw in reasoning? Are $j^{*}\mathscr{I}$ and $j^{*}(\mathscr{I} / \mathscr{I}^{2})$ somehow canonically isomorphic or what?

Answer 1

Yes, $j^* \mathscr{I}$ and $j^* \mathscr{I}/\mathscr{I}^2$ are canonically isomorphic in this case. Working affine locally we see $j^* \mathscr{I} = I \otimes_B B/I \cong I/I^2$. The last isomorphism is an iso of $B$-modules, but obviously $I \otimes_B B/I$ has the structure of a $C = B/I$-module, so we can give $I/I^2$ the induced $C$-module structure, and this is still an isomorphism of $C$-modules.

Similarly, locally we see $j^* I/I^2 = I/I^2 \otimes_B B/I \cong (I/I^2)/(I(I/I^2)) = (I/I^2)/0 = I/I^2$. I think this should be thought of as saying that $I/I^2$ has a canonical $C$-module structure.

More generally, if $j : Z \hookrightarrow X$ is a closed embedding with ideal sheaf $\mathscr{I}$, then quasicoherent sheaves on $Z$ are equivalent to qcoh $j_* \mathcal{O}_Z$-modules. I think the explicit correspondence is given by $j_*$ and $j^*$. Looking at the closed embedding SES we see $j_* \mathcal{O}_Z = \mathcal{O}_X/\mathscr{I}$. As $\mathscr{I}/\mathscr{I}^2$ is an $\mathcal{O}_X$-module, but also naturally an $\mathcal{O}_X/\mathscr{I}$-module, it can be thought of as a qcoh $\mathcal{O}_Z$-module. This is why the $j^*$ is sometimes ommited when defining the cornormal sheaf. So while $j^* \mathscr{I}$ is isomorphic to $j^* \mathscr{I}/\mathscr{I}^2$, it makes sense to think of $\mathscr{I}/\mathscr{I}^2$ as a sheaf on both $Z$ and $X$, but $\mathscr{I}$ has no natural intepretation as a sheaf on $Z$.

Some other possibly helpful answers are Pullback of ideal sheaf defining closed immersion and Conormal Sheaf (Morphisms of Schemes, Stacks Project).