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

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?