Lemma 30.23.10. Let $f : X' \to X$ be a morphism of Noetherian schemes. Let $Z \subset X$ be a closed subscheme and denote $Z' = f^{-1}Z$ the scheme theoretic inverse image. Let $\mathcal{I} \subset \mathcal{O}_ X$, $\mathcal{I}' \subset \mathcal{O}_{X'}$ be the corresponding quasi-coherent sheaves of ideals. If $f$ is flat and the induced morphism $Z' \to Z$ is an isomorphism, then the pullback functor $f^* : \textit{Coh}(X, \mathcal{I}) \to \textit{Coh}(X', \mathcal{I}')$ (Lemma 30.23.9) is an equivalence.

Proof. If $X$ and $X'$ are affine, then this follows immediately from More on Algebra, Lemma 15.89.3. To prove it in general we let $Z_ n \subset X$, $Z'_ n \subset X'$ be the $n$th infinitesimal neighbourhoods of $Z$, $Z'$. The induced morphism $Z_ n \to Z'_ n$ is a homeomorphism on underlying topological spaces. On the other hand, if $z' \in Z'$ maps to $z \in Z$, then the ring map $\mathcal{O}_{X, z} \to \mathcal{O}_{X', z'}$ is flat and induces an isomorphism $\mathcal{O}_{X, z}/\mathcal{I}_ z \to \mathcal{O}_{X', z'}/\mathcal{I}'_{z'}$. Hence it induces an isomorphism $\mathcal{O}_{X, z}/\mathcal{I}_ z^ n \to \mathcal{O}_{X', z'}/(\mathcal{I}'_{z'})^ n$ for all $n \geq 1$ for example by More on Algebra, Lemma 15.89.2. Thus $Z'_ n \to Z_ n$ is an isomorphism of schemes. Thus $f^*$ induces an equivalence between the category of coherent $\mathcal{O}_ X$-modules annihilated by $\mathcal{I}^ n$ and the category of coherent $\mathcal{O}_{X'}$-modules annihilated by $(\mathcal{I}')^ n$, see Lemma 30.9.8. This clearly implies the lemma. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).