The Stacks project

Proof. The restriction functor is fully faithful by Lemma 58.17.1.

Let $U_1 \to Y$ be a finite étale morphism. To finish the proof we will show that $U_1$ is in the essential image of the restriction functor.

For $n \geq 1$ let $Y_ n$ be the $n$th infinitesimal neighbourhood of $Y$. By Lemma 58.8.3 there is a unique finite étale morphism $\pi _ n : U_ n \to Y_ n$ whose base change to $Y = Y_1$ recovers $U_1 \to Y_1$. Consider the sheaves $\mathcal{F}_ n = \pi _{n, *}\mathcal{O}_{U_ n}$. We may and do view $\mathcal{F}_ n$ as an $\mathcal{O}_ X$-module on $X$ which is locally isomorphic to $(\mathcal{O}_ X/f^{n + 1}\mathcal{O}_ X)^{\oplus r}$. This $(\mathcal{F}_ n)$ is a finite locally free object of $\textit{Coh}(X, \mathcal{I})$. By assumption there exists a finite locally free $\mathcal{O}_ X$-module $\mathcal{F}$ and a compatible system of isomorphisms

of $\mathcal{O}_ X$-modules.

To construct an algebra structure on $\mathcal{F}$ consider the multiplication maps $\mathcal{F}_ n \otimes _{\mathcal{O}_ X} \mathcal{F}_ n \to \mathcal{F}_ n$ coming from the fact that $\mathcal{F}_ n = \pi _{n, *}\mathcal{O}_{U_ n}$ are sheaves of algebras. These define a map

in the category $\textit{Coh}(X, \mathcal{I})$. Hence by assumption we may assume there is a map $\mu : \mathcal{F}\otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{F}$ whose restriction to $Y_ n$ gives the multiplication maps above. By faithfulness of the functor in the statement of the lemma, we conclude that $\mu$ defines a commutative $\mathcal{O}_ X$-algebra structure on $\mathcal{F}$ compatible with the given algebra structures on $\mathcal{F}_ n$. Setting

we obtain a finite locally free scheme $\pi : U \to X$ whose restriction to $Y$ is isomorphic to $U_1$. The the discriminant of $\pi$ is the zero set of the section

constructed in Discriminants, Section 49.3. Since the restriction of this to $Y_ n$ is an isomorphism for all $n$ by Discriminants, Lemma 49.3.1 we conclude that it is an isomorphism. Thus $\pi$ is étale by Discriminants, Lemma 49.3.1. $\square$