The Stacks project

Lemma 58.20.3. Let $(A, \mathfrak m)$ be a Noetherian local ring. Set $X = \mathop{\mathrm{Spec}}(A)$ and let $U = X \setminus \{ \mathfrak m\} $. Assume $A$ is normal of dimension $\geq 2$. The functor

\[ \textit{FÉt}_ U \longrightarrow \left\{ \begin{matrix} \text{finite normal }A\text{-algebras }B\text{ such} \\ \text{that }\mathop{\mathrm{Spec}}(B) \to X\text{ is étale over }U \end{matrix} \right\} , \quad V \longmapsto \Gamma (V, \mathcal{O}_ V) \]

is an equivalence. Moreover, $V = Y \times _ X U$ for some $Y \to X$ finite étale if and only if $B = \Gamma (V, \mathcal{O}_ V)$ is finite étale over $A$.

Proof. Observe that $\text{depth}(A) \geq 2$ because $A$ is normal (Serre's criterion for normality, Algebra, Lemma 10.157.4). Thus the final statement follows from Lemma 58.20.2. Given $\pi : V \to U$ finite étale, set $B = \Gamma (V, \mathcal{O}_ V)$. If we can show that $B$ is normal and finite over $A$, then we obtain the displayed functor. Since there is an obvious quasi-inverse functor, this is also all that we have to show.

Since $A$ is normal, the scheme $V$ is normal (Descent, Lemma 35.18.2). Hence $V$ is a finite disjoint union of integral schemes (Properties, Lemma 28.7.6). Thus we may assume $V$ is integral. In this case the function field $L$ of $V$ (Morphisms, Section 29.49) is a finite separable extension of the fraction field of $A$ (because we get it by looking at the generic fibre of $V \to U$ and using Morphisms, Lemma 29.36.7). By Algebra, Lemma 10.161.8 the integral closure $B' \subset L$ of $A$ in $L$ is finite over $A$. By More on Algebra, Lemma 15.23.20 we see that $B'$ is a reflexive $A$-module, which in turn implies that $\text{depth}_ A(B') \geq 2$ by More on Algebra, Lemma 15.23.18.

Let $f \in \mathfrak m$. Then $B_ f = \Gamma (V \times _ U D(f), \mathcal{O}_ V)$ (Properties, Lemma 28.17.1). Hence $B'_ f = B_ f$ because $B_ f$ is normal (see above), finite over $A_ f$ with fraction field $L$. It follows that $V = \mathop{\mathrm{Spec}}(B') \times _ X U$. Then we conclude that $B = B'$ from Lemma 58.20.1 applied to $\mathop{\mathrm{Spec}}(B') \to X$. This lemma applies because the localizations $B'_{\mathfrak m'}$ of $B'$ at maximal ideals $\mathfrak m' \subset B'$ lying over $\mathfrak m$ have depth $\geq 2$ by Algebra, Lemma 10.72.11 and the remark on depth in the preceding paragraph. $\square$

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