Lemma 58.10.2. Let $(A, \mathfrak m)$ be a local ring. Set $X = \mathop{\mathrm{Spec}}(A)$ and let $U = X \setminus \{ \mathfrak m\}$. If the punctured spectrum of the strict henselization of $A$ is connected, then

$\textit{FÉt}_ X \longrightarrow \textit{FÉt}_ U,\quad Y \longmapsto Y \times _ X U$

is a fully faithful functor.

Proof. Assume $A$ is strictly henselian. In this case any finite étale cover $Y$ of $X$ is isomorphic to a finite disjoint union of copies of $X$. Thus it suffices to prove that any morphism $U \to U \amalg \ldots \amalg U$ over $U$, extends uniquely to a morphism $X \to X \amalg \ldots \amalg X$ over $X$. If $U$ is connected (in particular nonempty), then this is true.

The general case. Since the category of finite étale coverings has an internal hom (Lemma 58.5.4) it suffices to prove the following: Given $Y$ finite étale over $X$ any morphism $s : U \to Y$ over $X$ extends to a morphism $t : X \to Y$ over $X$. Let $A^{sh}$ be the strict henselization of $A$ and denote $X^{sh} = \mathop{\mathrm{Spec}}(A^{sh})$, $U^{sh} = U \times _ X X^{sh}$, $Y^{sh} = Y \times _ X X^{sh}$. By the first paragraph and our assumption on $A$, we can extend the base change $s^{sh} : U^{sh} \to Y^{sh}$ of $s$ to $t^{sh} : X^{sh} \to Y^{sh}$. Set $A' = A^{sh} \otimes _ A A^{sh}$. Then the two pullbacks $t'_1, t'_2$ of $t^{sh}$ to $X' = \mathop{\mathrm{Spec}}(A')$ are extensions of the pullback $s'$ of $s$ to $U' = U \times _ X X'$. As $A \to A'$ is flat we see that $U' \subset X'$ is (topologically) dense by going down for $A \to A'$ (Algebra, Lemma 10.39.19). Thus $t'_1 = t'_2$ by Lemma 58.10.1. Hence $t^{sh}$ descends to a morphism $t : X \to Y$ for example by Descent, Lemma 35.13.7. $\square$

Comment #5350 by Jackson on

Shouldn't the extension $t$ of $s$ be a morphism over $X$?

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