The Stacks project

Lemma 57.11.4. Let $(A, \mathfrak m)$ be a normal local ring. Set $X = \mathop{\mathrm{Spec}}(A)$. Let $A^{sh}$ be the strict henselization of $A$. Let $K$ and $K^{sh}$ be the fraction fields of $A$ and $A^{sh}$. Then the sequence

\[ \pi _1(\mathop{\mathrm{Spec}}(K^{sh})) \to \pi _1(\mathop{\mathrm{Spec}}(K)) \to \pi _1(X) \to 1 \]

is exact in the sense of Lemma 57.4.3 part (1).

Proof. Note that $A^{sh}$ is a normal domain, see More on Algebra, Lemma 15.45.6. The map $\pi _1(\mathop{\mathrm{Spec}}(K)) \to \pi _1(X)$ is surjective by Proposition 57.11.3.

Write $X^{sh} = \mathop{\mathrm{Spec}}(A^{sh})$. Let $Y \to X$ be a finite étale morphism. Then $Y^{sh} = Y \times _ X X^{sh} \to X^{sh}$ is a finite étale morphism. Since $A^{sh}$ is strictly henselian we see that $Y^{sh}$ is isomorphic to a disjoint union of copies of $X^{sh}$. Thus the same is true for $Y \times _ X \mathop{\mathrm{Spec}}(K^{sh})$. It follows that the composition $\pi _1(\mathop{\mathrm{Spec}}(K^{sh})) \to \pi _1(X)$ is trivial, see Lemma 57.4.2.

To finish the proof, it suffices according to Lemma 57.4.3 to show the following: Given a finite étale morphism $V \to \mathop{\mathrm{Spec}}(K)$ such that $V \times _{\mathop{\mathrm{Spec}}(K)} \mathop{\mathrm{Spec}}(K^{sh})$ is a disjoint union of copies of $\mathop{\mathrm{Spec}}(K^{sh})$, we can find a finite étale morphism $Y \to X$ with $V \cong Y \times _ X \mathop{\mathrm{Spec}}(K)$ over $\mathop{\mathrm{Spec}}(K)$. Write $V = \mathop{\mathrm{Spec}}(L)$, so $L$ is a finite product of finite separable extensions of $K$. Let $B \subset L$ be the integral closure of $A$ in $L$. If $A \to B$ is étale, then we can take $Y = \mathop{\mathrm{Spec}}(B)$ and the proof is complete. By Algebra, Lemma 10.147.4 (and a limit argument we omit) we see that $B \otimes _ A A^{sh}$ is the integral closure of $A^{sh}$ in $L^{sh} = L \otimes _ K K^{sh}$. Our assumption is that $L^{sh}$ is a product of copies of $K^{sh}$ and hence $B^{sh}$ is a product of copies of $A^{sh}$. Thus $A^{sh} \to B^{sh}$ is étale. As $A \to A^{sh}$ is faithfully flat it follows that $A \to B$ is étale (Descent, Lemma 35.20.29) as desired. $\square$


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