The Stacks project

Lemma 58.11.4. 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 58.4.3 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 58.11.3.

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 58.4.2.

To finish the proof, it suffices according to Lemma 58.4.3 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 35.23.29) as desired. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BSM. Beware of the difference between the letter 'O' and the digit '0'.