The Stacks project

Proof. Observe that $Y \to X$ is an integral morphism. In each case the morphism $Y \to X$ is locally of finite type by definition. Hence we find that in each case $Y \to X$ is finite by Morphisms, Lemma 29.44.4. In particular we see that (2) is equivalent to (3). An étale morphism is unramified, hence (2) implies (1).

Conversely, assume $Y \to X$ is unramified. Since a normal scheme is geometrically unibranch (Properties, Lemma 28.15.2), we see that the morphism $Y \to X$ is étale by More on Morphisms, Lemma 37.37.2. We also give a direct proof in the next paragraph.

Let $x \in X$. We can choose an étale neighbourhood $(U, u) \to (X, x)$ such that

is a disjoint union of closed immersions, see Étale Morphisms, Lemma 41.17.3. Shrinking we may assume $U$ is quasi-compact. Then $U$ has finitely many irreducible components (Descent, Lemma 35.16.3). Since $U$ is normal (Descent, Lemma 35.18.2) the irreducible components of $U$ are open and closed (Properties, Lemma 28.7.5) and we may assume $U$ is irreducible. Then $U$ is an integral scheme whose generic point $\xi $ maps to the generic point of $X$. On the other hand, we know that $Y \times _ X U$ is the normalization of $U$ in $\mathop{\mathrm{Spec}}(L) \times _ X U$ by More on Morphisms, Lemma 37.19.2. Every point of $\mathop{\mathrm{Spec}}(L) \times _ X U$ maps to $\xi $. Thus every $V_ j$ contains a point mapping to $\xi $ by Morphisms, Lemma 29.53.9. Thus $V_ j \to U$ is an isomorphism as $U = \overline{\{ \xi \} }$. Thus $Y \times _ X U \to U$ is étale. By Descent, Lemma 35.23.29 we conclude that $Y \to X$ is étale over the image of $U \to X$ (an open neighbourhood of $x$). $\square$

Proof. The scheme $Y$ is normal by Descent, Lemma 35.18.2. Since $Y \to X$ is flat every generic point of $Y$ maps to the generic point of $X$ by Morphisms, Lemma 29.25.9. Since $Y \to X$ is finite we see that $Y$ has a finite number of irreducible components. Thus $Y$ is the disjoint union of a finite number of integral normal schemes by Properties, Lemma 28.7.5. Thus if $Y$ is connected, then $Y$ is an integral normal scheme.

Let $L$ be the function field of $Y$ and let $Y' \to X$ be the normalization of $X$ in $L$. By Morphisms, Lemma 29.53.4 we obtain a factorization $Y' \to Y \to X$ and $Y' \to Y$ is the normalization of $Y$ in $L$. Since $Y$ is normal it is clear that $Y' = Y$ (this can also be deduced from Morphisms, Lemma 29.54.8). $\square$

Proof. The normal scheme $X$ is geometrically unibranch (Properties, Lemma 28.15.2). Hence Lemma 58.10.7 applies to $X$. Thus $\pi _1(\eta , \overline{\eta }) \to \pi _1(X, \overline{\eta })$ is surjective and top horizontal arrow of the commutative diagram

is fully faithful. The left vertical arrow is the equivalence of Theorem 58.6.2 and the right vertical arrow is the equivalence of Lemma 58.6.3. The lower horizontal arrow is induced by the map of the proposition. By Lemmas 58.11.1 and 58.11.2 we see that the essential image of $c$ consists of $\text{Gal}(K^{sep}/K)\textit{-Sets}$ isomorphic to sets of the form

with $L_ i/K$ finite separable such that $X$ is unramified in $L_ i$. Thus if $M \subset K^{sep}$ is as in the statement of the lemma, then $\text{Gal}(K^{sep}/M)$ is exactly the subgroup of $\text{Gal}(K^{sep}/K)$ acting trivially on every object in the essential image of $c$. On the other hand, the essential image of $c$ is exactly the category of $S$ such that the $\text{Gal}(K^{sep}/K)$-action factors through the surjection $\text{Gal}(K^{sep}/K) \to \pi _1(X, \overline{\eta })$. We conclude that $\text{Gal}(K^{sep}/M)$ is the kernel. Hence $\text{Gal}(K^{sep}/M)$ is a normal subgroup, $M/K$ is Galois, and we have a short exact sequence

by Galois theory (Fields, Theorem 9.22.4 and Lemma 9.22.5). The proof is done. $\square$

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. $\square$