**Proof.**
The equivalence of (2) and (3) follows from Lemma 58.13.1 which tells us that $I_ y$ is conjugate to $\text{Gal}(K^{sep}/K_ x^{sh})$ if $y$ lies over $x$. By Lemma 58.11.4 we see that $\text{Gal}(K^{sep}/K_ x^{sh})$ maps trivially to $\pi _1(\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}))$ and therefore the subgroup $N \subset G = \text{Gal}(K^{sep}/K)$ of (2) and (3) is contained in the kernel of $G \longrightarrow \pi _1(X)$.

To prove the other inclusion, since $N$ is normal, it suffices to prove: given $N \subset U \subset G$ with $U$ open normal, the quotient map $G \to G/U$ factors through $\pi _1(X)$. In other words, if $L/K$ is the Galois extension corresponding to $U$, then we have to show that $X$ is unramified in $L$ (Section 58.11, especially Proposition 58.11.3). It suffices to do this when $X$ is affine (we do this so we can refer to algebra results in the rest of the proof). Let $Y \to X$ be the normalization of $X$ in $L$. The inclusion $L \subset K^{sep}$ induces a morphism $\pi : X^{sep} \to Y$. For $y \in X^{sep}$ the inertia group of $\pi (y)$ in $\text{Gal}(L/K)$ is the image of $I_ y$ in $\text{Gal}(L/K)$; this follows from More on Algebra, Lemma 15.110.8. Since $N \subset U$ all these inertia groups are trivial. We conclude that $Y \to X$ is étale by applying Lemma 58.12.4. (Alternative: you can use Lemma 58.11.4 to see that the pullback of $Y$ to $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$ is étale for all $x \in X$ and then conclude from there with a bit more work.)
$\square$

## Comments (0)