Lemma 58.14.3. Let $k$ be a field with algebraic closure $\overline{k}$. Let $X$ be a quasi-compact and quasi-separated scheme over $k$. If the base change $X_{\overline{k}}$ is connected, then there is a short exact sequence

\[ 1 \to \pi _1(X_{\overline{k}}) \to \pi _1(X) \to \pi _1(\mathop{\mathrm{Spec}}(k)) \to 1 \]

of profinite topological groups.

**Proof.**
Connected objects of $\textit{FÉt}_{\mathop{\mathrm{Spec}}(k)}$ are of the form $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$ with $k'/k$ a finite separable extension. Then $X_{\mathop{\mathrm{Spec}}{k'}}$ is connected, as the morphism $X_{\overline{k}} \to X_{\mathop{\mathrm{Spec}}(k')}$ is surjective and $X_{\overline{k}}$ is connected by assumption. Thus $\pi _1(X) \to \pi _1(\mathop{\mathrm{Spec}}(k))$ is surjective by Lemma 58.4.1.

Before we go on, note that we may assume that $k$ is a perfect field. Namely, we have $\pi _1(X_{k^{perf}}) = \pi _1(X)$ and $\pi _1(\mathop{\mathrm{Spec}}(k^{perf})) = \pi _1(\mathop{\mathrm{Spec}}(k))$ by Lemma 58.14.2.

It is clear that the composition of the functors $\textit{FÉt}_{\mathop{\mathrm{Spec}}(k)} \to \textit{FÉt}_ X \to \textit{FÉt}_{X_{\overline{k}}}$ sends objects to disjoint unions of copies of $X_{\mathop{\mathrm{Spec}}(\overline{k})}$. Therefore the composition $\pi _1(X_{\overline{k}}) \to \pi _1(X) \to \pi _1(\mathop{\mathrm{Spec}}(k))$ is the trivial homomorphism by Lemma 58.4.2.

Let $U \to X$ be a finite étale morphism with $U$ connected. Observe that $U \times _ X X_{\overline{k}} = U_{\overline{k}}$. Suppose that $U_{\overline{k}} \to X_{\overline{k}}$ has a section $s : X_{\overline{k}} \to U_{\overline{k}}$. Then $s(X_{\overline{k}})$ is an open connected component of $U_{\overline{k}}$. For $\sigma \in \text{Gal}(\overline{k}/k)$ denote $s^\sigma $ the base change of $s$ by $\mathop{\mathrm{Spec}}(\sigma )$. Since $U_{\overline{k}} \to X_{\overline{k}}$ is finite étale it has only a finite number of sections. Thus

\[ \overline{T} = \bigcup s^\sigma (X_{\overline{k}}) \]

is a finite union and we see that $\overline{T}$ is a $\text{Gal}(\overline{k}/k)$-stable open and closed subset. By Varieties, Lemma 33.7.10 we see that $\overline{T}$ is the inverse image of a closed subset $T \subset U$. Since $U_{\overline{k}} \to U$ is open (Morphisms, Lemma 29.23.4) we conclude that $T$ is open as well. As $U$ is connected we see that $T = U$. Hence $U_{\overline{k}}$ is a (finite) disjoint union of copies of $X_{\overline{k}}$. By Lemma 58.4.5 we conclude that the image of $\pi _1(X_{\overline{k}}) \to \pi _1(X)$ is normal.

Let $V \to X_{\overline{k}}$ be a finite étale cover. Recall that $\overline{k}$ is the union of finite separable extensions of $k$. By Lemma 58.14.1 we find a finite separable extension $k'/k$ and a finite étale morphism $U \to X_{k'}$ such that $V = X_{\overline{k}} \times _{X_{k'}} U = U \times _{\mathop{\mathrm{Spec}}(k')} \mathop{\mathrm{Spec}}(\overline{k})$. Then the composition $U \to X_{k'} \to X$ is finite étale and $U \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(\overline{k})$ contains $V = U \times _{\mathop{\mathrm{Spec}}(k')} \mathop{\mathrm{Spec}}(\overline{k})$ as an open and closed subscheme. (Because $\mathop{\mathrm{Spec}}(\overline{k})$ is an open and closed subscheme of $\mathop{\mathrm{Spec}}(k') \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(\overline{k})$ via the multiplication map $k' \otimes _ k \overline{k} \to \overline{k}$.) By Lemma 58.4.4 we conclude that $\pi _1(X_{\overline{k}}) \to \pi _1(X)$ is injective.

Finally, we have to show that for any finite étale morphism $U \to X$ such that $U_{\overline{k}}$ is a disjoint union of copies of $X_{\overline{k}}$ there is a finite étale morphism $V \to \mathop{\mathrm{Spec}}(k)$ and a surjection $V \times _{\mathop{\mathrm{Spec}}(k)} X \to U$. See Lemma 58.4.3. Arguing as above using Lemma 58.14.1 we find a finite separable extension $k'/k$ such that there is an isomorphism $U_{k'} \cong \coprod _{i = 1, \ldots , n} X_{k'}$. Thus setting $V = \coprod _{i = 1, \ldots , n} \mathop{\mathrm{Spec}}(k')$ we conclude.
$\square$

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