The Stacks project

Proposition 58.30.2. Let $f : X \to S$ be a smooth proper morphism with geometrically connected fibres. Let $s' \leadsto s$ be a specialization. If the characteristic to $\kappa (s)$ is zero, then the specialization map

\[ sp : \pi _1(X_{\overline{s}'}) \to \pi _1(X_{\overline{s}}) \]

is an isomorphism.

Proof. The map is surjective by Lemma 58.30.1. Thus we have to show it is injective.

We may assume $S$ is affine. Then $S$ is a cofiltered limit of affine schemes of finite type over $\mathbf{Z}$. Hence we can assume $X \to S$ is the base change of $X_0 \to S_0$ where $S_0$ is the spectrum of a finite type $\mathbf{Z}$-algebra and $X_0 \to S_0$ is smooth and proper. See Limits, Lemma 32.10.1, 32.8.9, and 32.13.1. By Lemma 58.16.1 we reduce to the case where the base is Noetherian.

Applying Lemma 58.16.4 we reduce to the case where the base $S$ is the spectrum of a strictly henselian discrete valuation ring $A$ and we are looking at the specialization map over $A$. Let $K$ be the fraction field of $A$. Choose an algebraic closure $\overline{K}$ which corresponds to a geometric generic point $\overline{\eta }$ of $\mathop{\mathrm{Spec}}(A)$. For $\overline{K}/L/K$ finite separable, let $B \subset L$ be the integral closure of $A$ in $L$. This is a discrete valuation ring by More on Algebra, Remark 15.111.6.

Let $X \to \mathop{\mathrm{Spec}}(A)$ be as in the previous paragraph. To show injectivity of the specialization map it suffices to prove that every finite étale cover $V$ of $X_{\overline{\eta }}$ is the base change of a finite étale cover $Y \to X$. Namely, then $\pi _1(X_{\overline{\eta }}) \to \pi _1(X) = \pi _1(X_ s)$ is injective by Lemma 58.4.4.

Given $V$ we can first descend $V$ to $V' \to X_{K^{sep}}$ by Lemma 58.14.2 and then to $V'' \to X_ L$ by Lemma 58.14.1. Let $Z \to X_ B$ be the normalization of $X_ B$ in $V''$. Observe that $Z$ is normal and that $Z_ L = V''$ as schemes over $X_ L$. Hence $Z \to X_ B$ is finite étale over the generic fibre. The problem is that we do not know that $Z \to X_ B$ is everywhere étale. Since $X \to \mathop{\mathrm{Spec}}(A)$ has geometrically connected smooth fibres, we see that the special fibre $X_ s$ is geometrically irreducible. Hence the special fibre of $X_ B \to \mathop{\mathrm{Spec}}(B)$ is irreducible; let $\xi _ B$ be its generic point. Let $\xi _1, \ldots , \xi _ r$ be the points of $Z$ mapping to $\xi _ B$. Our first (and it will turn out only) problem is now that the extensions

\[ \mathcal{O}_{X_ B, \xi _ B} \subset \mathcal{O}_{Z, \xi _ i} \]

of discrete valuation rings may be ramified. Let $e_ i$ be the ramification index of this extension. Note that since the characteristic of $\kappa (s)$ is zero, the ramification is tame!

To get rid of the ramification we are going to choose a further finite separable extension $K^{sep}/L'/L/K$ such that the ramification index $e$ of the induced extensions $B'/B$ is divisible by $e_ i$. Consider the normalized base change $Z'$ of $Z$ with respect to $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(B)$, see discussion in More on Morphisms, Section 37.65. Let $\xi _{i, j}$ be the points of $Z'$ mapping to $\xi _{B'}$ and to $\xi _ i$ in $Z$. Then the local rings

\[ \mathcal{O}_{Z', \xi _{i, j}} \]

are localizations of the integral closure of $\mathcal{O}_{Z, \xi _ i}$ in $L' \otimes _ L F_ i$ where $F_ i$ is the fraction field of $\mathcal{O}_{Z, \xi _ i}$; details omitted. Hence Abhyankar's lemma (More on Algebra, Lemma 15.114.4) tells us that

\[ \mathcal{O}_{X_{B'}, \xi _{B'}} \subset \mathcal{O}_{Z', \xi _{i, j}} \]

is unramified. We conclude that the morphism $Z' \to X_{B'}$ is étale away from codimension $1$. Hence by purity of branch locus (Lemma 58.21.4) we see that $Z' \to X_{B'}$ is finite étale!

However, since the residue field extension induced by $A \to B'$ is trivial (as the residue field of $A$ is algebraically closed being separably closed of characteristic zero) we conclude that $Z'$ is the base change of a finite étale cover $Y \to X$ by applying Lemma 58.9.1 twice (first to get $Y$ over $A$, then to prove that the pullback to $B$ is isomorphic to $Z'$). This finishes the proof. $\square$

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