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 \input{preamble} % OK, start here. % \begin{document} \title{Fundamental Groups of Schemes} \maketitle \phantomsection \label{section-phantom} \tableofcontents \section{Introduction} \label{section-introduction} \noindent In this chapter we discuss Grothendieck's fundamental group of a scheme and applications. A foundational reference is \cite{SGA1}. A nice introduction is \cite{Lenstra}. Other references \cite{Murre-lectures} and \cite{Grothendieck-Murre}. \section{Schemes \'etale over a point} \label{section-schemes-etale-point} \noindent In this section we describe schemes \'etale over the spectrum of a field. Before we state the result we introduce the category of $G$-sets for a topological group $G$. \begin{definition} \label{definition-G-set-continuous} Let $G$ be a topological group. A {\it $G$-set}, sometime called a {\it discrete $G$-set}, is a set $X$ endowed with a left action $a : G \times X \to X$ such that $a$ is continuous when $X$ is given the discrete topology and $G \times X$ the product topology. A {\it morphism of $G$-sets} $f : X \to Y$ is simply any $G$-equivariant map from $X$ to $Y$. The category of $G$-sets is denoted {\it $G\textit{-Sets}$}. \end{definition} \noindent The condition that $a : G \times X \to X$ is continuous signifies simply that the stabilizer of any $x \in X$ is open in $G$. If $G$ is an abstract group $G$ (i.e., a group but not a topological group) then this agrees with our preceding definition (see for example Sites, Example \ref{sites-example-site-on-group}) provided we endow $G$ with the discrete topology. \medskip\noindent Recall that if $L/K$ is an infinite Galois extension then the Galois group $G = \text{Gal}(L/K)$ comes endowed with a canonical topology, see Fields, Section \ref{fields-section-infinite-galois}. \begin{lemma} \label{lemma-sheaves-point} Let $K$ be a field. Let $K^{sep}$ a separable closure of $K$. Consider the profinite group $G = \text{Gal}(K^{sep}/K)$. The functor $$\begin{matrix} \text{schemes \'etale over }K & \longrightarrow & G\textit{-Sets} \\ X/K & \longmapsto & \Mor_{\Spec(K)}(\Spec(K^{sep}), X) \end{matrix}$$ is an equivalence of categories. \end{lemma} \begin{proof} A scheme $X$ over $K$ is \'etale over $K$ if and only if $X \cong \coprod_{i\in I} \Spec(K_i)$ with each $K_i$ a finite separable extension of $K$ (Morphisms, Lemma \ref{morphisms-lemma-etale-over-field}). The functor of the lemma associates to $X$ the $G$-set $$\coprod\nolimits_i \Hom_K(K_i, K^{sep})$$ with its natural left $G$-action. Each element has an open stabilizer by definition of the topology on $G$. Conversely, any $G$-set $S$ is a disjoint union of its orbits. Say $S = \coprod S_i$. Pick $s_i \in S_i$ and denote $G_i \subset G$ its open stabilizer. By Galois theory (Fields, Theorem \ref{fields-theorem-inifinite-galois-theory}) the fields $(K^{sep})^{G_i}$ are finite separable field extensions of $K$, and hence the scheme $$\coprod\nolimits_i \Spec((K^{sep})^{G_i})$$ is \'etale over $K$. This gives an inverse to the functor of the lemma. Some details omitted. \end{proof} \begin{remark} \label{remark-covering-surjective} Under the correspondence of Lemma \ref{lemma-sheaves-point}, the coverings in the small \'etale site $\Spec(K)_\etale$ of $K$ correspond to surjective families of maps in $G\textit{-Sets}$. \end{remark} \section{Galois categories} \label{section-galois} \noindent In this section we discuss some of the material the reader can find in \cite[Expos\'e V, Sections 4, 5, and 6]{SGA1}. \medskip\noindent Let $F : \mathcal{C} \to \textit{Sets}$ be a functor. Recall that by our conventions categories have a set of objects and for any pair of objects a set of morphisms. There is a canonical injective map \begin{equation} \label{equation-embedding-product} \text{Aut}(F) \longrightarrow \prod\nolimits_{X \in \Ob(\mathcal{C})} \text{Aut}(F(X)) \end{equation} For a set $E$ we endow $\text{Aut}(E)$ with the compact open topology, see Topology, Example \ref{topology-example-automorphisms-of-a-set}. Of course this is the discrete topology when $E$ is finite, which is the case of interest in this section\footnote{When we discuss the pro-\'etale fundamental group the general case will be of interest.}. We endow $\text{Aut}(F)$ with the topology induced from the product topology on the right hand side of (\ref{equation-embedding-product}). In particular, the action maps $$\text{Aut}(F) \times F(X) \longrightarrow F(X)$$ are continuous when $F(X)$ is given the discrete topology because this is true for the action maps $\text{Aut}(E) \times E \to E$ for any set $E$. The universal property of our topology on $\text{Aut}(F)$ is the following: suppose that $G$ is a topological group and $G \to \text{Aut}(F)$ is a group homomorphism such that the induced actions $G \times F(X) \to F(X)$ are continuous for all $X \in \Ob(\mathcal{C})$ where $F(X)$ has the discrete topology. Then $G \to \text{Aut}(F)$ is continuous. \medskip\noindent The following lemma tells us that the group of automorphisms of a functor to the category of finite sets is automatically a profinite group. \begin{lemma} \label{lemma-aut-inverse-limit} Let $\mathcal{C}$ be a category and let $F : \mathcal{C} \to \textit{Sets}$ be a functor. The map (\ref{equation-embedding-product}) identifies $\text{Aut}(F)$ with a closed subgroup of $\prod_{X \in \Ob(\mathcal{C})} \text{Aut}(F(X))$. In particular, if $F(X)$ is finite for all $X$, then $\text{Aut}(F)$ is a profinite group. \end{lemma} \begin{proof} Let $\xi = (\gamma_X) \in \prod \text{Aut}(F(X))$ be an element not in $\text{Aut}(F)$. Then there exists a morphism $f : X \to X'$ of $\mathcal{C}$ and an element $x \in F(X)$ such that $F(f)(\gamma_X(x)) \not = \gamma_{X'}(F(f)(x))$. Consider the open neighbourhood $U = \{\gamma \in \text{Aut}(F(X)) \mid \gamma(x) = \gamma_X(x)\}$ of $\gamma_X$ and the open neighbourhood $U' = \{\gamma' \in \text{Aut}(F(X')) \mid \gamma'(F(f)(x)) = \gamma_{X'}(F(f)(x))\}$. Then $U \times U' \times \prod_{X'' \not = X, X'} \text{Aut}(F(X''))$ is an open neighbourhood of $\xi$ not meeting $\text{Aut}(F)$. The final statement is follows from the fact that $\prod \text{Aut}(F(X))$ is a profinite space if each $F(X)$ is finite. \end{proof} \begin{example} \label{example-galois-category-G-sets} Let $G$ be a topological group. An important example will be the forgetful functor \begin{equation} \label{equation-forgetful} \textit{Finite-}G\textit{-Sets} \longrightarrow \textit{Sets} \end{equation} where $\textit{Finite-}G\textit{-Sets}$ is the full subcategory of $G\textit{-Sets}$ whose objects are the finite $G$-sets. The category $G\textit{-Sets}$ of $G$-sets is defined in Definition \ref{definition-G-set-continuous}. \end{example} \noindent Let $G$ be a topological group. The {\it profinite completion} of $G$ will be the profinite group $$G^\wedge = \lim_{U \subset G\text{ open, normal, finite index}} G/U$$ with its profinite topology. Observe that the limit is cofiltered as a finite intersection of open, normal subgroups of finite index is another. The universal property of the profinite completion is that any continuous map $G \to H$ to a profinite group $H$ factors canonically as $G \to G^\wedge \to H$. \begin{lemma} \label{lemma-single-out-profinite} Let $G$ be a topological group. The automorphism group of the functor (\ref{equation-forgetful}) endowed with its profinite topology from Lemma \ref{lemma-aut-inverse-limit} is the profinite completion of $G$. \end{lemma} \begin{proof} Denote $F_G$ the functor (\ref{equation-forgetful}). Any morphism $X \to Y$ in $\textit{Finite-}G\textit{-Sets}$ commutes with the action of $G$. Thus any $g \in G$ defines an automorphism of $F_G$ and we obtain a canonical homomorphism $G \to \text{Aut}(F_G)$ of groups. Observe that any finite $G$-set $X$ is a finite disjoint union of $G$-sets of the form $G/H_i$ with canonical $G$-action where $H_i \subset G$ is an open subgroup of finite index. Then $U_i = \bigcap gH_ig^{-1}$ is open, normal, and has finite index. Moreover $U_i$ acts trivially on $G/H_i$ hence $U = \bigcap U_i$ acts trivially on $F_G(X)$. Hence the action $G \times F_G(X) \to F_G(X)$ is continuous. By the universal property of the topology on $\text{Aut}(F_G)$ the map $G \to \text{Aut}(F_G)$ is continuous. By Lemma \ref{lemma-aut-inverse-limit} and the universal property of profinite completion there is an induced continuous group homomorphism $$G^\wedge \longrightarrow \text{Aut}(F_G)$$ Moreover, since $G/U$ acts faithfully on $G/U$ this map is injective. If the image is dense, then the map is surjective and hence a homeomorphism by Topology, Lemma \ref{topology-lemma-bijective-map}. \medskip\noindent Let $\gamma \in \text{Aut}(F_G)$ and let $X \in \Ob(\mathcal{C})$. We will show there is a $g \in G$ such that $\gamma$ and $g$ induce the same action on $F_G(X)$. This will finish the proof. As before we see that $X$ is a finite disjoint union of $G/H_i$. With $U_i$ and $U$ as above, the finite $G$-set $Y = G/U$ surjects onto $G/H_i$ for all $i$ and hence it suffices to find $g \in G$ such that $\gamma$ and $g$ induce the same action on $F_G(G/U) = G/U$. Let $e \in G$ be the neutral element and say that $\gamma(eU) = g_0U$ for some $g_0 \in G$. For any $g_1 \in G$ the morphism $$R_{g_1} : G/U \longrightarrow G/U,\quad gU \longmapsto gg_1U$$ of $\textit{Finite-}G\textit{-Sets}$ commutes with the action of $\gamma$. Hence $$\gamma(g_1U) = \gamma(R_{g_1}(eU)) = R_{g_1}(\gamma(eU)) = R_{g_1}(g_0U) = g_0g_1U$$ Thus we see that $g = g_0$ works. \end{proof} \noindent Recall that an exact functor is one which commutes with all finite limits and finite colimits. In particular such a functor commutes with equalizers, coequalizers, fibred products, pushouts, etc. \begin{lemma} \label{lemma-second-fundamental-functor} Let $G$ be a topological group. Let $F : \textit{Finite-}G\textit{-Sets} \to \textit{Sets}$ be an exact functor with $F(X)$ finite for all $X$. Then $F$ is isomorphic to the functor (\ref{equation-forgetful}). \end{lemma} \begin{proof} Let $X$ be a nonempty object of $\textit{Finite-}G\textit{-Sets}$. The diagram $$\xymatrix{ X \ar[r] \ar[d] & \{*\} \ar[d] \\ \{*\} \ar[r] & \{*\} }$$ is cocartesian. Hence we conclude that $F(X)$ is nonempty. Let $U \subset G$ be an open, normal subgroup with finite index. Observe that $$G/U \times G/U = \coprod\nolimits_{gU \in G/U} G/U$$ where the summand corresponding to $gU$ corresponds to the orbit of $(eU, gU)$ on the left hand side. Then we see that $$F(G/U) \times F(G/U) = F(G/U \times G/U) = \coprod\nolimits_{gU \in G/U} F(G/U)$$ Hence $|F(G/U)| = |G/U|$ as $F(G/U)$ is nonempty. Thus we see that $$\lim_{U \subset G\text{ open, normal, finite idex}} F(G/U)$$ is nonempty (Categories, Lemma \ref{categories-lemma-nonempty-limit}). Pick $\gamma = (\gamma_U)$ an element in this limit. Denote $F_G$ the functor (\ref{equation-forgetful}). We can identify $F_G$ with the functor $$X \longmapsto \colim_U \Mor(G/U, X)$$ where $f : G/U \to X$ corresponds to $f(eU) \in X = F_G(X)$ (details omitted). Hence the element $\gamma$ determines a well defined map $$t : F_G \longrightarrow F$$ Namely, given $x \in X$ choose $U$ and $f : G/U \to X$ sending $eU$ to $x$ and then set $t_X(x) = F(f)(\gamma_U)$. We will show that $t$ induces a bijective map $t_{G/U} : F_G(G/U) \to F(G/U)$ for any $U$. This implies in a straightforward manner that $t$ is an isomorphism (details omitted). Since $|F_G(G/U)| = |F(G/U)|$ it suffices to show that $t_{G/U}$ is surjective. The image contains at least one element, namely $t_{G/U}(eU) = F(\text{id}_{G/U})(\gamma_U) = \gamma_U$. For $g \in G$ denote $R_g : G/U \to G/U$ right multiplication. Then set of fixed points of $F(R_g) : F(G/U) \to F(G/U)$ is equal to $F(\emptyset) = \emptyset$ if $g \not \in U$ because $F$ commutes with equalizers. It follows that if $g_1, \ldots, g_{|G/U|}$ is a system of representatives for $G/U$, then the elements $F(R_{g_i})(\gamma_U)$ are pairwise distinct and hence fill out $F(G/U)$. Then $$t_{G/U}(g_iU) = F(R_{g_i})(\gamma_U)$$ and the proof is complete. \end{proof} \begin{example} \label{example-from-C-F-to-G-sets} Let $\mathcal{C}$ be a category and let $F : \mathcal{C} \to \textit{Sets}$ be a functor such that $F(X)$ is finite for all $X \in \Ob(\mathcal{C})$. By Lemma \ref{lemma-aut-inverse-limit} we see that $G = \text{Aut}(F)$ comes endowed with the structure of a profinite topological group in a canonical manner. We obtain a functor \begin{equation} \label{equation-remember} \mathcal{C} \longrightarrow \textit{Finite-}G\textit{-Sets},\quad X \longmapsto F(X) \end{equation} where $F(X)$ is endowed with the induced action of $G$. This action is continuous by our construction of the topology on $\text{Aut}(F)$. \end{example} \noindent The purpose of defining Galois categories is to single out those pairs $(\mathcal{C}, F)$ for which the functor (\ref{equation-remember}) is an equivalence. Our definition of a Galois category is as follows. \begin{definition} \label{definition-galois-category} \begin{reference} Different from the definition in \cite[Expos\'e V, Definition 5.1]{SGA1}. Compare with \cite[Definition 7.2.1]{BS}. \end{reference} Let $\mathcal{C}$ be a category and let $F : \mathcal{C} \to \textit{Sets}$ be a functor. The pair $(\mathcal{C}, F)$ is a {\it Galois category} if \begin{enumerate} \item $\mathcal{C}$ has finite limits and finite colimits, \item \label{item-connected-components} every object of $\mathcal{C}$ is a finite (possibly empty) coproduct of connected objects, \item $F(X)$ is finite for all $X \in \Ob(\mathcal{C})$, and \item $F$ reflects isomorphisms and is exact. \end{enumerate} Here we say $X \in \Ob(\mathcal{C})$ is connected if it is not initial and for any monomorphism $Y \to X$ either $Y$ is initial or $Y \to X$ is an isomorphism. \end{definition} \noindent {\bf Warning:} This definition is not the same (although eventually we'll see it is equivalent) as the definition given in most references. Namely, in \cite[Expos\'e V, Definition 5.1]{SGA1} a Galois category is defined to be a category equivalent to $\textit{Finite-}G\textit{-Sets}$ for some profinite group $G$. Then Grothendieck characterizes Galois categories by a list of axioms (G1) -- (G6) which are weaker than our axioms above. The motivation for our choice is to stress the existence of finite limits and finite colimits and exactness of the functor $F$. The price we'll pay for this later is that we'll have to work a bit harder to apply the results of this section. \begin{lemma} \label{lemma-epi-mono} Let $(\mathcal{C}, F)$ be a Galois category. Let $X \to Y \in \text{Arrows}(\mathcal{C})$. Then \begin{enumerate} \item $F$ is faithful, \item $X \to Y$ is a monomorphism $\Leftrightarrow F(X) \to F(Y)$ is injective, \item $X \to Y$ is an epimorphism $\Leftrightarrow F(X) \to F(Y)$ is surjective, \item an object $A$ of $\mathcal{C}$ is initial if and only if $F(A) = \emptyset$, \item an object $Z$ of $\mathcal{C}$ is final if and only if $F(Z)$ is a singleton, \item if $X$ and $Y$ are connected, then $X \to Y$ is an epimorphism, \item \label{item-one-element} if $X$ is connected and $a, b : X \to Y$ are two morphisms then $a = b$ as soon as $F(a)$ and $F(b)$ agree on one element of $F(X)$, \item if $X = \coprod_{i = 1, \ldots, n} X_i$ and $Y = \coprod_{j = 1, \ldots, m} Y_j$ where $X_i$, $Y_j$ are connected, then there is map $\alpha : \{1, \ldots, n\} \to \{1, \ldots, m\}$ such that $X \to Y$ comes from a collection of morphisms $X_i \to Y_{\alpha(i)}$. \end{enumerate} \end{lemma} \begin{proof} Proof of (1). Suppose $a, b : X \to Y$ with $F(a) = F(b)$. Let $E$ be the equalizer of $a$ and $b$. Then $F(E) = F(X)$ and we see that $E = X$ because $F$ reflects isomorphisms. \medskip\noindent Proof of (2). This is true because $F$ turns the morphism $X \to X \times_Y X$ into the map $F(X) \to F(X) \times_{F(Y)} F(X)$ and $F$ reflects isomorphisms. \medskip\noindent Proof of (3). This is true because $F$ turns the morphism $Y \amalg_X Y \to Y$ into the map $F(Y) \amalg_{F(X)} F(Y) \to F(Y)$ and $F$ reflects isomorphisms. \medskip\noindent Proof of (4). There exists an initial object $A$ and certainly $F(A) = \emptyset$. On the other hand, if $X$ is an object with $F(X) = \emptyset$, then the unique map $A \to X$ induces a bijection $F(A) \to F(X)$ and hence $A \to X$ is an isomorphism. \medskip\noindent Proof of (5). There exists a final object $Z$ and certainly $F(Z)$ is a singleton. On the other hand, if $X$ is an object with $F(X)$ a singleton, then the unique map $X \to Z$ induces a bijection $F(X) \to F(Z)$ and hence $X \to Z$ is an isomorphism. \medskip\noindent Proof of (6). The equalizer $E$ of the two maps $Y \to Y \amalg_X Y$ is not an initial object of $\mathcal{C}$ because $X \to Y$ factors through $E$ and $F(X) \not = \emptyset$. Hence $E = Y$ and we conclude. \medskip\noindent Proof of (\ref{item-one-element}). The equalizer $E$ of $a$ and $b$ comes with a monomorphism $E \to X$ and $F(E) \subset F(X)$ is the set of elements where $F(a)$ and $F(b)$ agree. To finish use that either $E$ is initial or $E = X$. \medskip\noindent Proof of (8). For each $i, j$ we see that $E_{ij} = X_i \times_Y Y_j$ is either initial or equal to $X_i$. Picking $s \in F(X_i)$ we see that $E_{ij} = X_i$ if and only if $s$ maps to an element of $F(Y_j) \subset F(Y)$, hence this happens for a unique $j = \alpha(i)$. \end{proof} \noindent By the lemma above we see that, given a connected object $X$ of a Galois category $(\mathcal{C}, F)$, the automorphism group $\text{Aut}(X)$ has order at most $|F(X)|$. Namely, given $s \in F(X)$ and $g \in \text{Aut}(X)$ we see that $g(s) = s$ if and only if $g = \text{id}_X$ by (\ref{item-one-element}). We say $X$ is {\it Galois} if equality holds. Equivalently, $X$ is Galois if it is connected and $\text{Aut}(X)$ acts transitively on $F(X)$. \begin{lemma} \label{lemma-galois} Let $(\mathcal{C}, F)$ be a Galois category. For any connected object $X$ of $\mathcal{C}$ there exists a Galois object $Y$ and a morphism $Y \to X$. \end{lemma} \begin{proof} We will use the results of Lemma \ref{lemma-epi-mono} without further mention. Let $n = |F(X)|$. Consider $X^n$ endowed with its natural action of $S_n$. Let $$X^n = \coprod\nolimits_{t \in T} Z_t$$ be the decomposition into connected objects. Pick a $t$ such that $F(Z_t)$ contains $(s_1, \ldots, s_n)$ with $s_i$ pairwise distinct. If $(s'_1, \ldots, s'_n) \in F(Z_t)$ is another element, then we claim $s'_i$ are pairwise distinct as well. Namely, if not, say $s'_i = s'_j$, then $Z_t$ is the image of an connected component of $X^{n - 1}$ under the diagonal morphism $$\Delta_{ij} : X^{n - 1} \longrightarrow X^n$$ Since morphisms of connected objects are epimorphisms and induce surjections after applying $F$ it would follow that $s_i = s_j$ which is not the case. \medskip\noindent Let $G \subset S_n$ be the subgroup of elements with $g(Z_t) = Z_t$. Looking at the action of $S_n$ on $$F(X)^n = F(X^n) = \coprod\nolimits_{t' \in T} F(Z_{t'})$$ we see that $G = \{g \in S_n \mid g(s_1, \ldots, s_n) \in F(Z_t)\}$. Now pick a second element $(s'_1, \ldots, s'_n) \in F(Z_t)$. Above we have seen that $s'_i$ are pairwise distinct. Thus we can find a $g \in S_n$ with $g(s_1, \ldots, s_n) = (s'_1, \ldots, s'_n)$. In other words, the action of $G$ on $F(Z_t)$ is transitive and the proof is complete. \end{proof} \noindent Here is a key lemma. \begin{lemma} \label{lemma-tame} \begin{reference} Compare with \cite[Definition 7.2.4]{BS}. \end{reference} Let $(\mathcal{C}, F)$ be a Galois category. Let $G = \text{Aut}(F)$ be as in Example \ref{example-from-C-F-to-G-sets}. For any connected $X$ in $\mathcal{C}$ the action of $G$ on $F(X)$ is transitive. \end{lemma} \begin{proof} We will use the results of Lemma \ref{lemma-epi-mono} without further mention. Let $I$ be the set of isomorphism classes of Galois objects in $\mathcal{C}$. For each $i \in I$ let $X_i$ be a representative of the isomorphism class. Choose $\gamma_i \in F(X_i)$ for each $i \in I$. We define a partial ordering on $I$ by setting $i \geq i'$ if and only if there is a morphism $f_{ii'} : X_i \to X_{i'}$. Given such a morphism we can post-compose by an automorphism $X_{i'} \to X_{i'}$ to assure that $F(f_{ii'})(\gamma_i) = \gamma_{i'}$. With this normalization the morphism $f_{ii'}$ is unique. Observe that $I$ is a directed partially ordered set: (Categories, Definition \ref{categories-definition-directed-set}) if $i_1, i_2 \in I$ there exists a Galois object $Y$ and a morphism $Y \to X_{i_1} \times X_{i_2}$ by Lemma \ref{lemma-galois} applied to a connected component of $X_{i_1} \times X_{i_2}$. Then $Y \cong X_i$ for some $i \in I$ and $i \geq i_1$, $i \geq I_2$. \medskip\noindent We claim that the functor $F$ is isomorphic to the functor $F'$ which sends $X$ to $$F'(X) = \colim_I \Mor_\mathcal{C}(X_i, X)$$ via the transformation of functors $t : F' \to F$ defined as follows: given $f : X_i \to X$ we set $t_X(f) = F(f)(\gamma_i)$. Using (\ref{item-one-element}) we find that $t_X$ is injective. To show surjectivity, let $\gamma \in F(X)$. Then we can immediately reduce to the case where $X$ is connected by the definition of a Galois category. Then we may assume $X$ is Galois by Lemma \ref{lemma-galois}. In this case $X$ is isomorphic to $X_i$ for some $i$ and we can choose the isomorphism $X_i \to X$ such that $\gamma_i$ maps to $\gamma$ (by definition of Galois objects). We conclude that $t$ is an isomorphism. \medskip\noindent Set $A_i = \text{Aut}(X_i)$. We claim that for $i \geq i'$ there is a canonical map $h_{ii'} : A_i \to A_{i'}$ such that for all $a \in A_i$ the diagram $$\xymatrix{ X_i \ar[d]_a \ar[r]_{f_{ii'}} & X_{i'} \ar[d]^{h_{ii'}(a)} \\ X_i \ar[r]^{f_{ii'}} & X_{i'} }$$ commutes. Namely, just let $h_{ii'}(a) = a' : X_{i'} \to X_{i'}$ be the unique automorphism such that $F(a')(\gamma_{i'}) = F(f_{ii'} \circ a)(\gamma_i)$. As before this makes the diagram commute and moreover the choice is unique. It follows that $h_{i'i''} \circ h_{ii'} = h_{ii''}$ if $i \geq i' \geq i''$. Since $F(X_i) \to F(X_{i'})$ is surjective we see that $A_i \to A_{i'}$ is surjective. Taking the inverse limit we obtain a group $$A = \lim_I A_i$$ This is a profinite group since the automorphism groups are finite. The map $A \to A_i$ is surjective for all $i$ by Categories, Lemma \ref{categories-lemma-nonempty-limit}. \medskip\noindent Since elements of $A$ act on the inverse system $X_i$ we get an action of $A$ (on the right) on $F'$ by pre-composing. In other words, we get a homomorphism $A^{opp} \to G$. Since $A \to A_i$ is surjective we conclude that $G$ acts transitively on $F(X_i)$ for all $i$. Since every connected object is dominated by one of the $X_i$ we conclude the lemma is true. \end{proof} \begin{proposition} \label{proposition-galois} \begin{reference} This is a weak version of \cite[Expos\'e V]{SGA1}. The proof is borrowed from \cite[Theorem 7.2.5]{BS}. \end{reference} Let $(\mathcal{C}, F)$ be a Galois category. Let $G = \text{Aut}(F)$ be as in Example \ref{example-from-C-F-to-G-sets}. The functor $F : \mathcal{C} \to \textit{Finite-}G\textit{-Sets}$ (\ref{equation-remember}) an equivalence. \end{proposition} \begin{proof} We will use the results of Lemma \ref{lemma-epi-mono} without further mention. In particular we know the functor is faithful. By Lemma \ref{lemma-tame} we know that for any connected $X$ the action of $G$ on $F(X)$ is transitive. Hence $F$ preserves the decomposition into connected components (existence of which is an axiom of a Galois category). Let $X$ and $Y$ be objects and let $s : F(X) \to F(Y)$ be a map. Then the graph $\Gamma_s \subset F(X) \times F(Y)$ of $s$ is a union of connected components. Hence there exists a union of connected components $Z$ of $X \times Y$, which comes equipped with a monomorphism $Z \to X \times Y$, with $F(Z) = \Gamma_s$. Since $F(Z) \to F(X)$ is bijective we see that $Z \to X$ is an isomorphism and we conclude that $s = F(f)$ where $f : X \cong Z \to Y$ is the composition. Hence $F$ is fully faithful. \medskip\noindent To finish the proof we show that $F$ is essentially surjective. It suffices to show that $G/H$ is in the essential image for any open subgroup $H \subset G$ of finite index. By definition of the topology on $G$ there exists a finite collection of objects $X_i$ such that $$\Ker(G \longrightarrow \prod\nolimits_i \text{Aut}(F(X_i)))$$ is contained in $H$. We may assume $X_i$ is connected for all $i$. We can choose a Galois object $Y$ mapping to a connected component of $\prod X_i$ using Lemma \ref{lemma-galois}. Choose an isomorphism $F(Y) = G/U$ in $G\textit{-sets}$ for some open subgroup $U \subset G$. As $Y$ is Galois, the group $\text{Aut}(Y) = \text{Aut}_{G\textit{-Sets}}(G/U)$ acts transitively on $F(Y) = G/U$. This implies that $U$ is normal. Since $F(Y)$ surjects onto $F(X_i)$ for each $i$ we see that $U \subset H$. Let $M \subset \text{Aut}(Y)$ be the finite subgroup corresponding to $$(H/U)^{opp} \subset (G/U)^{opp} = \text{Aut}_{G\textit{-Sets}}(G/U) = \text{Aut}(Y).$$ Set $X = Y/M$, i.e., $X$ is the coequalizer of the arrows $m : Y \to Y$, $m \in M$. Since $F$ is exact we see that $F(X) = G/H$ and the proof is complete. \end{proof} \begin{lemma} \label{lemma-functoriality-galois} Let $(\mathcal{C}, F)$ and $(\mathcal{C}', F')$ be Galois categories. Let $H : \mathcal{C} \to \mathcal{C}'$ be an exact functor. There exists an isomorphism $t : F' \circ H \to F$. The choice of $t$ determines a continuous homomorphism $h : G' = \text{Aut}(F') \to \text{Aut}(F) = G$ and a $2$-commutative diagram $$\xymatrix{ \mathcal{C} \ar[r]_H \ar[d] & \mathcal{C}' \ar[d] \\ \textit{Finite-}G\textit{-Sets} \ar[r]^h & \textit{Finite-}G'\textit{-Sets} }$$ The map $h$ is independent of $t$ up to an inner automorphism of $G$. Conversely, given a continuous homomorphism $h : G' \to G$ there is an exact functor $H : \mathcal{C} \to \mathcal{C}'$ and an isomorphism $t$ recovering $h$ as above. \end{lemma} \begin{proof} By Proposition \ref{proposition-galois} and Lemma \ref{lemma-single-out-profinite} we may assume $\mathcal{C} = \textit{Finite-}G\textit{-Sets}$ and $F$ is the forgetful functor and similarly for $\mathcal{C}'$. Thus the existence of $t$ follows from Lemma \ref{lemma-second-fundamental-functor}. The map $h$ comes from transport of structure via $t$. The commutativity of the diagram is obvious. Uniqueness of $h$ up to inner conjugation by an element of $G$ comes from the fact that the choice of $t$ is unique up to an element of $G$. The final statement is straightforward. \end{proof} \section{Functors and homomorphisms} \label{section-translation} \noindent Let $(\mathcal{C}, F)$, $(\mathcal{C}', F')$, $(\mathcal{C}'', F'')$ be Galois categories. Set $G = \text{Aut}(F)$, $G' = \text{Aut}(F')$, and $G'' = \text{Aut}(F'')$. Let $H : \mathcal{C} \to \mathcal{C}'$ and $H' : \mathcal{C}' \to \mathcal{C}''$ be exact functors. Let $h : G' \to G$ and $h' : G'' \to G'$ be the corresponding continuous homomorphism as in Lemma \ref{lemma-functoriality-galois}. In this section we consider the corresponding $2$-commutative diagram \begin{equation} \label{equation-translation} \vcenter{ \xymatrix{ \mathcal{C} \ar[r]_H \ar[d] & \mathcal{C}' \ar[r]_{H'} \ar[d] & \mathcal{C}'' \ar[d] \\ \textit{Finite-}G\textit{-Sets} \ar[r]^h & \textit{Finite-}G'\textit{-Sets} \ar[r]^{h'} & \textit{Finite-}G''\textit{-Sets} } } \end{equation} and we relate exactness properties of the sequence $1 \to G'' \to G' \to G \to 1$ to properties of the functors $H$ and $H'$. \begin{lemma} \label{lemma-functoriality-galois-surjective} In diagram (\ref{equation-translation}) the following are equivalent \begin{enumerate} \item $h : G' \to G$ is surjective, \item $H : \mathcal{C} \to \mathcal{C}'$ is fully faithful, \item if $X \in \Ob(\mathcal{C})$ is connected, then $H(X)$ is connected, \item if $X \in \Ob(\mathcal{C})$ is connected and there is a morphism $*' \to H(X)$ in $\mathcal{C}'$, then there is a morphism $* \to X$, and \item for any object $X$ of $\mathcal{C}$ the map $\Mor_\mathcal{C}(*, X) \to \Mor_{\mathcal{C}'}(*', H(X))$ is bijective. \end{enumerate} Here $*$ and $*'$ are final objects of $\mathcal{C}$ and $\mathcal{C}'$. \end{lemma} \begin{proof} The implications (5) $\Rightarrow$ (4) and (2) $\Rightarrow$ (5) are clear. \medskip\noindent Assume (3). Let $X$ be a connected object of $\mathcal{C}$ and let $*' \to H(X)$ be a morphism. Since $H(X)$ is connected by (3) we see that $*' \to H(X)$ is an isomorphism. Hence the $G'$-set corresponding to $H(X)$ has exactly one element, which means the $G$-set corresponding to $X$ has one element which means $X$ is isomorphic to the final object of $\mathcal{C}$, in particular there is a map $* \to X$. In this way we see that (3) $\Rightarrow$ (4). \medskip\noindent If (1) is true, then the functor $\textit{Finite-}G\textit{-Sets} \to \textit{Finite-}G'\textit{-Sets}$ is fully faithful: in this case a map of $G$-sets commutes with the action of $G$ if and only if it commutes with the action of $G'$. Thus (1) $\Rightarrow$ (2). \medskip\noindent If (1) is true, then for a $G$-set $X$ the $G$-orbits and $G'$-orbits agree. Thus (1) $\Rightarrow$ (3). \medskip\noindent To finish the proof it suffices to show that (4) implies (1). If (1) is false, i.e., if $h$ is not surjective, then there is an open subgroup $U \subset G$ containing $h(G')$ which is not equal to $G$. Then the finite $G$-set $M = G/U$ has a transitive action but $G'$ has a fixed point. The object $X$ of $\mathcal{C}$ corresponding to $M$ would contradict (3). In this way we see that (3) $\Rightarrow$ (1) and the proof is complete. \end{proof} \begin{lemma} \label{lemma-composition-trivial} In diagram (\ref{equation-translation}) the following are equivalent \begin{enumerate} \item $h \circ h'$ is trivial, and \item the image of $H' \circ H$ consists of objects isomorphic to finite coproducts of final objects. \end{enumerate} \end{lemma} \begin{proof} We may replace $H$ and $H'$ by the canonical functors $\textit{Finite-}G\textit{-Sets} \to \textit{Finite-}G'\textit{-Sets} \to \textit{Finite-}G''\textit{-Sets}$ determined by $h$ and $h'$. Then we are saying that the action of $G''$ on every $G$-set is trivial if and only if the homomorphism $G'' \to G$ is trivial. This is clear. \end{proof} \begin{lemma} \label{lemma-functoriality-galois-ses} In diagram (\ref{equation-translation}) the following are equivalent \begin{enumerate} \item the sequence $G'' \xrightarrow{h'} G' \xrightarrow{h} G \to 1$ is exact in the following sense: $h$ is surjective, $h \circ h'$ is trivial, and $\Ker(h)$ is the smallest closed normal subgroup containing $\Im(h')$, \item $H$ is fully faithful and an object $X'$ of $\mathcal{C}'$ is in the essential image of $H$ if and only if $H'(X')$ is isomorphic to a finite coproduct of final objects, and \item $H$ is fully faithful, $H \circ H'$ sends every object to a finite coproduct of final objects, and for an object $X'$ of $\mathcal{C}'$ such that $H'(X')$ is a finite coproduct of final objects there exists an object $X$ of $\mathcal{C}$ and an epimorphism $H(X) \to X'$. \end{enumerate} \end{lemma} \begin{proof} By Lemmas \ref{lemma-functoriality-galois-surjective} and \ref{lemma-composition-trivial} we may assume that $H$ is fully faithful, $h$ is surjective, $H' \circ H$ maps objects to disjoint unions of the final object, and $h \circ h'$ is trivial. Let $N \subset G'$ be the smallest closed normal subgroup containing the image of $h'$. It is clear that $N \subset \Ker(h)$. We may assume the functors $H$ and $H'$ are the canonical functors $\textit{Finite-}G\textit{-Sets} \to \textit{Finite-}G'\textit{-Sets} \to \textit{Finite-}G''\textit{-Sets}$ determined by $h$ and $h'$. \medskip\noindent Suppose that (2) holds. This means that for a finite $G'$-set $X'$ such that $G''$ acts trivially, the action of $G'$ factors through $G$. Apply this to $X' = G'/U'N$ where $U'$ is a small open subgroup of $G'$. Then we see that $\Ker(h) \subset U'N$ for all $U'$. Since $N$ is closed this implies $\Ker(h) \subset N$, i.e., (1) holds. \medskip\noindent Suppose that (1) holds. This means that $N = \Ker(h)$. Let $X'$ be a finite $G'$-set such that $G''$ acts trivially. This means that $\Ker(G' \to \text{Aut}(X'))$ is a closed normal subgroup containing $\Im(h')$. Hence $N = \Ker(h)$ is contained in it and the $G'$-action on $X'$ factors through $G$, i.e., (2) holds. \medskip\noindent Suppose that (3) holds. This means that for a finite $G'$-set $X'$ such that $G''$ acts trivially, there is a surjection of $G'$-sets $X \to X'$ where $X$ is a $G$-set. Clearly this means the action of $G'$ on $X'$ factors through $G$, i.e., (2) holds. \medskip\noindent The implication (2) $\Rightarrow$ (3) is immediate. This finishes the proof. \end{proof} \begin{lemma} \label{lemma-functoriality-galois-injective} In diagram (\ref{equation-translation}) the following are equivalent \begin{enumerate} \item $h'$ is injective, and \item for every connected object $X''$ of $\mathcal{C}''$ there exists an object $X'$ of $\mathcal{C}'$ and a diagram $$X'' \leftarrow Y'' \rightarrow H(X')$$ in $\mathcal{C}''$ where $Y'' \to X''$ is an epimorphism and $Y'' \to H(X')$ is a monomorphism. \end{enumerate} \end{lemma} \begin{proof} We may replace $H'$ by the corresponding functor between the categories of finite $G'$-sets and finite $G''$-sets. \medskip\noindent Assume $h' : G'' \to G'$ is injective. Let $H'' \subset G''$ be an open subgroup. Since the topology on $G''$ is the induced topology from $G'$ there exists an open subgroup $H' \subset G'$ such that $(h')^{-1}(H') \subset H''$. Then the desired diagram is $$G''/H'' \leftarrow G''/(h')^{-1}(H') \rightarrow G'/H'$$ Conversely, assume (2) holds for the functor $\textit{Finite-}G'\textit{-Sets} \to \textit{Finite-}G''\textit{-Sets}$. Let $g'' \in \Ker(h')$. Pick any open subgroup $H'' \subset G''$. By assumption there exists a finite $G'$-set $X'$ and a diagram $$G''/H'' \leftarrow Y'' \rightarrow X'$$ of $G''$-sets with the left arrow surjective and the right arrow injective. Since $g''$ is in the kernel of $h'$ we see that $g''$ acts trivially on $X'$. Hence $g''$ acts trivially on $Y''$ and hence trivially on $G''/H''$. Thus $g'' \in H''$. As this holds for all open subgroups we conclude that $g''$ is the identity element as desired. \end{proof} \begin{lemma} \label{lemma-functoriality-galois-normal} In diagram (\ref{equation-translation}) the following are equivalent \begin{enumerate} \item the image of $h'$ is normal, and \item for every connected object $X'$ of $\mathcal{C}'$ such that there is a morphism from the final object of $\mathcal{C}''$ to $H'(X')$ we have that $H'(X')$ is isomorphic to a finite coproduct of final objects. \end{enumerate} \end{lemma} \begin{proof} This translates into the following statement for the continuous group homomorphism $h' : G'' \to G'$: the image of $h'$ is normal if and only if every open subgroup $U' \subset G'$ which contains $h'(G'')$ also contains every conjugate of $h'(G'')$. The result follows easily from this; some details omitted. \end{proof} \section{Finite \'etale morphisms} \label{section-finite-etale} \noindent In this section we prove enough basic results on finite \'etale morphisms to be able to construct the \'etale fundamental group. \medskip\noindent Let $X$ be a scheme. We will use the notation $\textit{F\'Et}_X$ to denote the category of schemes finite and \'etale over $X$. Thus \begin{enumerate} \item an object of $\textit{F\'Et}_X$ is a finite \'etale morphism $Y \to X$ with target $X$, and \item a morphism in $\textit{F\'Et}_X$ from $Y \to X$ to $Y' \to X$ is a morphism $Y \to Y'$ making the diagram $$\xymatrix{ Y \ar[rr] \ar[rd] & & Y' \ar[ld] \\ & X }$$ commute. \end{enumerate} We will often call an object of $\textit{F\'Et}_X$ a {\it finite \'etale cover} of $X$ (even if $Y$ is empty). It turns out that there is a stack $p : \textit{F\'Et} \to \Sch$ over the category of schemes whose fibre over $X$ is the category $\textit{F\'Et}_X$ just defined. See Examples of Stacks, Section \ref{examples-stacks-section-finite-etale}. \begin{example} \label{example-finite-etale-geometric-point} Let $k$ be an algebraically closed field and $X = \Spec(k)$. In this case $\textit{F\'Et}_X$ is equivalent to the category of finite sets. This works more generally when $k$ is separably algebraically closed. The reason is that a scheme \'etale over $k$ is the disjoint union of spectra of fields finite separable over $k$, see Morphisms, Lemma \ref{morphisms-lemma-etale-over-field}. \end{example} \begin{lemma} \label{lemma-finite-etale-covers-limits-colimits} Let $X$ be a scheme. The category $\textit{F\'Et}_X$ has finite limits and finite colimits and for any morphism $X' \to X$ the base change functor $\textit{F\'Et}_X \to \textit{F\'Et}_{X'}$ is exact. \end{lemma} \begin{proof} Finite limits and left exactness. By Categories, Lemma \ref{categories-lemma-finite-limits-exist} it suffices to show that $\textit{F\'Et}_X$ has a final object and fibred products. This is clear because the category of all schemes over $X$ has a final object (namely $X$) and fibred products and fibred products of schemes finite \'etale over $X$ are finite \'etale over $X$. Moreover, it is clear that base change commutes with these operations and hence base change is left exact (Categories, Lemma \ref{categories-lemma-characterize-left-exact}). \medskip\noindent Finite colimits and right exactness. By Categories, Lemma \ref{categories-lemma-colimits-exist} it suffices to show that $\textit{F\'Et}_X$ has finite coproducts and coequalizers. Finite coproducts are given by disjoint unions (the empty coproduct is the empty scheme). Let $a, b : Z \to Y$ be two morphisms of $\textit{F\'Et}_X$. Since $Z \to X$ and $Y \to X$ are finite \'etale we can write $Z = \underline{\Spec}(\mathcal{C})$ and $Y = \underline{\Spec}(\mathcal{B})$ for some finite locally free $\mathcal{O}_X$-algebras $\mathcal{C}$ and $\mathcal{B}$. The morphisms $a, b$ induce two maps $a^\sharp, b^\sharp : \mathcal{B} \to \mathcal{C}$. Let $\mathcal{A} = \text{Eq}(a^\sharp, b^\sharp)$ be their equalizer. If $$\underline{\Spec}(\mathcal{A}) \longrightarrow X$$ is finite \'etale, then it is clear that this is the coequalizer (after all we can write any object of $\textit{F\'Et}_X$ as the relative spectrum of a sheaf of $\mathcal{O}_X$-algebras). This we may do after replacing $X$ by the members of an \'etale covering (Descent, Lemmas \ref{descent-lemma-descending-property-finite} and \ref{descent-lemma-descending-property-separated}). Thus by \'Etale Morphisms, Lemma \ref{etale-lemma-finite-etale-etale-local} we may assume that $Y = \coprod_{i = 1, \ldots, n} X$ and $Z = \coprod_{j = 1, \ldots, m} X$. Then $$\mathcal{C} = \prod\nolimits_{1 \leq j \leq m} \mathcal{O}_X \quad\text{and}\quad \mathcal{B} = \prod\nolimits_{1 \leq i \leq n} \mathcal{O}_X$$ After a further replacement by the members of an open covering we may assume that $a, b$ correspond to maps $a_s, b_s : \{1, \ldots, m\} \to \{1, \ldots, n\}$, i.e., the summand $X$ of $Z$ corresponding to the index $j$ maps into the summand $X$ of $Y$ corresponding to the index $a_s(j)$, resp.\ $b_s(j)$ under the morphism $a$, resp.\ $b$. Let $\{1, \ldots, n\} \to T$ be the coequalizer of $a_s, b_s$. Then we see that $$\mathcal{A} = \prod\nolimits_{t \in T} \mathcal{O}_X$$ whose spectrum is certainly finite \'etale over $X$. We omit the verification that this is compatible with base change. Thus base change is a right exact functor. \end{proof} \begin{remark} \label{remark-colimits-commute-forgetful} Let $X$ be a scheme. Consider the natural functors $F_1 : \textit{F\'Et}_X \to \Sch$ and $F_2 : \textit{F\'Et}_X \to \Sch/X$. Then \begin{enumerate} \item The functors $F_1$ and $F_2$ commute with finite colimits. \item The functor $F_2$ commutes with finite limits, \item The functor $F_1$ commutes with connected finite limits, i.e., with equalizers and fibre products. \end{enumerate} The results on limits are immediate from the discussion in the proof of Lemma \ref{lemma-finite-etale-covers-limits-colimits} and Categories, Lemma \ref{categories-lemma-connected-limit-over-X}. It is clear that $F_1$ and $F_2$ commute with finite coproducts. By the dual of Categories, Lemma \ref{categories-lemma-characterize-left-exact} we need to show that $F_1$ and $F_2$ commute with coequalizers. In the proof of Lemma \ref{lemma-finite-etale-covers-limits-colimits} we saw that coequalizers in $\textit{F\'Et}_X$ look \'etale locally like this $$\xymatrix{ \coprod_{j \in J} U \ar@<1ex>[r]^a \ar@<-1ex>[r]_b & \coprod_{i \in I} U \ar[r] & \coprod_{t \in \text{Coeq}(a, b)} U }$$ which is certainly a coequalizer in the category of schemes. Hence the statement follows from the fact that being a coequalizer is fpqc local as formulate precisely in Descent, Lemma \ref{descent-lemma-coequalizer-fpqc-local}. \end{remark} \begin{lemma} \label{lemma-internal-hom-finite-etale} Let $X$ be a scheme. Given $U, V$ finite \'etale over $X$ there exists a scheme $W$ finite \'etale over $X$ such that $$\Mor_X(X, W) = \Mor_X(U, V)$$ and such that the same remains true after any base change. \end{lemma} \begin{proof} By More on Morphisms, Lemma \ref{more-morphisms-lemma-hom-from-finite-locally-free-separated-lqf} there exists a scheme $W$ representing $\mathit{Mor}_X(U, V)$. (Use that an \'etale morphism is locally quasi-finite by Morphisms, Lemmas \ref{morphisms-lemma-etale-locally-quasi-finite} and that a finite morphism is separated.) This scheme clearly satisfies the formula after any base change. To finish the proof we have to show that $W \to X$ is finite \'etale. This we may do after replacing $X$ by the members of an \'etale covering (Descent, Lemmas \ref{descent-lemma-descending-property-finite} and \ref{descent-lemma-descending-property-separated}). Thus by \'Etale Morphisms, Lemma \ref{etale-lemma-finite-etale-etale-local} we may assume that $U = \coprod_{i = 1, \ldots, n} X$ and $V = \coprod_{j = 1, \ldots, m} X$. In this case $W = \coprod_{\alpha : \{1, \ldots, n\} \to \{1, \ldots, m\}} X$ by inspection (details omitted) and the proof is complete. \end{proof} \noindent Let $X$ be a scheme. A {\it geometric point} of $X$ is a morphism $\Spec(k) \to X$ where $k$ is algebraically closed. Such a point is usually denoted $\overline{x}$, i.e., by an overlined small case letter. We often use $\overline{x}$ to denote the scheme $\Spec(k)$ as well as the morphism, and we use $\kappa(\overline{x})$ to denote $k$. We say $\overline{x}$ {\it lies over} $x$ to indicate that $x \in X$ is the image of $\overline{x}$. We will discuss this further in \'Etale Cohomology, Section \ref{etale-cohomology-section-stalks}. Given $\overline{x}$ and an \'etale morphism $U \to X$ we can consider $$|U_{\overline{x}}| : \text{the underlying set of points of the scheme }U_{\overline{x}} = U \times_X \overline{x}$$ Since $U_{\overline{x}}$ as a scheme over $\overline{x}$ is a disjoint union of copies of $\overline{x}$ (Morphisms, Lemma \ref{morphisms-lemma-etale-over-field}) we can also describe this set as $$|U_{\overline{x}}| = \left\{ \begin{matrix} \text{commutative} \\ \text{diagrams} \end{matrix} \vcenter{ \xymatrix{ \overline{x} \ar[rd]_{\overline{x}} \ar[r]_{\overline{u}} & U \ar[d] \\ & X } } \right\}$$ The assignment $U \mapsto |U_{\overline{x}}|$ is a functor which is often denoted $F_{\overline{x}}$. \begin{lemma} \label{lemma-finite-etale-connected-galois-category} Let $X$ be a connected scheme. Let $\overline{x}$ be a geometric point. The functor $$F_{\overline{x}} : \textit{F\'Et}_X \longrightarrow \textit{Sets},\quad Y \longmapsto |Y_{\overline{x}}|$$ defines a Galois category (Definition \ref{definition-galois-category}). \end{lemma} \begin{proof} After identifying $\textit{F\'Et}_{\overline{x}}$ with the category of finite sets (Example \ref{example-finite-etale-geometric-point}) we see that our functor $F_{\overline{x}}$ is nothing but the base change functor for the morphism $\overline{x} \to X$. Thus we see that $\textit{F\'Et}_X$ has finite limits and finite colimits and that $F_{\overline{x}}$ is exact by Lemma \ref{lemma-finite-etale-covers-limits-colimits}. We will also use that finite limits in $\textit{F\'Et}_X$ agree with the corresponding finite limits in the category of schemes over $X$, see Remark \ref{remark-colimits-commute-forgetful}. \medskip\noindent If $Y' \to Y$ is a monomorphism in $\textit{F\'Et}_X$ then we see that $Y' \to Y' \times_Y Y'$ is an isomorphism, and hence $Y' \to Y$ is a monomorphism of schemes. It follows that $Y' \to Y$ is an open immersion (\'Etale Morphisms, Theorem \ref{etale-theorem-etale-radicial-open}). Since $Y'$ is finite over $X$ and $Y$ separated over $X$, the morphism $Y' \to Y$ is finite (Morphisms, Lemma \ref{morphisms-lemma-finite-permanence}), hence closed (Morphisms, Lemma \ref{morphisms-lemma-finite-proper}), hence it is the inclusion of an open and closed subscheme of $Y$. It follows that $Y$ is a connected objects of the category $\textit{F\'Et}_X$ (as in Definition \ref{definition-galois-category}) if and only if $Y$ is connected as a scheme. Then it follows from Topology, Lemma \ref{topology-lemma-finite-fibre-connected-components} that $Y$ is a finite coproduct of its connected components both as a scheme and in the sense of Definition \ref{definition-galois-category}. \medskip\noindent Let $Y \to Z$ be a morphism in $\textit{F\'Et}_X$ which induces a bijection $F_{\overline{x}}(Y) \to F_{\overline{x}}(Z)$. We have to show that $Y \to Z$ is an isomorphism. By the above we may assume $Z$ is connected. Since $Y \to Z$ is finite \'etale and hence finite locally free it suffices to show that $Y \to Z$ is finite locally free of degree $1$. This is true in a neighbourhood of any point of $Z$ lying over $\overline{x}$ and since $Z$ is connected and the degree is locally constant we conclude. \end{proof} \section{Fundamental groups} \label{section-fundamental-groups} \noindent In this section we define Grothendieck's algebraic fundamental group. The following definition makes sense thanks to Lemma \ref{lemma-finite-etale-connected-galois-category}. \begin{definition} \label{definition-fundamental-group} Let $X$ be a connected scheme. Let $\overline{x}$ be a geometric point of $X$. The {\it fundamental group} of $X$ with {\it base point} $\overline{x}$ is the group $$\pi_1(X, \overline{x}) = \text{Aut}(F_{\overline{x}})$$ of automorphisms of the fibre functor $F_{\overline{x}} : \textit{F\'Et}_X \to \textit{Sets}$ endowed with its canonical profinite topology from Lemma \ref{lemma-aut-inverse-limit}. \end{definition} \noindent Combining the above with the material from Section \ref{section-galois} we obtain the following theorem. \begin{theorem} \label{theorem-fundamental-group} Let $X$ be a connected scheme. Let $\overline{x}$ be a geometric point of $X$. \begin{enumerate} \item The fibre functor $F_{\overline{x}}$ defines an equivalence of categories $$\textit{F\'Et}_X \longrightarrow \textit{Finite-}\pi_1(X, \overline{x})\textit{-Sets}$$ \item Given a second geometric point $\overline{x}'$ of $X$ there exists an isomorphism $t : F_{\overline{x}} \to F_{\overline{x}'}$. This gives an isomorphism $\pi_1(X, \overline{x}) \to \pi_1(X, \overline{x}')$ compatible with the equivalences in (1). This isomorphism is independent of $t$ up to inner conjugation. \item Given a morphism $f : X \to Y$ of connected schemes denote $\overline{y} = f \circ \overline{x}$. There is a canonical continuous homomorphism $$f_* : \pi_1(X, \overline{x}) \to \pi_1(Y, \overline{y})$$ such that the diagram $$\xymatrix{ \textit{F\'Et}_Y \ar[r]_{\text{base change}} \ar[d]_{F_{\overline{y}}} & \textit{F\'Et}_X \ar[d]^{F_{\overline{x}}} \\ \textit{Finite-}\pi_1(Y, \overline{y})\textit{-Sets} \ar[r]^{f_*} & \textit{Finite-}\pi_1(X, \overline{x})\textit{-Sets} }$$ is commutative. \end{enumerate} \end{theorem} \begin{proof} Part (1) follows from Lemma \ref{lemma-finite-etale-connected-galois-category} and Proposition \ref{proposition-galois}. Part (2) is a special case of Lemma \ref{lemma-functoriality-galois}. For part (3) observe that the diagram $$\xymatrix{ \textit{F\'Et}_Y \ar[r] \ar[d]_{F_{\overline{y}}} & \textit{F\'Et}_X \ar[d]^{F_{\overline{x}}} \\ \textit{Sets} \ar@{=}[r] & \textit{Sets} }$$ is commutative (actually commutative, not just $2$-commutative) because $\overline{y} = f \circ \overline{x}$. Hence we can apply Lemma \ref{lemma-functoriality-galois} with the implied transformation of functors to get (3). \end{proof} \begin{lemma} \label{lemma-fundamental-group-Galois-group} Let $K$ be a field and set $X = \Spec(K)$. Let $\overline{K}$ be an algebraic closure and denote $\overline{x} : \Spec(\overline{K}) \to X$ the corresponding geometric point. Let $K^{sep} \subset \overline{K}$ be the separable algebraic closure. \begin{enumerate} \item The functor of Lemma \ref{lemma-sheaves-point} induces an equivalence $$\textit{F\'Et}_X \longrightarrow \textit{Finite-}\text{Gal}(K^{sep}/K)\textit{-Sets}.$$ compatible with $F_{\overline{x}}$ and the functor $\textit{Finite-}\text{Gal}(K^{sep}/K)\textit{-Sets} \to \textit{Sets}$. \item This induces a canonical isomorphism $$\text{Gal}(K^{sep}/K) \longrightarrow \pi_1(X, \overline{x})$$ of profinite topological groups. \end{enumerate} \end{lemma} \begin{proof} The functor of Lemma \ref{lemma-sheaves-point} is the same as the functor $F_{\overline{x}}$ because for any $Y$ \'etale over $X$ we have $$\Mor_X(\Spec(\overline{K}), Y) = \Mor_X(\Spec(K^{sep}), Y)$$ Namely, as seen in the proof of Lemma \ref{lemma-sheaves-point} we have $Y = \coprod_{i \in I} \Spec(L_i)$ with $L_i/K$ finite separable over $K$. Hence any $K$-algebra homomorphism $L_i \to \overline{K}$ factors through $K^{sep}$. Also, note that $F_{\overline{x}}(Y)$ is finite if and only if $I$ is finite if and only if $Y \to X$ is finite \'etale. This proves (1). \medskip\noindent Part (2) is a formal consequence of (1), Lemma \ref{lemma-functoriality-galois}, and Lemma \ref{lemma-single-out-profinite}. (Please also see the remark below.) \end{proof} \begin{remark} \label{remark-variance} In the situation of Lemma \ref{lemma-fundamental-group-Galois-group} let us give a more explicit construction of the isomorphism $\text{Gal}(K^{sep}/K) \to \pi_1(X, \overline{x}) = \text{Aut}(F_{\overline{x}})$. Observe that $\text{Gal}(K^{sep}/K) = \text{Aut}(\overline{K}/K)$ as $\overline{K}$ is the perfection of $K^{sep}$. Since $F_{\overline{x}}(Y) = \Mor_X(\Spec(\overline{K}), Y)$ we may consider the map $$\text{Aut}(\overline{K}/K) \times F_{\overline{x}}(Y) \to F_{\overline{x}}(Y), \quad (\sigma, \overline{y}) \mapsto \sigma \cdot \overline{y} = \overline{y} \circ \Spec(\sigma)$$ This is an action because $$\sigma\tau \cdot \overline{y} = \overline{y} \circ \Spec(\sigma\tau) = \overline{y} \circ \Spec(\tau) \circ \Spec(\sigma) = \sigma \cdot (\tau \cdot \overline{y})$$ The action is functorial in $Y \in \textit{F\'Et}_X$ and we obtain the desired map. \end{remark} \section{Topological invariance of the fundamental group} \label{section-topological-invariance} \noindent The main result of this section is that a universal homeomorphism of connected schemes induces an isomorphism on fundamental groups. See Proposition \ref{proposition-universal-homeomorphism}. \medskip\noindent Instead of directly proving two schemes have the same fundamental group, we often prove that their categories of finite \'etale coverings are the same. This of course implies that their fundamental groups are equal provided they are connected. \begin{lemma} \label{lemma-what-equivalence-gives} Let $f : X \to Y$ be a morphism of quasi-compact and quasi-separated schemes such that the base change functor $\textit{F\'Et}_Y \to \textit{F\'Et}_X$ is an equivalence of categories. In this case \begin{enumerate} \item $f$ induces a homeomorphism $\pi_0(X) \to \pi_0(Y)$, \item if $X$ or equivalently $Y$ is connected, then $\pi_1(X, \overline{x}) = \pi_1(Y, \overline{y})$. \end{enumerate} \end{lemma} \begin{proof} Let $Y = Y_0 \amalg Y_1$ be a decomposition into nonempty open and closed subschemes. We claim that $f(X)$ meets both $Y_i$. Namely, if not, say $f(X) \subset Y_1$, then we can consider the finite \'etale morphism $V = Y_1 \to Y$. This is not an isomorphism but $V \times_Y X \to X$ is an isomorphism, which is a contradiction. \medskip\noindent Suppose that $X = X_0 \amalg X_1$ is a decomposition into open and closed subschemes. Consider the finite \'etale morphism $U = X_1 \to X$. Then $U = X \times_Y V$ for some finite \'etale morphism $V \to Y$. The degree of the morphism $V \to Y$ is locally constant, hence we obtain a decomposition $Y = \coprod_{d \geq 0} Y_d$ into open and closed subschemes such that $V \to Y$ has degree $d$ over $Y_d$. Since $f^{-1}(Y_d) = \emptyset$ for $d > 1$ we conclude that $Y_d = \emptyset$ for $d > 1$ by the above. And we conclude that $f^{-1}(Y_i) = X_i$ for $i = 0, 1$. \medskip\noindent It follows that $f^{-1}$ induces a bijection between the set of open and closed subsets of $Y$ and the set of open and closed subsets of $X$. Note that $X$ and $Y$ are spectral spaces, see Properties, Lemma \ref{properties-lemma-quasi-compact-quasi-separated-spectral}. By Topology, Lemma \ref{topology-lemma-connected-component-intersection} the lattice of open and closed subsets of a spectral space determines the set of connected components. Hence $\pi_0(X) \to \pi_0(Y)$ is bijective. Since $\pi_0(X)$ and $\pi_0(Y)$ are profinite spaces (Topology, Lemma \ref{topology-lemma-pi0-profinite}) we conclude that $\pi_0(X) \to \pi_0(Y)$ is a homeomorphism by Topology, Lemma \ref{topology-lemma-bijective-map}. This proves (1). Part (2) is immediate. \end{proof} \noindent The following lemma tells us that the fundamental group of a henselian pair is the fundamental group of the closed subset. \begin{lemma} \label{lemma-gabber} Let $(A, I)$ be a henselian pair. Set $X = \Spec(A)$ and $Z = \Spec(A/I)$. The functor $$\textit{F\'Et}_X \longrightarrow \textit{F\'Et}_Z,\quad U \longmapsto U \times_X Z$$ is an equivalence of categories. \end{lemma} \begin{proof} This is a translation of More on Algebra, Lemma \ref{more-algebra-lemma-finite-etale-equivalence}. \end{proof} \noindent The following lemma tells us that the fundamental group of a thickening is the same as the fundamental group of the original. We will use this in the proof of the strong proposition concerning universal homeomorphisms below. \begin{lemma} \label{lemma-thickening} Let $X \subset X'$ be a thickening of schemes. The functor $$\textit{F\'Et}_{X'} \longrightarrow \textit{F\'Et}_X,\quad U' \longmapsto U' \times_{X'} X$$ is an equivalence of categories. \end{lemma} \begin{proof} For a discussion of thickenings see More on Morphisms, Section \ref{more-morphisms-section-thickenings}. Let $U' \to X'$ be an \'etale morphism such that $U = U' \times_{X'} X \to X$ is finite \'etale. Then $U' \to X'$ is finite \'etale as well. This follows for example from More on Morphisms, Lemma \ref{more-morphisms-lemma-properties-that-extend-over-thickenings}. Now, if $X \subset X'$ is a finite order thickening then this remark combined with \'Etale Morphisms, Theorem \ref{etale-theorem-remarkable-equivalence} proves the lemma. Below we will prove the lemma for general thickenings, but we suggest the reader skip the proof. \medskip\noindent Let $X' = \bigcup X_i'$ be an affine open covering. Set $X_i = X \times_{X'} X_i'$, $X_{ij}' = X'_i \cap X'_j$, $X_{ij} = X \times_{X'} X_{ij}'$, $X_{ijk}' = X'_i \cap X'_j \cap X'_k$, $X_{ijk} = X \times_{X'} X_{ijk}'$. Suppose that we can prove the theorem for each of the thickenings $X_i \subset X'_i$, $X_{ij} \subset X_{ij}'$, and $X_{ijk} \subset X_{ijk}'$. Then the result follows for $X \subset X'$ by relative glueing of schemes, see Constructions, Section \ref{constructions-section-relative-glueing}. Observe that the schemes $X_i'$, $X_{ij}'$, $X_{ijk}'$ are each separated as open subschemes of affine schemes. Repeating the argument one more time we reduce to the case where the schemes $X'_i$, $X_{ij}'$, $X_{ijk}'$ are affine. \medskip\noindent In the affine case we have $X' = \Spec(A')$ and $X = \Spec(A'/I')$ where $I'$ is a locally nilpotent ideal. Then $(A', I')$ is a henselian pair (More on Algebra, Lemma \ref{more-algebra-lemma-locally-nilpotent-henselian}) and the result follows from Lemma \ref{lemma-gabber} (which is much easier in this case). \end{proof} \noindent The correct'' way to prove the following proposition would be to deduce it from the invariance of the \'etale site, see \'Etale Cohomology, Theorem \ref{etale-cohomology-theorem-topological-invariance}. \begin{proposition} \label{proposition-universal-homeomorphism} Let $f : X \to Y$ be a universal homeomorphism of schemes. Then $$\textit{F\'Et}_Y \longrightarrow \textit{F\'Et}_X,\quad V \longmapsto V \times_Y X$$ is an equivalence. Thus if $X$ and $Y$ are connected, then $f$ induces an isomorphism $\pi_1(X, \overline{x}) \to \pi_1(Y, \overline{y})$ of fundamental groups. \end{proposition} \begin{proof} Recall that a universal homeomorphism is the same thing as an integral, universally injective, surjective morphism, see Morphisms, Lemma \ref{morphisms-lemma-universal-homeomorphism}. In particular, the diagonal $\Delta : X \to X \times_Y X$ is a thickening by Morphisms, Lemma \ref{morphisms-lemma-universally-injective}. Thus by Lemma \ref{lemma-thickening} we see that given a finite \'etale morphism $U \to X$ there is a unique isomorphism $$\varphi : U \times_Y X \to X \times_Y U$$ of schemes finite \'etale over $X \times_Y X$ which pulls back under $\Delta$ to $\text{id} : U \to U$ over $X$. Since $X \to X \times_Y X \times_Y X$ is a thickening as well (it is bijective and a closed immersion) we conclude that $(U, \varphi)$ is a descent datum relative to $X/Y$. By \'Etale Morphisms, Proposition \ref{etale-proposition-effective} we conclude that $U = X \times_Y V$ for some $V \to Y$ quasi-compact, separated, and \'etale. We omit the proof that $V \to Y$ is finite (hints: the morphism $U \to V$ is surjective and $U \to Y$ is integral). We conclude that $\textit{F\'Et}_Y \to \textit{F\'Et}_X$ is essentially surjective. \medskip\noindent Arguing in the same manner as above we see that given $V_1 \to Y$ and $V_2 \to Y$ in $\textit{F\'Et}_Y$ any morphism $a : X \times_Y V_1 \to X \times_Y V_2$ over $X$ is compatible with the canonical descent data. Thus $a$ descends to a morphism $V_1 \to V_2$ over $Y$ by \'Etale Morphisms, Lemma \ref{etale-lemma-fully-faithful-cases}. \end{proof} \section{Finite \'etale covers of proper schemes} \label{section-finite-etale-over-proper} \noindent In this section we show that the fundamental group of a connected proper scheme over a henselian local ring is the same as the fundamental group of its special fibre. We also show that the fundamental group of a connected proper scheme over an algebraically closed field $k$ does not change if we replace $k$ by an algebraically closed extension. Instead of stating and proving the results in the connected case we prove the results in general and we leave it to the reader to deduce the result for fundamental groups using Lemma \ref{lemma-what-equivalence-gives}. \begin{lemma} \label{lemma-finite-etale-on-proper-over-henselian} Let $A$ be a henselian local ring. Let $X$ be a proper scheme over $A$ with closed fibre $X_0$. Then the functor $$\textit{F\'Et}_X \to \textit{F\'Et}_{X_0},\quad U \longmapsto U_0 = U \times_X X_0$$ is an equivalence of categories. \end{lemma} \begin{proof} The proof given here is an example of applying algebraization and approximation. We proceed in a number of stages. \medskip\noindent Essential surjectivity when $A$ is a complete local Noetherian ring. Let $X_n = X \times_{\Spec(A)} \Spec(A/\mathfrak m^{n + 1})$. By \'Etale Morphisms, Theorem \ref{etale-theorem-remarkable-equivalence} the inclusions $$X_0 \to X_1 \to X_2 \to \ldots$$ induce equivalence of categories between the category of schemes \'etale over $X_0$ and the category of schemes \'etale over $X_n$. Moreover, if $U_n \to X_n$ corresponds to a finite \'etale morphism $U_0 \to X_0$, then $U_n \to X_n$ is finite too, for example by More on Morphisms, Lemma \ref{more-morphisms-lemma-thicken-property-morphisms-cartesian}. In this case the morphism $U_0 \to \Spec(A/\mathfrak m)$ is proper as $X_0$ is proper over $A/\mathfrak m$. Thus we may apply Grothendieck's algebraization theorem (in the form of Cohomology of Schemes, Lemma \ref{coherent-lemma-algebraize-formal-scheme-finite-over-proper}) to see that there is a finite morphism $U \to X$ whose restriction to $X_0$ recovers $U_0$. By More on Morphisms, Lemma \ref{more-morphisms-lemma-check-smoothness-on-infinitesimal-nbhds} we see that $U \to X$ is \'etale at every point of $U_0$. However, since every point of $U$ specializes to a point of $U_0$ (as $U$ is proper over $A$), we conclude that $U \to X$ is \'etale. In this way we conclude the functor is essentially surjective. \medskip\noindent Fully faithfulness when $A$ is a complete local Noetherian ring. Let $U \to X$ and $V \to X$ be finite \'etale morphisms and let $\varphi_0 : U_0 \to V_0$ be a morphism over $X_0$. Look at the morphism $$\Gamma_{\varphi_0} : U_0 \longrightarrow U_0 \times_{X_0} V_0$$ This morphism is both finite \'etale and a closed immersion. By essential surjectivity applied to $X = U \times_X V$ we find a finite \'etale morphism $W \to U \times_X V$ whose special fibre is isomorphic to $\Gamma_{\varphi_0}$. Consider the projection $W \to U$. It is finite \'etale and an isomorphism over $U_0$ by construction. By \'Etale Morphisms, Lemma \ref{etale-lemma-finite-etale-one-point} $W \to U$ is an isomorphism in an open neighbourhood of $U_0$. Thus it is an isomorphism and the composition $\varphi : U \cong W \to V$ is the desired lift of $\varphi_0$. \medskip\noindent Essential surjectivity when $A$ is a henselian local Noetherian G-ring. Let $U_0 \to X_0$ be a finite \'etale morphism. Let $A^\wedge$ be the completion of $A$ with respect to the maximal ideal. Let $X^\wedge$ be the base change of $X$ to $A^\wedge$. By the result above there exists a finite \'etale morphism $V \to X^\wedge$ whose special fibre is $U_0$. Write $A^\wedge = \colim A_i$ with $A \to A_i$ of finite type. By Limits, Lemma \ref{limits-lemma-descend-finite-presentation} there exists an $i$ and a finitely presented morphism $U_i \to X_{A_i}$ whose base change to $X^\wedge$ is $V$. After increasing $i$ we may assume that $U_i \to X_{A_i}$ is finite and \'etale (Limits, Lemmas \ref{limits-lemma-descend-finite-finite-presentation} and \ref{limits-lemma-descend-etale}). Writing $$A_i = A[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$$ the ring map $A_i \to A^\wedge$ can be reinterpreted as a solution $(a_1, \ldots, a_n)$ in $A^\wedge$ for the system of equations $f_j = 0$. By Smoothing Ring Maps, Theorem \ref{smoothing-theorem-approximation-property} we can approximate this solution (to order $11$ for example) by a solution $(b_1, \ldots, b_n)$ in $A$. Translating back we find an $A$-algebra map $A_i \to A$ which gives the same closed point as the original map $A_i \to A^\wedge$ (as $11 > 1$). The base change $U \to X$ of $V \to X_{A_i}$ by this ring map will therefore be a finite \'etale morphism whose special fibre is isomorphic to $U_0$. \medskip\noindent Fully faithfulness when $A$ is a henselian local Noetherian G-ring. This can be deduced from essential surjectivity in exactly the same manner as was done in the case that $A$ is complete Noetherian. \medskip\noindent General case. Let $(A, \mathfrak m)$ be a henselian local ring. Set $S = \Spec(A)$ and denote $s \in S$ the closed point. By Limits, Lemma \ref{limits-lemma-proper-limit-of-proper-finite-presentation-noetherian} we can write $X \to \Spec(A)$ as a cofiltered limit of proper morphisms $X_i \to S_i$ with $S_i$ of finite type over $\mathbf{Z}$. For each $i$ let $s_i \in S_i$ be the image of $s$. Since $S = \lim S_i$ and $A = \mathcal{O}_{S, s}$ we have $A = \colim \mathcal{O}_{S_i, s_i}$. The ring $A_i = \mathcal{O}_{S_i, s_i}$ is a Noetherian local G-ring (More on Algebra, Proposition \ref{more-algebra-proposition-ubiquity-G-ring}). By More on Algebra, Lemma \ref{more-algebra-lemma-henselization-colimit} we see that $A = \colim A_i^h$. By More on Algebra, Lemma \ref{more-algebra-lemma-henselization-G-ring} the rings $A_i^h$ are G-rings. Thus we see that $A = \colim A_i^h$ and $$X = \lim (X_i \times_{S_i} \Spec(A_i^h))$$ as schemes. The category of schemes finite \'etale over $X$ is the limit of the category of schemes finite \'etale over $X_i \times_{S_i} \Spec(A_i^h)$ (by Limits, Lemmas \ref{limits-lemma-descend-finite-presentation}, \ref{limits-lemma-descend-finite-finite-presentation}, and \ref{limits-lemma-descend-etale}) The same thing is true for schemes finite \'etale over $X_0 = \lim (X_i \times_{S_i} s_i)$. Thus we formally deduce the result for $X / \Spec(A)$ from the result for the $(X_i \times_{S_i} \Spec(A_i^h)) / \Spec(A_i^h)$ which we dealt with above. \end{proof} \begin{lemma} \label{lemma-finite-etale-invariant-over-proper} Let $k \subset k'$ be an extension of algebraically closed fields. Let $X$ be a proper scheme over $k$. Then the functor $$U \longmapsto U_{k'}$$ is an equivalence of categories between schemes finite \'etale over $X$ and schemes finite \'etale over $X_{k'}$. \end{lemma} \begin{proof} Let us prove the functor is essentially surjective. Let $U' \to X_{k'}$ be a finite \'etale morphism. Write $k' = \colim A_i$ as a filtered colimit of finite type $k$-algebras. By Limits, Lemma \ref{limits-lemma-descend-finite-presentation} there exists an $i$ and a finitely presented morphism $U_i \to X_{A_i}$ whose base change to $X_{k'}$ is $U'$. After increasing $i$ we may assume that $U_i \to X_{A_i}$ is finite and \'etale (Limits, Lemmas \ref{limits-lemma-descend-finite-finite-presentation} and \ref{limits-lemma-descend-etale}). Since $k$ is algebraically closed we can find a $k$-valued point $t$ in $\Spec(A_i)$. Let $U = (U_i)_t$ be the fibre of $U_i$ over $t$. Let $A_i^h$ be the henselization of $(A_i)_{\mathfrak m}$ where $\mathfrak m$ is the maximal ideal corresponding to the point $t$. By Lemma \ref{lemma-finite-etale-on-proper-over-henselian} we see that $(U_i)_{A_i^h} = U \times \Spec(A_i^h)$ as schemes over $X_{A_i^h}$. Now since $A_i^h$ is algebraic over $A_i$ (see for example discussion in Smoothing Ring Maps, Example \ref{smoothing-example-describe-henselian}) and since $k'$ is algebraically closed we can find a ring map $A_i^h \to k'$ extending the given inclusion $A_i \subset k'$. Hence we conclude that $U'$ is isomorphic to the base change of $U$. The proof of fully faithfulness is exactly the same. \end{proof} \section{Local connectedness} \label{section-unibranch} \noindent In this section we ask when $\pi_1(U) \to \pi_1(X)$ is surjective for $U$ a dense open of a scheme $X$. We will see that this is the case (roughly) when $U \cap B$ is connected for any small ball'' $B$ around a point $x \in X \setminus U$. \begin{lemma} \label{lemma-dense-faithful} Let $f : X \to Y$ be a morphism of schemes. If $f(X)$ is dense in $Y$ then the base change functor $\textit{F\'Et}_Y \to \textit{F\'Et}_X$ is faithful. \end{lemma} \begin{proof} Since the category of finite \'etale coverings has an internal hom (Lemma \ref{lemma-internal-hom-finite-etale}) it suffices to prove the following: Given $W$ finite \'etale over $Y$ and a morphism $s : X \to W$ over $X$ there is at most one section $t : Y \to W$ such that $s = t \circ f$. Consider two sections $t_1, t_2 : Y \to W$ such that $s = t_1 \circ f = t_2 \circ f$. Since the equalizer of $t_1$ and $t_2$ is closed in $Y$ (Schemes, Lemma \ref{schemes-lemma-where-are-they-equal}) and since $f(X)$ is dense in $Y$ we see that $t_1$ and $t_2$ agree on $Y_{red}$. Then it follows that $t_1$ and $t_2$ have the same image which is an open and closed subscheme of $W$ mapping isomorphically to $Y$ (\'Etale Morphisms, Proposition \ref{etale-proposition-properties-sections}) hence they are equal. \end{proof} \noindent The condition in the following lemma that the punctured spectrum of the strict henselization is connected follows for example from the assumption that the local ring is geometrically unibranch, see More on Algebra, Lemma \ref{more-algebra-lemma-geometrically-unibranch}. There is a partial converse in Properties, Lemma \ref{properties-lemma-geometrically-unibranch}. \begin{lemma} \label{lemma-same-etale-extensions} Let $(A, \mathfrak m)$ be a local ring. Set $X = \Spec(A)$ and let $U = X \setminus \{\mathfrak m\}$. If the punctured spectrum of the strict henselization of $A$ is connected, then $$\textit{F\'Et}_X \longrightarrow \textit{F\'Et}_U,\quad Y \longmapsto Y \times_X U$$ is a fully faithful functor. \end{lemma} \begin{proof} Assume $A$ is strictly henselian. In this case any finite \'etale cover $Y$ of $X$ is isomorphic to a finite disjoint union of copies of $X$. Thus it suffices to prove that any morphism $U \to U \amalg \ldots \amalg U$ over $U$, extends uniquely to a morphism $X \to X \amalg \ldots \amalg X$ over $X$. If $U$ is connected (in particular nonempty), then this is true. \medskip\noindent The general case. Since the category of finite \'etale coverings has an internal hom (Lemma \ref{lemma-internal-hom-finite-etale}) it suffices to prove the following: Given $Y$ finite \'etale over $X$ any morphism $s : U \to Y$ over $X$ extends to a morphism $t : X \to Y$ over $Y$. Let $A^{sh}$ be the strict henselization of $A$ and denote $X^{sh} = \Spec(A^{sh})$, $U^{sh} = U \times_X X^{sh}$, $Y^{sh} = Y \times_X X^{sh}$. By the first paragraph and our assumption on $A$, we can extend the base change $s^{sh} : U^{sh} \to Y^{sh}$ of $s$ to $t^{sh} : X^{sh} \to Y^{sh}$. Set $A' = A^{sh} \otimes_A A^{sh}$. Then the two pullbacks $t'_1, t'_2$ of $t^{sh}$ to $X' = \Spec(A')$ are extensions of the pullback $s'$ of $s$ to $U' = U \times_X X'$. As $A \to A'$ is flat we see that $U' \subset X'$ is (topologically) dense by going down for $A \to A'$ (Algebra, Lemma \ref{algebra-lemma-flat-going-down}). Thus $t'_1 = t'_2$ by Lemma \ref{lemma-dense-faithful}. Hence $t^{sh}$ descends to a morphism $t : X \to Y$ for example by Descent, Lemma \ref{descent-lemma-fpqc-universal-effective-epimorphisms}. \end{proof} \noindent In view of Lemma \ref{lemma-same-etale-extensions} it is interesting to know when the punctured spectrum of a ring (and of its strict henselization) is connected. The following famous lemma due to Hartshorne gives a sufficient condition. \begin{lemma} \label{lemma-depth-2-connected-punctured-spectrum} \begin{reference} \cite[Proposition 2.1]{Hartshorne-connectedness} \end{reference} \begin{slogan} Hartshorne's connectedness \end{slogan} Let $A$ be a Noetherian local ring of depth $\geq 2$. Then the punctured spectra of $A$, $A^h$, and $A^{sh}$ are connected. \end{lemma} \begin{proof} Let $U$ be the punctured spectrum of $A$. If $U$ is disconnected then we see that $\Gamma(U, \mathcal{O}_U)$ has a nontrivial idempotent. But $A$, being local, does not have a nontrivial idempotent. Hence $A \to \Gamma(U, \mathcal{O}_U)$ is not an isomorphism. By Dualizing Complexes, Lemma \ref{dualizing-lemma-finiteness-pushforwards-and-H1-local} we conclude that either $H^0_\mathfrak m(A)$ or $H^1_\mathfrak m(A)$ is nonzero. Thus $\text{depth}(A) \leq 1$ by Dualizing Complexes, Lemma \ref{dualizing-lemma-depth}. To see the result for $A^h$ and $A^{sh}$ use More on Algebra, Lemma \ref{more-algebra-lemma-henselization-depth}. \end{proof} \begin{lemma} \label{lemma-quasi-compact-dense-open-connected-at-infinity-Noetherian} Let $X$ be a scheme. Let $U \subset X$ be a dense open. Assume \begin{enumerate} \item the underlying topological space of $X$ is Noetherian, and \item for every $x \in X \setminus U$ the punctured spectrum of the strict henselization of $\mathcal{O}_{X, x}$ is connected. \end{enumerate} Then $\textit{F\'Et}_X \to \textit{F\'et}_U$ is fully faithful. \end{lemma} \begin{proof} Let $Y_1, Y_2$ be finite \'etale over $X$ and let $\varphi : (Y_1)_U \to (Y_2)_U$ be a morphism over $U$. We have to show that $\varphi$ lifts uniquely to a morphism $Y_1 \to Y_2$ over $X$. Uniqueness follows from Lemma \ref{lemma-dense-faithful}. \medskip\noindent Let $x \in X \setminus U$ be a generic point of an irreducible component of $X \setminus U$. Set $V = U \times_X \Spec(\mathcal{O}_{X, x})$. By our choice of $x$ this is the punctured spectrum of $\Spec(\mathcal{O}_{X, x})$. By Lemma \ref{lemma-same-etale-extensions} we can extend the morphism $\varphi_V : (Y_1)_V \to (Y_2)_V$ uniquely to a morphism $(Y_1)_{\Spec(\mathcal{O}_{X, x})} \to (Y_2)_{\Spec(\mathcal{O}_{X, x})}$. By Limits, Lemma \ref{limits-lemma-glueing-near-point} we find an open $U \subset U'$ containing $x$ and an extension $\varphi' : (Y_1)_{U'} \to (Y_2)_{U'}$ of $\varphi$. Since the underlying topological space of $X$ is Noetherian this finishes the proof by Noetherian induction on the complement of the open over which $\varphi$ is defined. \end{proof} \begin{lemma} \label{lemma-retrocompact-dense-open-connected-at-infinity-closed} Let $X$ be a scheme. Let $U \subset X$ be a dense open. Assume \begin{enumerate} \item $U \to X$ is quasi-compact, \item every point of $X \setminus U$ is closed, and \item for every $x \in X \setminus U$ the punctured spectrum of the strict henselization of $\mathcal{O}_{X, x}$ is connected. \end{enumerate} Then $\textit{F\'Et}_X \to \textit{F\'et}_U$ is fully faithful. \end{lemma} \begin{proof} Let $Y_1, Y_2$ be finite \'etale over $X$ and let $\varphi : (Y_1)_U \to (Y_2)_U$ be a morphism over $U$. We have to show that $\varphi$ lifts uniquely to a morphism $Y_1 \to Y_2$ over $X$. Uniqueness follows from Lemma \ref{lemma-dense-faithful}. \medskip\noindent Let $x \in X \setminus U$. Set $V = U \times_X \Spec(\mathcal{O}_{X, x})$. Since every point of $X \setminus U$ is closed $V$ is the punctured spectrum of $\Spec(\mathcal{O}_{X, x})$. By Lemma \ref{lemma-same-etale-extensions} we can extend the morphism $\varphi_V : (Y_1)_V \to (Y_2)_V$ uniquely to a morphism $(Y_1)_{\Spec(\mathcal{O}_{X, x})} \to (Y_2)_{\Spec(\mathcal{O}_{X, x})}$. By Limits, Lemma \ref{limits-lemma-glueing-near-point} (this uses that $U$ is retrocompact in $X$) we find an open $U \subset U'_x$ containing $x$ and an extension $\varphi'_x : (Y_1)_{U'_x} \to (Y_2)_{U'_x}$ of $\varphi$. Note that given two points $x, x' \in X \setminus U$ the morphisms $\varphi'_x$ and $\varphi'_{x'}$ agree over $U'_x \cap U'_{x'}$ as $U$ is dense in that open (Lemma \ref{lemma-dense-faithful}). Thus we can extend $\varphi$ to $\bigcup U'_x = X$ as desired. \end{proof} \begin{lemma} \label{lemma-quasi-compact-dense-open-connected-at-infinity} Let $X$ be a scheme. Let $U \subset X$ be a dense open. Assume \begin{enumerate} \item every quasi-compact open of $X$ has finitely many irreducible components, \item for every $x \in X \setminus U$ the punctured spectrum of the strict henselization of $\mathcal{O}_{X, x}$ is connected. \end{enumerate} Then $\textit{F\'Et}_X \to \textit{F\'et}_U$ is fully faithful. \end{lemma} \begin{proof} Let $Y_1, Y_2$ be finite \'etale over $X$ and let $\varphi : (Y_1)_U \to (Y_2)_U$ be a morphism over $U$. We have to show that $\varphi$ lifts uniquely to a morphism $Y_1 \to Y_2$ over $X$. Uniqueness follows from Lemma \ref{lemma-dense-faithful}. We will prove existence by showing that we can enlarge $U$ if $U \not = X$ and using Zorn's lemma to finish the proof. \medskip\noindent Let $x \in X \setminus U$ be a generic point of an irreducible component of $X \setminus U$. Set $V = U \times_X \Spec(\mathcal{O}_{X, x})$. By our choice of $x$ this is the punctured spectrum of $\Spec(\mathcal{O}_{X, x})$. By Lemma \ref{lemma-same-etale-extensions} we can extend the morphism $\varphi_V : (Y_1)_V \to (Y_2)_V$ (uniquely) to a morphism $(Y_1)_{\Spec(\mathcal{O}_{X, x})} \to (Y_2)_{\Spec(\mathcal{O}_{X, x})}$. Choose an affine neighbourhood $W \subset X$ of $x$. Since $U \cap W$ is dense in $W$ it contains the generic points $\eta_1, \ldots, \eta_n$ of $W$. Choose an affine open $W' \subset W \cap U$ containing $\eta_1, \ldots, \eta_n$. Set $V' = W' \times_X \Spec(\mathcal{O}_{X, x})$. By Limits, Lemma \ref{limits-lemma-glueing-near-point} applied to $W' \subset W \ni x$ we find an open $W' \subset W'' \subset W$ with $x \in W''$ and a morphism $\varphi'' : (Y_1)_{W''} \to (Y_2)_{W''}$ agreeing with $\varphi$ over $W'$. Since $W'$ is dense in $W'' \cap U$, we see by Lemma \ref{lemma-dense-faithful} that $\varphi$ and $\varphi''$ agree over $U \cap W'$. Thus $\varphi$ and $\varphi''$ glue to a morphism $\varphi'$ over $U' = U \cup W''$ agreeing with $\varphi$ over $U$. Observe that $x \in U'$ so that we've extended $\varphi$ to a strictly larger open. \medskip\noindent Consider the set $\mathcal{S}$ of pairs $(U', \varphi')$ where $U \subset U'$ and $\varphi'$ is an extension of $\varphi$. We endow $\mathcal{S}$ with a partial ordering in the obvious manner. If $(U'_i, \varphi'_i)$ is a totally ordered subset, then it has a maximum $(U', \varphi')$. Just take $U' = \bigcup U'_i$ and let $\varphi' : (Y_1)_{U'} \to (Y_2)_{U'}$ be the morphism agreeing with $\varphi'_i$ over $U'_i$. Thus Zorn's lemma applies and $\mathcal{S}$ has a maximal element. By the argument above we see that this maximal element is an extension of $\varphi$ over all of $X$. \end{proof} \begin{lemma} \label{lemma-local-exact-sequence} Let $(A, \mathfrak m)$ be a local ring. Set $X = \Spec(A)$ and $U = X \setminus \{\mathfrak m\}$. Let $U^{sh}$ be the punctured spectrum of the strict henselization $A^{sh}$ of $A$. Assume $U$ is quasi-compact and $U^{sh}$ is connected. Then the sequence $$\pi_1(U^{sh}, \overline{u}) \to \pi_1(U, \overline{u}) \to \pi_1(X, \overline{u}) \to 1$$ is exact in the sense of Lemma \ref{lemma-functoriality-galois-ses} part (1). \end{lemma} \begin{proof} The map $\pi_1(U) \to \pi_1(X)$ is surjective by Lemmas \ref{lemma-same-etale-extensions} and \ref{lemma-functoriality-galois-surjective}. \medskip\noindent Write $X^{sh} = \Spec(A^{sh})$. Let $Y \to X$ be a finite \'etale morphism. Then $Y^{sh} = Y \times_X X^{sh} \to X^{sh}$ is a finite \'etale 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 $Y \times_X U^{sh}$. It follows that the composition $\pi_1(U^{sh}) \to \pi_1(U) \to \pi_1(X)$ is trivial, see Lemma \ref{lemma-composition-trivial}. \medskip\noindent To finish the proof, it suffices according to Lemma \ref{lemma-functoriality-galois-ses} to show the following: Given a finite \'etale morphism $V \to U$ such that $V \times_U U^{sh}$ is a disjoint union of copies of $U^{sh}$, we can find a finite \'etale morphism $Y \to X$ with $V \cong Y \times_X U$ over $U$. The assumption implies that there exists a finite \'etale morphism $Y^{sh} \to X^{sh}$ and an isomorphism $V \times_U U^{sh} \cong Y^{sh} \times_{X^{sh}} U^{sh}$. Consider the following diagram $$\xymatrix{ U \ar[d] & U^{sh} \ar[d] \ar[l] & U^{sh} \times_U U^{sh} \ar[d] \ar@<1ex>[l] \ar@<-1ex>[l] & U^{sh} \times_U U^{sh} \times_U U^{sh} \ar[d] \ar@<1ex>[l] \ar[l] \ar@<-1ex>[l] \\ X & X^{sh} \ar[l] & X^{sh} \times_X X^{sh} \ar@<1ex>[l] \ar@<-1ex>[l] & X^{sh} \times_X X^{sh} \times_X X^{sh} \ar@<1ex>[l] \ar[l] \ar@<-1ex>[l] }$$ Since $U \subset X$ is quasi-compact by assumption, all the downward arrows are quasi-compact open immersions. Let $\xi \in X^{sh} \times_X X^{sh}$ be a point not in $U^{sh} \times_U U^{sh}$. Then $\xi$ lies over the closed point $x^{sh}$ of $X^{sh}$. Consider the local ring homomorphism $$A^{sh} = \mathcal{O}_{X^{sh}, x^{sh}} \to \mathcal{O}_{X^{sh} \times_X X^{sh}, \xi}$$ determined by the first projection $X^{sh} \times_X X^{sh}$. This is a filtered colimit of local homomorphisms which are localizations \'etale ring maps. Since $A^{sh}$ is strictly henselian, we conclude that it is an isomorphism. Since this holds for every $\xi$ in the complement it follows there are no specializations among these points and hence every such $\xi$ is a closed point (you can also prove this directly). As the local ring at $\xi$ is isomorphic to $A^{sh}$, it is strictly henselian and has connected punctured spectrum. Similarly for points $\xi$ of $X^{sh} \times_X X^{sh} \times_X X^{sh}$ not in $U^{sh} \times_U U^{sh} \times_U U^{sh}$. It follows from Lemma \ref{lemma-retrocompact-dense-open-connected-at-infinity-closed} that pullback along the vertical arrows induce fully faithful functors on the categories of finite \'etale schemes. Thus the canonical descent datum on $V \times_U U^{sh}$ relative to the fpqc covering $\{U^{sh} \to U\}$ translates into a descent datum for $Y^{sh}$ relative to the fpqc covering $\{X^{sh} \to X\}$. Since $Y^{sh} \to X^{sh}$ is finite hence affine, this descent datum is effective (Descent, Lemma \ref{descent-lemma-affine}). Thus we get an affine morphism $Y \to X$ and an isomorphism $Y \times_X X^{sh} \to Y^{sh}$ compatible with descent data. By fully faithfulness of descent data (as in Descent, Lemma \ref{descent-lemma-refine-coverings-fully-faithful}) we get an isomorphism $V \to U \times_X Y$. Finally, $Y \to X$ is finite \'etale as $Y^{sh} \to X^{sh}$ is, see Descent, Lemmas \ref{descent-lemma-descending-property-etale} and \ref{descent-lemma-descending-property-finite}. \end{proof} \noindent Let $X$ be an irreducible scheme. Let $\eta \in X$ be the geometric point. The canonical morphism $\eta \to X$ induces a canonical map \begin{equation} \label{equation-inclusion-generic-point} \text{Gal}(\kappa(\eta)^{sep}/\kappa(\eta)) = \pi_1(\eta, \overline{\eta}) \longrightarrow \pi_1(X, \overline{\eta}) \end{equation} The identification on the left hand side is Lemma \ref{lemma-fundamental-group-Galois-group}. \begin{lemma} \label{lemma-irreducible-geometrically-unibranch} Let $X$ be an irreducible, geometrically unibranch scheme. For any nonempty open $U \subset X$ the canonical map $$\pi_1(U, \overline{u}) \longrightarrow \pi_1(X, \overline{u})$$ is surjective. The map (\ref{equation-inclusion-generic-point}) $\pi_1(\eta, \overline{\eta}) \to \pi_1(X, \overline{\eta})$ is surjective as well. \end{lemma} \begin{proof} By Lemma \ref{lemma-thickening} we may replace $X$ by its reduction. Thus we may assume that $X$ is an integral scheme. By Lemma \ref{lemma-functoriality-galois-surjective} the assertion of the lemma translates into the statement that the functors $\textit{F\'Et}_X \to \textit{F\'Et}_U$ and $\textit{F\'Et}_X \to \textit{F\'Et}_\eta$ are fully faithful. \medskip\noindent The result for $\textit{F\'Et}_X \to \textit{F\'Et}_U$ follows from Lemma \ref{lemma-quasi-compact-dense-open-connected-at-infinity} and the fact that for a local ring $A$ which is geometrically unibranch its strict henselization has an irreducible spectrum. See More on Algebra, Lemma \ref{more-algebra-lemma-geometrically-unibranch}. \medskip\noindent Observe that the residue field $\kappa(\eta) = \mathcal{O}_{X, \eta}$ is the filtered colimit of $\mathcal{O}_X(U)$ over $U \subset X$ nonempty open affine. Hence $\textit{F\'Et}_\eta$ is the colimit of the categories $\textit{F\'Et}_U$ over such $U$, see Limits, Lemmas \ref{limits-lemma-descend-finite-presentation}, \ref{limits-lemma-descend-finite-finite-presentation}, and \ref{limits-lemma-descend-etale}. A formal argument then shows that fully faithfulness for $\textit{F\'Et}_X \to \textit{F\'Et}_\eta$ follows from the fully faithfulness of the functors $\textit{F\'Et}_X \to \textit{F\'Et}_U$. \end{proof} \begin{lemma} \label{lemma-exact-sequence-finite-nr-closed-pts} Let $X$ be a scheme. Let $x_1, \ldots, x_n \in X$ be a finite number of closed points such that \begin{enumerate} \item $U = X \setminus \{x_1, \ldots, x_n\}$ is connected and is a retrocompact open of $X$, and \item for each $i$ the punctured spectrum $U_i^{sh}$ of the strict henselization of $\mathcal{O}_{X, x_i}$ is connected. \end{enumerate} Then the map $\pi_1(U) \to \pi_1(X)$ is surjective and the kernel is the smallest closed normal subgroup of $\pi_1(U)$ containing the image of $\pi_1(U_i^{sh}) \to \pi_1(U)$ for $i = 1, \ldots, n$. \end{lemma} \begin{proof} Surjectivity follows from Lemmas \ref{lemma-retrocompact-dense-open-connected-at-infinity-closed} and \ref{lemma-functoriality-galois-surjective}. We can consider the sequence of maps $$\pi_1(U) \to \ldots \to \pi_1(X \setminus \{x_1, x_2\}) \to \pi_1(X \setminus \{x_1\}) \to \pi_1(X)$$ A group theory argument then shows it suffices to prove the statement on the kernel in the case $n = 1$ (details omitted). Write $x = x_1$, $U^{sh} = U_1^{sh}$, set $A = \mathcal{O}_{X, x}$, and let $A^{sh}$ be the strict henselization. Consider the diagram $$\xymatrix{ U \ar[d] & \Spec(A) \setminus \{\mathfrak m\} \ar[l] \ar[d] & U^{sh} \ar[d] \ar[l] \\ X & \Spec(A) \ar[l] & \Spec(A^{sh}) \ar[l] }$$ By Lemma \ref{lemma-functoriality-galois-ses} we have to show finite \'etale morphisms $V \to U$ which pull back to trivial coverings of $U^{sh}$ extend to finite \'etale schemes over $X$. By Lemma \ref{lemma-local-exact-sequence} we know the corresponding statement for finite \'etale schemes over the punctured spectrum of $A$. However, by Limits, Lemma \ref{limits-lemma-glueing-near-closed-point} schemes of finite presentation over $X$ are the same thing as schemes of finite presentation over $U$ and $A$ glued over the punctured spectrum of $A$. This finishes the proof. \end{proof} \section{Fundamental groups of normal schemes} \label{section-normal} \noindent Let $X$ be an integral, geometrically unibranch scheme. In the previous section we have seen that the fundamental group of $X$ is a quotient of the Galois group of the function field $K$ of $X$. Since the map is continuous the kernel is a normal closed subgroup of the Galois group. Hence this kernel corresponds to a Galois extension $M/K$ by Galois theory (Fields, Theorem \ref{fields-theorem-inifinite-galois-theory}). In this section we will determine $M$ when $X$ is a normal integral scheme. \medskip\noindent Let $X$ be an integral normal scheme with function field $K$. Let $K \subset L$ be a finite extension. Consider the normalization $Y \to X$ of $X$ in the morphism $\Spec(L) \to X$ as defined in Morphisms, Section \ref{morphisms-section-normalization-X-in-Y}. We will say (in this setting) that {\it $X$ is unramified in $L$} if $Y \to X$ is an unramified morphism of schemes. In Lemma \ref{lemma-unramified} we will elucidate this condition. Observe that the scheme theoretic fibre of $Y \to X$ over $\Spec(K)$ is $\Spec(L)$. Hence the field extension $L/K$ is separable if $X$ is unramified in $L$, see Morphisms, Lemmas \ref{morphisms-lemma-unramified-over-field}. \begin{lemma} \label{lemma-unramified-in-L} In the situation above the following are equivalent \begin{enumerate} \item $X$ is unramified in $L$, \item $Y \to X$ is \'etale, and \item $Y \to X$ is finite \'etale. \end{enumerate} \end{lemma} \begin{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 the lemma is finite by Morphisms, Lemma \ref{morphisms-lemma-finite-integral}. In particular we see that (2) is equivalent to (3). An \'etale morphism is unramified, hence (2) implies (1). \medskip\noindent Conversely, assume $Y \to X$ is unramified. Let $x \in X$. We can choose an \'etale neighbourhood $(U, u) \to (X, x)$ such that $$Y \times_X U = \coprod V_j \longrightarrow U$$ is a disjoint union of closed immersions, see \'Etale Morphisms, Lemma \ref{etale-lemma-finite-unramified-etale-local}. Shrinking we may assume $U$ is quasi-compact. Then $U$ has finitely many irreducible components (Descent, Lemma \ref{descent-lemma-locally-finite-nr-irred-local-fppf}). Since $U$ is normal (Descent, Lemma \ref{descent-lemma-normal-local-smooth}) the irreducible components of $U$ are open and closed (Properties, Lemma \ref{properties-lemma-normal-locally-finite-nr-irreducibles}) 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 $\Spec(L) \times_X U$ by More on Morphisms, Lemma \ref{more-morphisms-lemma-normalization-smooth-localization}. Every point of $\Spec(L) \times_X U$ maps to $\xi$. Thus every $V_j$ contains a point mapping to $\xi$ by Morphisms, Lemma \ref{morphisms-lemma-normalization-generic}. Thus $V_j \to U$ is an isomorphism as $U = \overline{\{\xi\}}$. Thus $Y \times_X U \to U$ is \'etale. By Descent, Lemma \ref{descent-lemma-descending-property-etale} we conclude that $Y \to X$ is \'etale over the image of $U \to X$ (an open neighbourhood of $x$). \end{proof} \begin{lemma} \label{lemma-finite-etale-covering-normal-unramified} Let $X$ be a normal integral scheme with function field $K$. Let $Y \to X$ be a finite \'etale morphism. If $Y$ is connected, then $Y$ is an integral normal scheme and $Y$ is the normalization of $X$ in the function field of $Y$. \end{lemma} \begin{proof} The scheme $Y$ is normal by Descent, Lemma \ref{descent-lemma-normal-local-smooth}. Since $Y \to X$ is flat every generic point of $Y$ maps to the generic point of $X$ by Morphisms, Lemma \ref{morphisms-lemma-generalizations-lift-flat}. 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 \ref{properties-lemma-normal-locally-finite-nr-irreducibles}. Thus if $Y$ is connected, then $Y$ is an integral normal scheme. \medskip\noindent Let $L$ be the function field of $Y$ and let $Y' \to X$ be the normalization of $X$ in $L$. By Morphisms, Lemma \ref{morphisms-lemma-characterize-normalization} 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 \ref{morphisms-lemma-finite-birational-over-normal}). \end{proof} \begin{proposition} \label{proposition-normal} Let $X$ be a normal integral scheme with function field $K$. Then the canonical map (\ref{equation-inclusion-generic-point}) $$\text{Gal}(K^{sep}/K) = \pi_1(\eta, \overline{\eta}) \longrightarrow \pi_1(X, \overline{\eta})$$ is identified with the quotient map $\text{Gal}(K^{sep}/K) \to \text{Gal}(M/K)$ where $M \subset K^{sep}$ is the union of the finite subextensions $L$ such that $X$ is unramified in $L$. \end{proposition} \begin{proof} The normal scheme $X$ is geometrically unibranch (Properties, Lemma \ref{properties-lemma-normal-geometrically-unibranch}). Hence Lemma \ref{lemma-irreducible-geometrically-unibranch} 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 $$\xymatrix{ \textit{F\'Et}_X \ar[r] \ar[d] \ar[rd]_c & \textit{F\'Et}_\eta \ar[d] \\ \textit{Finite-}\pi_1(X, \overline{\eta})\textit{-sets} \ar[r] & \textit{Finite-}\text{Gal}(K^{sep}/K)\textit{-sets} }$$ is fully faithful. The left vertical arrow is the equivalence of Theorem \ref{theorem-fundamental-group} and the right vertical arrow is the equivalence of Lemma \ref{lemma-fundamental-group-Galois-group}. The lower horizontal arrow is induced by the map of the proposition. By Lemmas \ref{lemma-unramified-in-L} and \ref{lemma-finite-etale-covering-normal-unramified} we see that the essential image of $c$ consists of $\text{Gal}(K^{sep}/K)\textit{-Sets}$ isomorphic to sets of the form $$S = \Hom_K(\prod\nolimits_{i = 1, \ldots, n} L_i, K^{sep}) = \coprod\nolimits_{i = 1, \ldots, n} \Hom_K(L_i, K^{sep})$$ 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 $$1 \to \text{Gal}(K^{sep}/M) \to \text{Gal}(K^{sep}/K) \to \text{Gal}(M/K) \to 1$$ by Galois theory (Fields, Theorem \ref{fields-theorem-inifinite-galois-theory} and Lemma \ref{fields-lemma-ses-infinite-galois}). The proof is done. \end{proof} \begin{lemma} \label{lemma-local-exact-sequence-normal} Let $(A, \mathfrak m)$ be a normal local ring. Set $X = \Spec(A)$. 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 $$\pi_1(\Spec(K^{sh})) \to \pi_1(\Spec(K)) \to \pi_1(X) \to 1$$ is exact in the sense of Lemma \ref{lemma-functoriality-galois-ses} part (1). \end{lemma} \begin{proof} Note that $A^{sh}$ is a normal domain, see More on Algebra, Lemma \ref{more-algebra-lemma-henselization-normal}. The map $\pi_1(\Spec(K)) \to \pi_1(X)$ is surjective by Proposition \ref{proposition-normal}. \medskip\noindent Write $X^{sh} = \Spec(A^{sh})$. Let $Y \to X$ be a finite \'etale morphism. Then $Y^{sh} = Y \times_X X^{sh} \to X^{sh}$ is a finite \'etale 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 $Y \times_X \Spec(K^{sh})$. It follows that the composition $\pi_1(\Spec(K^{sh})) \to \pi_1(X)$ is trivial, see Lemma \ref{lemma-composition-trivial}. \medskip\noindent To finish the proof, it suffices according to Lemma \ref{lemma-functoriality-galois-ses} to show the following: Given a finite \'etale morphism $V \to \Spec(K)$ such that $V \times_{\Spec(K)} \Spec(K^{sh})$ is a disjoint union of copies of $\Spec(K^{sh})$, we can find a finite \'etale morphism $Y \to X$ with $V \cong Y \times_X \Spec(K)$ over $\Spec(K)$. Write $V = \Spec(L)$, 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 \'etale, then we can take $Y = \Spec(B)$ and the proof is complete. By Algebra, Lemma \ref{algebra-lemma-integral-closure-commutes-smooth} (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 \'etale. As $A \to A^{sh}$ is faithfully flat it follows that $A \to B$ is \'etale (Descent, Lemma \ref{descent-lemma-descending-property-etale}) as desired. \end{proof} \section{Group actions and integral closure} \label{section-group-actions-integral} \noindent In this section we continue the discussion of More on Algebra, Section \ref{more-algebra-section-group-actions-integral}. Recall that a normal local ring is a domain by definition. \begin{lemma} \label{lemma-get-algebraic-closure} Let $A$ be a normal domain whose fraction field is separably algebraically closed. Let $\mathfrak p \subset A$ be a nonzero prime ideal. Then the residue field $\kappa(\mathfrak p)$ is algebraically closed. \end{lemma} \begin{proof} Assume the lemma is not true to get a contradiction. Then there exists a monic irreducible polynomial $P(T) \in \kappa(\mathfrak p)[T]$ of degree $d > 1$. After replacing $P$ by $a^d P(a^{-1}T)$ for suitable $a \in A$ (to clear denominators) we may assume that $P$ is the image of a monic polynomial $Q$ in $A[T]$. Observe that $Q$ is irreducible in $f.f.(A)[T]$. Namely a factorization over $f.f.(A)$ leads to a factorization over $A$ by Algebra, Lemma \ref{algebra-lemma-polynomials-divide} which we could reduce modulo $\mathfrak p$ to get a factorization of $P$. As $f.f.(A)$ is separably closed, $Q$ is not a separable polynomial (Fields, Definition \ref{fields-definition-separable}). Then the characteristic of $f.f.(A)$ is $p > 0$ and $Q$ has vanishing linear term (Fields, Definition \ref{fields-definition-separable}). However, then we can replace $Q$ by $Q + a T$ where $a \in \mathfrak p$ is nonzero to get a contradiction. \end{proof} \begin{lemma} \label{lemma-normal-local-domain-separablly-closed-fraction-field} A normal local ring with separably closed fraction field is strictly henselian. \end{lemma} \begin{proof} Let $(A, \mathfrak m, \kappa)$ be normal local with separably closed fraction field $K$. If $A = K$, then we are done. If not, then the residue field $\kappa$ is algebraically closed by Lemma \ref{lemma-get-algebraic-closure} and it suffices to check that $A$ is henselian. Let $f \in A[T]$ be monic and let $a_0 \in \kappa$ be a root of multiplicity $1$ of the reduction $\overline{f} \in \kappa[T]$. Let $f = \prod f_i$ be the factorization in $K[T]$. By Algebra, Lemma \ref{algebra-lemma-polynomials-divide} we have $f_i \in A[T]$. Thus $a_0$ is a root of $f_i$ for some $i$. After replacing $f$ by $f_i$ we may assume $f$ is irreducible. Then, since the derivative $f'$ cannot be zero in $A[T]$ as $a_0$ is a single root, we conclude that $f$ is linear due to the fact that $K$ is separably algebraically closed. Thus $A$ is henselian, see Algebra, Definition \ref{algebra-definition-henselian}. \end{proof} \begin{lemma} \label{lemma-inertia-base-change} Let $G$ be a finite group acting on a ring $R$. Let $R^G \to A$ be a ring map. Let $\mathfrak q' \subset A \otimes_{R^G} R$ be a prime lying over the prime $\mathfrak q \subset R$. Then $$I_\mathfrak q = \{\sigma \in G \mid \sigma(\mathfrak q) = \mathfrak q\text{ and } \sigma \bmod \mathfrak q = \text{id}_{\kappa(\mathfrak q)}\}$$ is equal to $$I_{\mathfrak q'} = \{\sigma \in G \mid \sigma(\mathfrak q') = \mathfrak q'\text{ and } \sigma \bmod \mathfrak q' = \text{id}_{\kappa(\mathfrak q')}\}$$ \end{lemma} \begin{proof} Since $\mathfrak q$ is the inverse image of $\mathfrak q'$ and since $\kappa(\mathfrak q) \subset \kappa(\mathfrak q')$, we get $I_{\mathfrak q'} \subset I_\mathfrak q$. Conversely, if $\sigma \in I_\mathfrak q$, the $\sigma$ acts trivially on the fibre ring $A \otimes_{R^G} \kappa(\mathfrak q)$. Thus $\sigma$ fixes all the primes lying over $\mathfrak q$ and induces the identity on their residue fields. \end{proof} \begin{lemma} \label{lemma-inertia-invariants-etale} Let $G$ be a finite group acting on a ring $R$. Let $\mathfrak q \subset R$ be a prime. Set $$I = \{\sigma \in G \mid \sigma(\mathfrak q) = \mathfrak q \text{ and } \sigma \bmod \mathfrak q = \text{id}_\mathfrak q\}$$ Then $R^G \to R^I$ is \'etale at $R^I \cap \mathfrak q$. \end{lemma} \begin{proof} The strategy of the proof is to use \'etale localization to reduce to the case where $R \to R^I$ is a local isomorphism at $R^I \cap \mathfrak p$. Let $R^G \to A$ be an \'etale ring map. We claim that if the result holds for the action of $G$ on $A \otimes_{R^G} R$ and some prime $\mathfrak q'$ of $A \otimes_{R^G} R$ lying over $\mathfrak q$, then the result is true. \medskip\noindent To check this, note that since $R^G \to A$ is flat we have $A = (A \otimes_{R^G} R)^G$, see More on Algebra, Lemma \ref{more-algebra-lemma-base-change-invariants}. By Lemma \ref{lemma-inertia-base-change} the group $I$ does not change. Then a second application of More on Algebra, Lemma \ref{more-algebra-lemma-base-change-invariants} shows that $A \otimes_{R^G} R^I = (A \otimes_{R^G} R)^I$ (because $R^I \to A \otimes_{R^G} R^I$ is flat). Thus $$\xymatrix{ \Spec((A \otimes_{R^G} R)^I) \ar[d] \ar[r] & \Spec(R^I) \ar[d] \\ \Spec(A) \ar[r] & \Spec(R^G) }$$ is cartesian and the horizontal arrows are \'etale. Thus if the left vertical arrow is \'etale in some open neighbourhood $W$ of $(A \otimes_{R^G} R)^I \cap \mathfrak q'$, then the right vertical arrow is \'etale at the points of the (open) image of $W$ in $\Spec(R^I)$, see Descent, Lemma \ref{descent-lemma-smooth-permanence}. In particular the morphism $\Spec(R^I) \to \Spec(R^G)$ is \'etale at $R^I \cap \mathfrak q$. \medskip\noindent Let $\mathfrak p = R^G \cap \mathfrak q$. By More on Algebra, Lemma \ref{more-algebra-lemma-one-orbit} the fibre of $\Spec(R) \to \Spec(R^G)$ over $\mathfrak p$ is finite. Moreover the residue field extensions at these points are algebraic, normal, with finite automorphism groups by More on Algebra, Lemma \ref{more-algebra-lemma-one-orbit-geometric}. Thus we may apply More on Morphisms, Lemma \ref{more-morphisms-lemma-etale-makes-integral-split} to the integral ring map $R^G \to R$ and the prime $\mathfrak p$. Combined with the claim above we reduce to the case where $R = A_1 \times \ldots \times A_n$ with each $A_i$ having a single prime $\mathfrak q_i$ lying over $\mathfrak p$ such that the residue field extensions $\kappa(\mathfrak q_i)/\kappa(\mathfrak p)$ are purely inseparable. Of course $\mathfrak q$ is one of these primes, say $\mathfrak q = \mathfrak q_1$. \medskip\noindent It may not be the case that $G$ permutes the factors $A_i$ (this would be true if the spectrum of $A_i$ were connected, for example if $R^G$ was local). This we can fix as follows; we suggest the reader think this through for themselves, perhaps using idempotents instead of topology. Recall that the product decomposition gives a corresponding disjoint union decomposition of $\Spec(R)$ by open and closed subsets $U_i$. Since $G$ is finite, we can refine this covering by a finite disjoint union decomposition $\Spec(R) = \coprod_{j \in J} W_j$ by open and closed subsets $W_j$, such that for all $j \in J$ there exists a $j' \in J$ with $\sigma(W_j) = W_{j'}$. The union of the $W_j$ not meeting $\{\mathfrak q_1, \ldots, \mathfrak q_n\}$ is a closed subset not meeting the fibre over $\mathfrak p$ hence maps to a closed subset of $\Spec(R^G)$ not meeting $\mathfrak p$ as $\Spec(R) \to \Spec(R^G)$ is closed. Hence after replacing $R^G$ by a principal localization (permissible by the claim) we may assume each $W_j$ meets one of the points $\mathfrak q_i$. Then we set $U_i = W_j$ if $\mathfrak q_i \in W_j$. The corresponding product decomposition $R = A_1 \times \ldots \times A_n$ is one where $G$ permutes the factors $A_i$. \medskip\noindent Thus we may assume we have a product decomposition $R = A_1 \times \ldots \times A_n$ compatible with $G$-action, where each $A_i$ has a single prime $\mathfrak q_i$ lying over $\mathfrak p$ and the field extensions $\kappa(\mathfrak q_i)/\kappa(\mathfrak p)$ are purely inseparable. Write $A' = A_2 \times \ldots \times A_n$ so that $$R = A_1 \times A'$$ Since $\mathfrak q = \mathfrak q_1$ we find that every $\sigma \in I$ preserves the product decomposition above. Hence $$R^I = (A_1)^I \times (A')^I$$ Observe that $I = D = \{\sigma \in G \mid \sigma(\mathfrak q) = \mathfrak q\}$ because $\kappa(\mathfrak q)/\kappa(\mathfrak p)$ is purely inseparable. Since the action of $G$ on primes over $\mathfrak p$ is transitive (More on Algebra, Lemma \ref{more-algebra-lemma-one-orbit}) we conclude that, the index of $I$ in $G$ is $n$ and we can write $G = eI \amalg \sigma_2I \amalg \ldots \amalg \sigma_nI$ so that $A_i = \sigma_i(A_1)$ for $i = 2, \ldots, n$. It follows that $$R^G = (A_1)^I.$$ Thus the map $R^G \to R^I$ is \'etale at $R^I \cap \mathfrak q$ and the proof is complete. \end{proof} \noindent The following lemma generalizes More on Algebra, Lemma \ref{more-algebra-lemma-inertial-invariants-unramified}. \begin{lemma} \label{lemma-inertial-invariants-unramified} Let $A$ be a normal domain with fraction field $K$. Let $L/K$ be a (possibly infinite) Galois extension. Let $G = \text{Gal}(L/K)$ and let $B$ be the integral closure of $A$ in $L$. Let $\mathfrak q \subset B$. Set $$I = \{\sigma \in G \mid \sigma(\mathfrak q) = \mathfrak q \text{ and } \sigma \bmod \mathfrak q = \text{id}_{\kappa(\mathfrak q)}\}$$ Then $(B^I)_{B^I \cap \mathfrak q}$ is a filtered colimit of \'etale $A$-algebras. \end{lemma} \begin{proof} We can write $L$ as the filtered colimit of finite Galois extensions of $K$. Hence it suffices to prove this lemma in case $L/K$ is a finite Galois extension, see Algebra, Lemma \ref{algebra-lemma-colimit-colimit-etale}. Since $A = B^G$ as $A$ is integrally closed in $K = L^G$ the result follows from Lemma \ref{lemma-inertia-invariants-etale}. \end{proof} \section{Ramification theory} \label{section-ramification} \noindent In this section we continue the discussion of More on Algebra, Section \ref{more-algebra-section-ramification} and we relate it to our discussion of the fundamental groups of schemes. \medskip\noindent Let $(A, \mathfrak m, \kappa)$ be a normal local ring with fraction field $K$. Choose a separable algebraic closure $K^{sep}$. Let $A^{sep}$ be the integral closure of $A$ in $K^{sep}$. Choose maximal ideal $\mathfrak m^{sep} \subset A^{sep}$. Let $A \subset A^h \subset A^{sh}$ be the henselization and strict henselization. Observe that $A^h$ and $A^{sh}$ are normal rings as well (More on Algebra, Lemma \ref{more-algebra-lemma-henselization-normal}). Denote $K^h$ and $K^{sh}$ their fraction fields. Since $(A^{sep})_{\mathfrak m^{sep}}$ is strictly henselian by Lemma \ref{lemma-normal-local-domain-separablly-closed-fraction-field} we can choose an $A$-algebra map $A^{sh} \to (A^{sep})_{\mathfrak m^{sep}}$. Namely, first choose a $\kappa$-embedding\footnote{This is possible because $\kappa(\mathfrak m^{sh})$ is a separable algebraic closure of $\kappa$ and $\kappa(\mathfrak m^{sep})$ is an algebraic closure of $\kappa$ by Lemma \ref{lemma-get-algebraic-closure}.} $\kappa(\mathfrak m^{sh}) \to \kappa(\mathfrak m^{sep})$ and then extend (uniquely) to an $A$-algebra homomorphism by Algebra, Lemma \ref{algebra-lemma-strictly-henselian-functorial}. We get the following diagram $$\xymatrix{ K^{sep} & K^{sh} \ar[l] & K^h \ar[l] & K \ar[l] \\ (A^{sep})_{\mathfrak m^{sep}} \ar[u] & A^{sh} \ar[u] \ar[l] & A^h \ar[u] \ar[l] & A \ar[u] \ar[l] }$$ We can take the fundamental groups of the spectra of these rings. Of course, since $K^{sep}$, $(A^{sep})_{\mathfrak m^{sep}}$, and $A^{sh}$ are strictly henselian, for them we obtain trivial groups. Thus the interesting part is the following \begin{equation} \label{equation-inertia-diagram-pione} \vcenter{ \xymatrix{ \pi_1(U^{sh}) \ar[r] \ar[rd]_1 & \pi_1(U^h) \ar[d] \ar[r] & \pi_1(U) \ar[d] \\ & \pi_1(X^h) \ar[r] & \pi_1(X) } } \end{equation} Here $X^h$ and $X$ are the spectra of $A^h$ and $A$ and $U^{sh}$, $U^h$, $U$ are the spectra of $K^{sh}$, $K^h$, and $K$. The label $1$ means that the map is trivial; this follows as it factors through the trivial group $\pi_1(X^{sh})$. On the other hand, the profinite group $G = \text{Gal}(K^{sep}/K)$ acts on $A^{sep}$ and we can make the following definitions $$D = \{\sigma \in G \mid \sigma(\mathfrak m^{sep}) = \mathfrak m^{sep}\} \supset I = \{\sigma \in D \mid \sigma \bmod \mathfrak m^{sep} = \text{id}_{\kappa(\mathfrak m^{sep})}\}$$ These groups are sometimes called the {\it decomposition group} and the {\it inertia group} especially when $A$ is a discrete valuation ring. \begin{lemma} \label{lemma-identify-inertia} In the situation described above, via the isomorphism $\pi_1(U) = \text{Gal}(K^{sep}/K)$ the diagram (\ref{equation-inertia-diagram-pione}) translates into the diagram $$\xymatrix{ I \ar[r] \ar[rd]_1 & D \ar[d] \ar[r] & \text{Gal}(K^{sep}/K) \ar[d] \\ & \text{Gal}(\kappa(\mathfrak m^{sh})/\kappa) \ar[r] & \text{Gal}(M/K) }$$ where $K^{sep}/M/K$ is the maximal subextension unramified with respect to $A$. Moreover, the vertical arrows are surjective, the kernel of the left vertical arrow is $I$ and the kernel of the right vertical arrow is the smallest closed normal subgroup of $\text{Gal}(K^{sep}/K)$ containing $I$. \end{lemma} \begin{proof} By construction the group $D$ acts on $(A^{sep})_{\mathfrak m^{sep}}$ over $A$. By the uniqueness of $A^{sh} \to (A^{sep})_{\mathfrak m^{sep}}$ given the map on residue fields (Algebra, Lemma \ref{algebra-lemma-strictly-henselian-functorial}) we see that the image of $A^{sh} \to (A^{sep})_{\mathfrak m^{sep}}$ is contained in $((A^{sep})_{\mathfrak m^{sep}})^I$. On the other hand, Lemma \ref{lemma-inertial-invariants-unramified} shows that $((A^{sep})_{\mathfrak m^{sep}})^I$ is a filtered colimit of \'etale extensions of $A$. Since $A^{sh}$ is the maximal such extension, we conclude that $A^{sh} = ((A^{sep})_{\mathfrak m^{sep}})^I$. Hence $K^{sh} = (K^{sep})^I$. \medskip\noindent Recall that $I$ is the kernel of a surjective map $D \to \text{Aut}(\kappa(\mathfrak m^{sep})/\kappa)$, see More on Algebra, Lemma \ref{more-algebra-lemma-one-orbit-geometric-galois}. We have $\text{Aut}(\kappa(\mathfrak m^{sep})/\kappa) = \text{Gal}(\kappa(\mathfrak m^{sh})/\kappa)$ as we have seen above that these fields are the algebraic and separable algebraic closures of $\kappa$. On the other hand, any automorphism of $A^{sh}$ over $A$ is an automorphism of $A^{sh}$ over $A^h$ by the uniqueness in Algebra, Lemma \ref{algebra-lemma-henselian-functorial}. Furthermore, $A^{sh}$ is the colimit of finite \'etale extensions $A^h \subset A'$ which correspond $1$-to-$1$ with finite separable extension $\kappa'/\kappa$, see Algebra, Remark \ref{algebra-remark-construct-sh-from-h}. Thus $$\text{Aut}(A^{sh}/A) = \text{Aut}(A^{sh}/A^h) = \text{Gal}(\kappa(\mathfrak m^{sh})/\kappa)$$ Let $\kappa \subset \kappa'$ be a finite Galois extension with Galois group $G$. Let $A^h \subset A'$ be the finite \'etale extension corresponding to $\kappa \subset \kappa'$ by Algebra, Lemma \ref{algebra-lemma-henselian-cat-finite-etale}. Then it follows that $(A')^G = A^h$ by looking at fraction fields and degrees (small detail omitted). Taking the colimit we conclude that $(A^{sh})^{\text{Gal}(\kappa(\mathfrak m^{sh})/\kappa)} = A^h$. Combining all of the above, we find $A^h = ((A^{sep})_{\mathfrak m^{sep}})^D$. Hence $K^h = (K^{sep})^D$. \medskip\noindent Since $U$, $U^h$, $U^{sh}$ are the spectra of the fields $K$, $K^h$, $K^{sh}$ we see that the top lines of the diagrams correspond via Lemma \ref{lemma-fundamental-group-Galois-group}. By Lemma \ref{lemma-gabber} we have $\pi_1(X^h) = \text{Gal}(\kappa(\mathfrak m^{sh})/\kappa)$. The exactness of the sequence $1 \to I \to D \to \text{Gal}(\kappa(\mathfrak m^{sh})/\kappa) \to 1$ was pointed out above. By Proposition \ref{proposition-normal} we see that $\pi_1(X) = \text{Gal}(M/K)$. Finally, the statement on the kernel of $\text{Gal}(K^{sep}/K) \to \text{Gal}(M/K) = \pi_1(X)$ follows from Lemma \ref{lemma-local-exact-sequence-normal}. This finishes the proof. \end{proof} \noindent Let $X$ be a normal integral scheme with function field $K$. Let $K^{sep}$ be a separable algebraic closure of $K$. Let $X^{sep} \to X$ be the normalization of $X$ in $K^{sep}$. Since $G = \text{Gal}(K^{sep}/K)$ acts on $K^{sep}$ we obtain a right action of $G$ on $X^{sep}$. For $y \in X^{sep}$ define $$D_y = \{\sigma \in G \mid \sigma(y) = y\} \supset I_y = \{\sigma \in D \mid \sigma \bmod \mathfrak m_y = \text{id}_{\kappa(y)} \}$$ similarly to the above. On the other hand, for $x \in X$ let $\mathcal{O}_{X, x}^{sh}$ be a strict henselization, let $K_x^{sh}$ be the fraction field of $\mathcal{O}_{X, x}^{sh}$ and choose a $K$-embedding $K_x^{sh} \to K^{sep}$. \begin{lemma} \label{lemma-normal-pione-quotient-inertia} Let $X$ be a normal integral scheme with function field $K$. With notation as above, the following three subgroups of $\text{Gal}(K^{sep}/K) = \pi_1(\Spec(K))$ are equal \begin{enumerate} \item the kernel of the surjection $\text{Gal}(K^{sep}/K) \longrightarrow \pi_1(X)$, \item the smallest normal closed subgroup containing $I_y$ for all $y \in X^{sep}$, and \item the smallest normal closed subgroup containing $\text{Gal}(K^{sep}/K_x^{sh})$ for all $x \in X$. \end{enumerate} \end{lemma} \begin{proof} The equivalence of (2) and (3) follows from Lemma \ref{lemma-identify-inertia} which tells us that $I_y$ is conjugate to $\text{Gal}(K^{sep}/K_x^{sh})$ if $y$ lies over $x$. By Lemma \ref{lemma-local-exact-sequence-normal} we see that $\text{Gal}(K^{sep}/K_x^{sh})$ maps trivially to $\pi_1(\Spec(\mathcal{O}_{X, x}))$ and therefore the subgroup $N \subset G = \text{Gal}(K^{sep}/K)$ of (2) and (3) is contained in the kernel of $G \longrightarrow \pi_1(X)$. \medskip\noindent 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 \ref{section-normal}, especially Proposition \ref{proposition-normal}). 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 \ref{more-algebra-lemma-one-orbit-geometric-galois-compare}. Since $N \subset U$ all these inertia groups are trivial. We conclude that $Y \to X$ is \'etale by applying Lemma \ref{lemma-inertia-invariants-etale}. (Alternative: you can use Lemma \ref{lemma-local-exact-sequence-normal} to see that the pullback of $Y$ to $\Spec(\mathcal{O}_{X, x})$ is \'etale for all $x \in X$ and then conclude from there with a bit more work.) \end{proof} \begin{example} \label{example-bigger-codim} Let $X$ be a normal integral Noetherian scheme with function field $K$. Purity of branch locus (see below) tells us that if $X$ is regular, then it suffices in Lemma \ref{lemma-normal-pione-quotient-inertia} to consider the inertia groups $I = \pi_1(\Spec(K_x^{sh}))$ for points $x$ of codimension $1$ in $X$. In general this is not enough however. Namely, let $Y = \mathbf{A}_k^n = \Spec(k[t_1, \ldots, t_n])$ where $k$ is a field not of characteristic $2$. Let $G = \{\pm 1\}$ be the group of order $2$ acting on $Y$ by multiplication on the coordinates. Set $$X = \Spec(k[t_it_j, i, j \in \{1, \ldots, n\}])$$ The embedding $k[t_it_j] \subset k[t_1, \ldots, t_n]$ defines a degree $2$ morphism $Y \to X$ which is unramified everywhere except over the maximal ideal $\mathfrak m = (t_it_j)$ which is a point of codimension $n$ in $X$. \end{example} \begin{lemma} \label{lemma-unramified} Let $X$ be an integral normal scheme with function field $K$. Let $L/K$ be a finite extension. Let $Y \to X$ be the normalization of $X$ in $L$. The following are equivalent \begin{enumerate} \item $X$ is unramified in $L$ as defined in Section \ref{section-normal}, \item $Y \to X$ is an unramified morphism of schemes, \item $Y \to X$ is an \'etale morphism of schemes, \item $Y \to X$ is a finite \'etale morphism of schemes, \item for $x \in X$ the projection $Y \times_X \Spec(\mathcal{O}_{X, x}) \to \Spec(\mathcal{O}_{X, x})$ is unramified, \item same as in (5) but with $\mathcal{O}_{X, x}^h$, \item same as in (5) but with $\mathcal{O}_{X, x}^{sh}$, \item for $x \in X$ the scheme theoretic fibre $Y_x$ is \'etale over $x$ of degree $\geq [L : K]$. \end{enumerate} If $L/K$ is Galois with Galois group $G$, then these are also equivalent to \begin{enumerate} \item[(9)] for $y \in Y$ the group $I_y = \{g \in G \mid g(y) = y\text{ and } g \bmod \mathfrak m_y = \text{id}_{\kappa(y)}\}$ is trivial. \end{enumerate} \end{lemma} \begin{proof} The equivalence of (1) and (2) is the definition of (1). The equivalence of (2), (3), and (4) is Lemma \ref{lemma-unramified-in-L}. It is straightforward to prove that (4) $\Rightarrow$ (5), (5) $\Rightarrow$ (6), (6) $\Rightarrow$ (7). \medskip\noindent Assume (7). Observe that $\mathcal{O}_{X, x}^{sh}$ is a normal local domain (More on Algebra, Lemma \ref{more-algebra-lemma-henselization-normal}). Let $L^{sh} = L \otimes_K K_x^{sh}$ where $K_x^{sh}$ is the fraction field of $\mathcal{O}_{X, x}^{sh}$. Then $L^{sh} = \prod_{i = 1, \ldots, n} L_i$ with $L_i/K_x^{sh}$ finite separable. By Algebra, Lemma \ref{algebra-lemma-integral-closure-commutes-smooth} (and a limit argument we omit) we see that $Y \times_X \Spec(\mathcal{O}_{X, x}^{sh})$ is the integral closure of $\Spec(\mathcal{O}_{X, x}^{sh})$ in $L^{sh}$. Hence by Lemma \ref{lemma-unramified-in-L} (applied to the factors $L_i$ of $L^{sh}$) we see that $Y \times_X \Spec(\mathcal{O}_{X, x}^{sh}) \to \Spec(\mathcal{O}_{X, x}^{sh})$ is finite \'etale. Looking at the generic point we see that the degree is equal to $[L : K]$ and hence we see that (8) is true. \medskip\noindent Assume (8). Assume that $x \in X$ and that the scheme theoretic fibre $Y_x$ is \'etale over $x$ of degree $\geq [L : K]$. Observe that this means that $Y$ has $\geq [L : K]$ geometric points lying over $x$. We will show that $Y \to X$ is finite \'etale over a neighbourhood of $x$. This will prove (1) holds. To prove this we may assume $X = \Spec(R)$, the point $x$ corresponds to the prime $\mathfrak p \subset R$, and $Y = \Spec(S)$. We apply More on Morphisms, Lemma \ref{more-morphisms-lemma-etale-makes-integral-split} and we find an \'etale neighbourhood $(U, u) \to (X, x)$ such that $Y \times_X U = V_1 \amalg \ldots \amalg V_m$ such that $V_i$ has a unique point $v_i$ lying over $u$ with $\kappa(v_i)/\kappa(u)$ purely inseparable. Shrinking $U$ if necessary we may assume $U$ is a normal integral scheme with generic point $\xi$ (use Descent, Lemmas \ref{descent-lemma-locally-finite-nr-irred-local-fppf} and \ref{descent-lemma-normal-local-smooth} and Properties, Lemma \ref{properties-lemma-normal-locally-finite-nr-irreducibles}). By our remark on geometric points we see that $m \geq [L : K]$. On the other hand, by More on Morphisms, Lemma \ref{more-morphisms-lemma-normalization-smooth-localization} we see that $\coprod V_i \to U$ is the normalization of $U$ in $\Spec(L) \times_X U$. As $K \subset \kappa(\xi)$ is finite separable, we can write $\Spec(L) \times_X U = \Spec(\prod_{i = 1, \ldots, n} L_i)$ with $L_i/\kappa(\xi)$ finite and $[L : K] = \sum [L_i : \kappa(\xi)]$. Since $V_j$ is nonempty for each $j$ and $m \geq [L : K]$ we conclude that $m = n$ and $[L_i : \kappa(\xi)] = 1$ for all $i$. Then $V_j \to U$ is an isomorphism in particular \'etale, hence $Y \times_X U \to U$ is \'etale. By Descent, Lemma \ref{descent-lemma-descending-property-etale} we conclude that $Y \to X$ is \'etale over the image of $U \to X$ (an open neighbourhood of $x$). \medskip\noindent Assume $L/K$ is Galois and (9) holds. Then $Y \to X$ is \'etale by Lemma \ref{lemma-inertial-invariants-unramified}. We omit the proof that (1) implies (9). \end{proof} \noindent In the case of infinite Galois extensions of discrete valuation rings we can say a tiny bit more. To do so we introduce the following notation. A subset $S \subset \mathbf{N}$ of integers is {\it multiplicativity directed} if $1 \in S$ and for $n, m \in S$ there exists $k \in S$ with $n | k$ and $m | k$. Define a partial ordering on $S$ by the rule $n \geq_S m$ if and only if $m | n$. Given a field $\kappa$ we obtain an inverse system of finite groups $\{\mu_n(\kappa)\}_{n \in S}$ with transition maps $$\mu_n(\kappa) \longrightarrow \mu_m(\kappa),\quad \zeta \longmapsto \zeta^{n/m}$$ for $n \geq_S m$. Then we can form the profinite group $$\lim_{n \in S} \mu_n(\kappa)$$ Observe that the limit is cofiltered (as $S$ is directed). The construction is functorial in $\kappa$. In particular $\text{Aut}(\kappa)$ acts on this profinite group. For example, if $S = \{1, n\}$, then this gives $\mu_n(\kappa)$. If $S = \{1, \ell, \ell^2, \ell^3, \ldots\}$ for some prime $\ell$ different from the characteristic of $\kappa$ this produces $\lim_n \mu_{\ell^n}(\kappa)$ which is sometimes called the $\ell$-adic Tate module of the multiplicative group of $\kappa$ (compare with More on Algebra, Example \ref{more-algebra-example-spectral-sequence-principal}). \begin{lemma} \label{lemma-structure-decomposition} Let $A$ be a discrete valuation ring with fraction field $K$. Let $L/K$ be a (possibly infinite) Galois extension. Let $B$ be the integral closure of $A$ in $L$. Let $\mathfrak m$ be a maximal ideal of $B$. Let $G = \text{Gal}(L/K)$, $D = \{\sigma \in G \mid \sigma(\mathfrak m) = \mathfrak m\}$, and $I = \{\sigma \in D \mid \sigma \bmod \mathfrak m = \text{id}_{\kappa(\mathfrak m)}\}$. The decomposition group $D$ fits into a canonical exact sequence $$1 \to I \to D \to \text{Aut}(\kappa(\mathfrak m)/\kappa_A) \to 1$$ The inertia group $I$ fits into a canonical exact sequence $$1 \to P \to I \to I_t \to 1$$ such that \begin{enumerate} \item $P$ is a normal subgroup of $D$, \item $P$ is a pro-p-group if the characteristic of $\kappa_A$ is $p > 1$ and $P = \{1\}$ if the characteristic of $\kappa_A$ is zero, \item there is a multiplicatively directed $S \subset \mathbf{N}$ such that $\kappa(\mathfrak m)$ contains a primitive $n$th root of unity for each $n \in S$ (elements of $S$ are prime to $p$), \item there exists a canonical surjective map $$\theta_{can} : I \to \lim_{n \in S} \mu_n(\kappa(\mathfrak m))$$ whose kernel is $P$, which satisfies $\theta_{can}(\tau \sigma \tau^{-1}) = \tau(\theta_{can}(\sigma))$ for $\tau \in D$, $\sigma \in I$, and which induces an isomorphism $I_t \to \lim_{n \in S} \mu_n(\kappa(\mathfrak m))$. \end{enumerate} \end{lemma} \begin{proof} This is mostly a reformulation of the results on finite Galois extensions proved in More on Algebra, Section \ref{more-algebra-section-ramification}. The surjectivity of the map $D \to \text{Aut}(\kappa(\mathfrak m)/\kappa)$ is More on Algebra, Lemma \ref{more-algebra-lemma-one-orbit-geometric-galois}. This gives the first exact sequence. \medskip\noindent To construct the second short exact sequence let $\Lambda$ be the set of finite Galois subextensions, i.e., $\lambda \in \Lambda$ corresponds to $L/L_\lambda/K$. Set $G_\lambda = \text{Gal}(L_\lambda/K)$. Recall that $G_\lambda$ is an inverse system of finite groups with surjective transition maps and that $G = \lim_{\lambda \in \Lambda} G_\lambda$, see Fields, Lemma \ref{fields-lemma-infinite-galois-limit}. We let $B_\lambda$ be the integral closure of $A$ in $L_\lambda$. Then we set $\mathfrak m_\lambda = \mathfrak m \cap B_\lambda$ and we denote $P_\lambda, I_\lambda, D_\lambda$ the wild inertia, inertia, and decomposition group of $\mathfrak m_\lambda$, see More on Algebra, Lemma \ref{more-algebra-lemma-galois-inertia}. For $\lambda \geq \lambda'$ the restriction defines a commutative diagram $$\xymatrix{ P_\lambda \ar[d] \ar[r] & I_\lambda \ar[d] \ar[r] & D_\lambda \ar[d] \ar[r] & G_\lambda \ar[d] \\ P_{\lambda'} \ar[r] & I_{\lambda'} \ar[r] & D_{\lambda'} \ar[r] & G_{\lambda'} }$$ with surjective vertical maps, see More on Algebra, Lemma \ref{more-algebra-lemma-compare-inertia}. \medskip\noindent From the definitions it follows immediately that $I = \lim I_\lambda$ and $D = \lim D_\lambda$ under the isomorphism $G = \lim G_\lambda$ above. Since $L = \colim L_\lambda$ we have $B = \colim B_\lambda$ and $\kappa(\mathfrak m) = \colim \kappa(\mathfrak m_\lambda)$. Since the transition maps of the system $D_\lambda$ are compatible with the maps $D_\lambda \to \text{Aut}(\kappa(\mathfrak m_\lambda)/\kappa)$ (see More on Algebra, Lemma \ref{more-algebra-lemma-compare-inertia}) we see that the map $D \to \text{Aut}(\kappa(\mathfrak m)/\kappa)$ is the limit of the maps $D_\lambda \to \text{Aut}(\kappa(\mathfrak m_\lambda)/\kappa)$. \medskip\noindent There exist canonical maps $$\theta_{\lambda, can} : I_\lambda \longrightarrow \mu_{n_\lambda}(\kappa(\mathfrak m_\lambda))$$ where $n_\lambda = |I_\lambda|/|P_\lambda|$, where $\mu_{n_\lambda}(\kappa(\mathfrak m_\lambda))$ has order $n_\lambda$, such that $\theta_{\lambda, can}(\tau \sigma \tau^{-1}) = \tau(\theta_{\lambda, can}(\sigma))$ for $\tau \in D_\lambda$ and $\sigma \in I_\lambda$, and such that we get commutative diagrams $$\xymatrix{ I_\lambda \ar[r]_-{\theta_{\lambda, can}} \ar[d] & \mu_{n_\lambda}(\kappa(\mathfrak m_\lambda)) \ar[d]^{(-)^{n_\lambda/n_{\lambda'}}} \\ I_{\lambda'} \ar[r]^-{\theta_{\lambda', can}} & \mu_{n_{\lambda'}}(\kappa(\mathfrak m_{\lambda'})) }$$ see More on Algebra, Remark \ref{more-algebra-remark-canonical-inertia-character}. \medskip\noindent Let $S \subset \mathbf{N}$ be the collection of integers $n_\lambda$. Since $\Lambda$ is directed, we see that $S$ is multiplicatively directed. By the displayed commutative diagrams above we can take the limits of the maps $\theta_{\lambda, can}$ to obtain $$\theta_{can} : I \to \lim_{n \in S} \mu_n(\kappa(\mathfrak m)).$$ This map is continuous (small detail omitted). Since the transition maps of the system of $I_\lambda$ are surjective and $\Lambda$ is directed, the projections $I \to I_\lambda$ are surjective. For every $\lambda$ the diagram $$\xymatrix{ I \ar[d] \ar[r]_-{\theta_{can}} & \lim_{n \in S} \mu_n(\kappa(\mathfrak m)) \ar[d] \\ I_{\lambda} \ar[r]^-{\theta_{\lambda, can}} & \mu_{n_\lambda}(\kappa(\mathfrak m_\lambda)) }$$ commutes. Hence the image of $\theta_{can}$ surjects onto the finite group $\mu_{n_\lambda}(\kappa(\mathfrak m)) = \mu_{n_\lambda}(\kappa(\mathfrak m_\lambda))$ of order $n_\lambda$ (see above). It follows that the image of $\theta_{can}$ is dense. On the other hand $\theta_{can}$ is continuous and the source is a profinite group. Hence $\theta_{can}$ is surjective by a topological argument. \medskip\noindent The property $\theta_{can}(\tau \sigma \tau^{-1}) = \tau(\theta_{can}(\sigma))$ for $\tau \in D$, $\sigma \in I$ follows from the corresponding properties of the maps $\theta_{\lambda, can}$ and the compatibility of the map $D \to \text{Aut}(\kappa(\mathfrak m))$ with the maps $D_\lambda \to \text{Aut}(\kappa(\mathfrak m_\lambda))$. Setting $P = \Ker(\theta_{can})$ this implies that $P$ is a normal subgroup of $D$. Setting $I_t = I/P$ we obtain the isomorphism $I_t \to \lim_{n \in S} \mu_n(\kappa(\mathfrak m))$ from the surjectivity of $\theta_{can}$. \medskip\noindent To finish the proof we show that $P = \lim P_\lambda$ which proves that $P$ is a pro-p-group. Recall that the tame inertia group $I_{\lambda, t} = I_\lambda/P_\lambda$ has order $n_\lambda$. Since the transition maps $P_\lambda \to P_{\lambda'}$ are surjective and $\Lambda$ is directed, we obtain a short exact sequence $$1 \to \lim P_\lambda \to I \to \lim I_{\lambda, t} \to 1$$ (details omitted). Since for each $\lambda$ the map $\theta_{\lambda, can}$ induces an isomorphism $I_{\lambda, t} \cong \mu_{n_\lambda}(\kappa(\mathfrak m))$ the desired result follows. \end{proof} \begin{lemma} \label{lemma-structure-decomposition-separable-closure} Let $A$ be a discrete valuation ring with fraction field $K$. Let $K^{sep}$ be a separable closure of $K$. Let $A^{sep}$ be the integral closure of $A$ in $K^{sep}$. Let $\mathfrak m^{sep}$ be a maximal ideal of $A^{sep}$. Let $\mathfrak m = \mathfrak m^{sep} \cap A$, let $\kappa = A/\mathfrak m$, and let $\overline{\kappa} = A^{sep}/\mathfrak m^{sep}$. Then $\overline{\kappa}$ is an algebraic closure of $\kappa$. Let $G = \text{Gal}(K^{sep}/K)$, $D = \{\sigma \in G \mid \sigma(\mathfrak m^{sep}) = \mathfrak m^{sep}\}$, and $I = \{\sigma \in D \mid \sigma \bmod \mathfrak m^{sep} = \text{id}_{\kappa(\mathfrak m^{sep})}\}$. The decomposition group $D$ fits into a canonical exact sequence $$1 \to I \to D \to \text{Gal}(\kappa^{sep}/\kappa) \to 1$$ where $\kappa^{sep} \subset \overline{\kappa}$ is the separable closure of $\kappa$. The inertia group $I$ fits into a canonical exact sequence $$1 \to P \to I \to I_t \to 1$$ such that \begin{enumerate} \item $P$ is a normal subgroup of $D$, \item $P$ is a pro-p-group if the characteristic of $\kappa_A$ is $p > 1$ and $P = \{1\}$ if the characteristic of $\kappa_A$ is zero, \item there exists a canonical surjective map $$\theta_{can} : I \to \lim_{n\text{ prime to }p} \mu_n(\kappa^{sep})$$ whose kernel is $P$, which satisfies $\theta_{can}(\tau \sigma \tau^{-1}) = \tau(\theta_{can}(\sigma))$ for $\tau \in D$, $\sigma \in I$, and which induces an isomorphism $I_t \to \lim_{n\text{ prime to }p} \mu_n(\kappa^{sep})$. \end{enumerate} \end{lemma} \begin{proof} The field $\overline{\kappa}$ is the algebraic closure of $\kappa$ by Lemma \ref{lemma-get-algebraic-closure}. Most of the statements immediately follow from the corresponding parts of Lemma \ref{lemma-structure-decomposition}. For example because $\text{Aut}(\overline{\kappa}/\kappa) = \text{Gal}(\kappa^{sep}/\kappa)$ we obtain the first sequence. Then the only other assertion that needs a proof is the fact that with $S$ as in Lemma \ref{lemma-structure-decomposition} the limit $\lim_{n \in S} \mu_n(\overline{\kappa})$ is equal to $\lim_{n\text{ prime to }p} \mu_n(\kappa^{sep})$. To see this it suffices to show that every integer $n$ prime to $p$ divides an element of $S$. Let $\pi \in A$ be a uniformizer and consider the splitting field $L$ of the polynomial $X^n - \pi$. Since the polynomial is separable we see that $L$ is a finite Galois extension of $K$. Choose an embedding $L \to K^{sep}$. Observe that if $B$ is the integral closure of $A$ in $L$, then the ramification index of $A \to B_{\mathfrak m^{sep} \cap B}$ is divisible by $n$ (because $\pi$ has an $n$th root in $B$; in fact the ramification index equals $n$ but we do not need this). Then it follows from the construction of the $S$ in the proof of Lemma \ref{lemma-structure-decomposition} that $n$ divides an element of $S$. \end{proof}