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\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}