\documentclass[12pt]{amsart}
\title{Lecture 32}
\author{}

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\begin{document}
\maketitle

\PP{32.0}{Generalities on topological groups}
%-------------------------------------------
A topological group is a group $G$, together with a topology
$\CalT\subset 2^G$ (that is, (i) $\emptyset, X \in\CalT$, and
(ii) $\CalT$ is closed under arbitrary unions and finite intersections),
such that multiplication and inverse 
\[
m:G\times G\to G,\qquad
i:G\to G
\]
(here, $m(a,b)=ab$ and $i(a)=a^{-1}$) are continuous maps.\\

For $a\in G$, we will denote by $\CalT(a) = \{U\in\CalT:a\in U\}$.
A subset $\CalB(a)\subset\CalT(a)$ is said to be a {\em fundamental
system of neighbourhoods} if for every $U\in\CalT(a)$, there
exists $V\in\CalB(a)$ such that $V\subset U$.\\

The continuity of multiplication can be written more explicitly
as the following condition on $\CalT$: for any $a,b\in G$ and
$U\in\CalT(ab)$, there exist $U_1\in\CalT(a)$ and $U_2\in\CalT(b)$
such that $U_1U_2\subset U$. Similarly, inverse being a homeomorphism
(as $i\circ i = \Id$) can be written as: $\CalT(a)^{-1} = \CalT(a^{-1})$.
Here, by a little abuse of notation, for any subset $V\subset G$,
we write $V^{-1}=\{a^{-1}:a\in V\}$. Thus,
\[
\CalT(a)^{-1} = \{V^{-1} : V\in\CalT(a)\}.
\]

The following properties of a topological group follow directly from
definitions:

\begin{enumerate}
\item For $g\in G$, let $L_g : G\to G$ be the multiplication on the left
by $g$. {\em Then $L_g$ is a homeomorphism}. Note that it is enough to show
that $L_g$ is continuous, since $L_g\circ L_{g^{-1}} = \Id_G$. That is,
for $h\in G$ and $U\in\CalT(gh)$, there exists $V\in\CalT(h)$ such that
$gV\subset U$. This is clear, since we have $U_1\in\CalT(g)$ and
$U_2\in\CalT(h)$, such that $U_1U_2\subset U$, we can take $V=U_2$.\\

\item Similarly, right multiplication by $g\in G$ is a homeomorphism.\\

\item Let $H\subset G$ be a subgroup. {\em If $H$ is open, then $H$ is closed}.
This is because, we have the following disjoint union:
\[
G = \bigsqcup_{g\in G/H} gH,\qquad H-\text{cosets in } G.
\]
Thus, $H$ is the complement of $\ds \bigcup_{g\not\in H} gH$ which
is open, being a union of open sets.\\

\item Again let $H\subset G$ be a subgroup. {\em Then $\ol{H}$ is
a subgroup}. To see this, let $g,h\in\ol{H}$. We need to show that
$gh\in\ol{H}$. That is, for every $U\in\CalT(gh)$, we have $U\cap H\neq\emptyset$.
By definition, we have $U_1\in\CalT(g)$ and $U_2\in\CalT(h)$
such that $U_1U_2\subset U$. Since $g,h\in\ol{H}$, we have
$U_1\cap H\neq\emptyset$ and $U_2\cap H\neq\emptyset$. Since $H$
is a group, we get $U_1U_2\cap H\neq\emptyset$. Same argument
can be used to show that $\ol{H}$ is closed under inverse.\\

\end{enumerate}

\PP{32.1}{Topology of the Galois group}
%-----------------------------------
Let $L/K$ be an algebraic, Galois extension. We simplify the notation
from the previous lecture.\\

\noindent {\em Hypothesis.}
Let $(I,\leq)$ be a {\em right directed} partially ordered set.
Assume that we have a direct system of {\em finite, Galois} subextensions:
$K\subset L_i\subset L$ ($L_i/K$ is finite and Galois); 
$L_i\subset L_j$ for every $i\leq j$,
such that
\[
\Dlim{i\in I} L_i \isom L.
\]
Note that such direct systems always exist - for instance, we can take $I$ to be
the indexing set labelling {\em all finite, Galois} subextensions, ordered
by inclusion.\\

Since each $L_i/K$ is a normal extension, we get restriction homomorphisms:
\[
\rho_{ij}:\Gal{L_j}{K}\to \Gal{L_i}{K},\ \forall i\leq j,
\]
which form an inverse system of (finite) groups. We also have:
\[
\phi_i:\Gal{L}{K}\to\Gal{L_i}{K},\ \forall i\in I,
\]
such that each
$\phi_i$ is surjective, and for $i\leq j$, we have: $\rho_{ij}\phi_j
=\phi_i$. Thus there is a group homomorphism
$\phi:\Gal{L}{K}\to \Ilim{i\in I}\Gal{L_i}{K}$, 
uniquely determined by: $\pi_i\circ \phi = \phi_i$, for every
$i\in I$. Here $\pi_i : \Ilim{j\in I}\Gal{L_j}{K} \to \Gal{L_i}{K}$
is the canonical homomorphism.\\

\begin{prop}
$\phi:\Gal{L}{K}\to \Ilim{i\in I}\Gal{L_i}{K}$ is an isomorphism of
groups. 
\end{prop}

\begin{pf}
It is clear that $\Ker(\phi)$ is trivial, since if $\phi_i(g)=\Id_{L_i}$
for each $i\in I$, and $L\cong \Dlim{i\in I} L_i$, then $g=\Id_L$.
Conversely, if $(g_i)_{i\in I} \in \Ilim{i\in I} \Gal{L_i}{K}$,
then we can define $g\in\Gal{L}{K}$ via $g|_{L_i}=g_i$.
This definition is unambiguous since $\rho_{ij}(g_j)=g_i$.
\end{pf}
\ \\

\begin{defn}
The Galois group $\Gal{L}{K}$ is assumed to be equipped with the
coarsest topology which makes each of the restriction homomorphisms
$\phi_i : \Gal{L}{K}\to \Gal{L_i}{K}$ continuous, when $\Gal{L_i}{K}$
is given the discrete topology.
\end{defn}
\ \\

\PP{32.2}{Explicit description of open sets}
%--------------------------------------------------------
We retain the set up of the previous paragraph. Let us use the notation:
\[
G = \Gal{L}{K},\qquad G_i = \Gal{L_i}{K}.
\]
Then, we have the following inverse system of finite, discrete groups:
\[
\mathcal{G}= \lp \{G_i\}_{i\in I}; \{\rho_{ij}:G_j\to G_i\}_{i\leq j} \rp.
\]
Also, for each $i\in I$ we have $\phi_i:G\to G_i$ which induces
an isomorphism
$\phi:G\isom \Ilim{i\in I} G_i$. Thus
$G$ can be viewed as a subgroup of $\wt{G}=\prod_{i\in I} G_i$.
The topology on $G$ and $\wt{G}$ is the coarsest
one that makes the following homomorphisms continuous.
\[
\xy
{\ar@{^(->} (0,0)*{G\ }="g"; (20,0)*{\wt{G}}="tg"};
{\ar_{\phi_i} "g"; (10,-10)*{G_i}="gi"};
{\ar^{\pi_i} "tg"; "gi"};
\endxy
\]
It makes it clear that the topology of $G\subset \wt{G}$ is that of a
subspace.\\

Recall that, by our definition, the following are open sets
containing the identity element $e\in \wt{G}$:
\[
\text{For each } i\in I,\ \wt{U}_i(e) := \Ker(\pi_i)
= \{\sigma\in \wt{G} : \pi_i(\sigma)=e_i\}
\subset \wt{G}.
\]
This is because $\{e_i\}\subset G_i$ is open.\\

Note that we are assuming $(I,\leq)$ to be {\em right directed}.
So, for finite intersections, for every $i_1,\ldots,i_n\in I$, choose
$i\in I$ such that $i_j\leq i$ for each $j$. Then we have:
\[
\wt{U}_i(e) \subset \bigcap_{j=1}^n \wt{U}_j(e).
\]
Thus, $\wt{\CalB}(e) := \{\wt{U}_i(e):i\in I\}\subset 2^{\wt{G}}$ is a fundamental
system of neighbourhoods of the identity element.\\

It is also clear that $G\subset\wt{G}$ is closed. This is because,
\[
G = \bigcap_{i\leq j} F_{ij},\ \text{ where }
F_{ij} = \{\sigma\in\wt{G} : \rho_{ij}(\pi_j(\sigma)) = \pi_i(\sigma)\}
\subset \wt{G}.
\]
Each $F_{ij}$ is closed, hence so must be $G$.\\

Let $U_i(e) = G\cap \wt{U}_i$, for each $i\in I$. Note that 
$U_i(e) = \Ker(\phi_i)$ is a normal subgroup, which
is open and hence closed. It is naturally identified with
$U_i(e) = \Gal{L}{L_i} \subset \Gal{L}{K}$.\\

For any $\sigma\in G$, we get a fundamental system of neighbourhoods
$\CalB(\sigma) = \sigma\CalB(e) = \CalB(e)\sigma$. If $\sigma\neq\tau$
are two elements of $G$, then there exists $i\in I$ such
that $\phi_i(\sigma)\neq \phi_i(\tau)$. Thus, we obtain
\[
\lp \sigma U_i(e)\rp \cap \lp \tau U_i(e)\rp = \emptyset.
\]
We have shown that {\em any two points can be separated by
sets which are both open and closed}. That is, $G$ is totally disconnected
(two distinct points belong to different connected components).\\

\PP{32.3}{Proof of the fundamental theorem}
%-----------------------------------------
Recall the set maps 
\[
\mathcal{F}:\{\text{Subgroups of } \Gal{L}{K}\}
\to \{\text{Subextensions of } L/K\},
\]
\[
\mathcal{G}:\{\text{Subextensions of } L/K\}\to
\{\text{Subgroups of } \Gal{L}{K}\}
\]

given by $\mathcal{F}(H) = L^H$ and $\mathcal{G}(E) = \Gal{L}{E}$.\\

\noindent {\bf 1.} For each subextension $K\subset E\subset L$,
$\Gal{L}{E}\subset\Gal{L}{K}$ is a closed subgroup.\\

Let $H=\Gal{L}{E}$ and $G=\Gal{L}{K}$.
To show that $H=\ol{H}$, it is enough to prove that every $\sigma\in \ol{H}$
is an element of $H$. For $\sigma$ to be in the closure of $H$,
it is necessary and sufficient that for each $i\in I$,
$\sigma U_i(e) \cap H \neq\emptyset$. For $z\in E$, choose $i\in I$
so that $z\in L_i$. Let $h=\sigma g \in \sigma U_i(e) \cap H$ (recall
$U_i(e)$ is normal). Then $z=h(z)=\sigma(g(z)) = \sigma(z)$.
Thus $\sigma|_E = \Id_E$ proving that $\sigma\in H$
as claimed.\\

\noindent {\bf 2.} For any subgroup $H$ of $\Gal{L}{K}$,
we have $L^H = L^{\ol{H}}$.\\

(It is the same argument as for the proof of {\bf 1} above).
Since $H\subset\ol{H}$, it is clear that $L^{\ol{H}}\subset L^H$.
Conversely, let $z\in L$ be fixed by every element of $H$,
and let $\sigma\in \ol{H}$. We want to show that $\sigma(z)=z$.
Choose $i\in I$ such that $z\in L_i$. As $\sigma U_i(e)\cap H
\neq \emptyset$, we can write $\sigma=gh$, where $g\in U_i(e)$
and $h\in H$. Now $h(z)=z=g(z)$, which implies $\sigma(z)=z$.\\

\noindent {\bf 3.} Finally, if $H\subset\Gal{L}{K}$ is a
subgroup, and $E=L^H$, then $\Gal{L}{E}=\ol{H}$.\\

We already know that $L/E$ is a Galois extension, and
$\Gal{L}{E}$ contains $H$ and is closed. Therefore, $\ol{H}\subset
\Gal{L}{E}$. Now assume that $\sigma\in\Gal{L}{E}$. In order
to show that $\sigma\in\ol{H}$, we need to prove that for
every $i\in I$, $\sigma U_i(e)\cap H\neq \emptyset$. That is,
there exists $h\in H$ such that $\phi_i(h)=\phi_i(\sigma)$.

Now let $N_i = L_iE \subset L$ be the smallest field containing
both $L_i$ and $E$. Note that $N_i/E$ is a finite Galois extension:
(i) if $L_i$ is the splitting extension of a set $P_i\subset K[x]$,
then $L_iE$ is the splitting extension of the same set of polynomials,
viewed as elements of $E[x]$, (ii) $[N_i:E] \leq [L_i:K]<\infty$.

Let $p_i : \Gal{L}{E} \to \Gal{N_i}{E}$ be the restriction map.
Note that $N_i^{p_i(H)} = E$ since $L^H=E$. From the theorem for
the finite case (Theorem 31.1), we conclude that $p_i(H)=\Gal{N_i}{E}$. So, there
exists $h\in H$ such that $p_i(h)=p_i(\sigma)$. Since $\phi_i$
restricted to $\Gal{L}{E}$ can be written as the composition:
\[
\Gal{L}{E} \morph{p_i} \Gal{N_i}{E} \hookrightarrow \Gal{N_i}{K}
\to \Gal{L_i}{K},
\]
we conclude that $\phi_i(g)=\phi_i(\sigma)$, as required.

\end{document}

