In the finite extension case, the size of the Galois group is nearly trivial:
\[|\operatorname{Gal}(E/F)|=|\operatorname{Hom}_F(E,\overline{F})|=[E:F],\]where the second equality follows from separability of $E/F$.
The infinite case is not so friendly. Still, the Galois group has the following “nice” description.
Proposition. Given any Galois extension $E/F$, we have
\[\operatorname{Gal}(E/F)\xrightarrow{\sim}\varprojlim_{K/F}\operatorname{Gal}(K/F),\]where $K/F$ runs through every finite Galois subextension.
Proof. $\sigma\in\operatorname{Gal}(E/F)$ is mapped to $(\sigma\vert_K)_K$. Conversely, the preimage of $(s_K)_K$ maps $x\in E$ to $s_K(x)$, where $K$ is any Galois subextension containing $x$.
Endow each finite $\operatorname{Gal}(K/F)$ with the discrete topology. $\operatorname{Gal}(E/F)$ then becomes a profinite group, and this topology is called the Krull topology. Recall that:
Theorem. Let $G$ be a topological group. The following are equivalent:
Proof. A profinite group $G$ has the structure
\[G=\left\{(a_i)_i\in\prod_{i\in I} G_i:\varphi_{ij}(a_j)=a_i,\forall i\leq j\right\}.\]This is a closed subset of product of Hausdorff, compact, totally disconnected spaces, hence has these properties as well. The topology is given by the neighborhood basis at $1$, which is simply $\{\ker(\pi_i):i\in I\}$, since $I$ is filtered and $G_i$’s are discrete. So we see that 1. implies 2. and 3.
Suppose 2. Let $H_i$ be the given neighborhood basis, indexed by $(I,\leq)$, ordered by inverse inclusion. By definition, $I$ is filtered, and each $H_i$ is a neighborhood of $1$, hence an clopen subgroup with finite index. Consequently we have a continuous homomorphism
\[\varphi:G\longrightarrow G':=\varprojlim_{i\in I} G/H_i\]from $G$ to a profinite group. Note that both sides are compact and Hausdorff. For it to be a homeomorphism, we just need $\varphi$ to be injective and $\operatorname{im}\varphi$ to be dense in $G’$.
But this is easy. Injectivity follows from Hausdorff property of $G$ and $\ker\varphi=\bigcap_iH_i$. “Surjectivity” follows from the above characterization of the topology of $G’$. This proves 2. implies 1.
Now suppose 3. We first prove a clopen basis exists. Indeed, $G$ is Hausdorff: the diagonal $\Delta$ is the preimage of $1$ under $(g,h)\mapsto g^{-1}h$, and totally disconnected spaces are $T_1$. Therefore Lemma 1 applies:
\[\{g\}=\bigcap_{U\in N}U.\]Given any open neighborhood of $g$, by compactness, it must contain some finite intersection in $N$, i.e. a clopen neighborhood of $g$. So we have a clopen basis.
Now fix $U\in\mathcal{N}_{\text{clopen}}(1)$. To conclude the proof, we need to find a open normal subgroup in $U$.
Since $U$ is compact, so is $U\times U$ and hence $U^2$ and $U^2\setminus U$. Therefore $A:=U^2\setminus U$ is closed. For every $g\in U$, $g\notin A$. So there exists open neighborhoods $U_g$ of $g$ and $V_g$ of $1$, such that $V_g\subset U$ and $U_gV_g\subset G\setminus A$. Since $U$ is compact, a finite number of the $U_g$’s covers $U$. Let $S$ be the finite intersection of the respective $V_g$’s.
Consequently $S$ is an open neighborhood of $1$, contained in $U$. By definition, $US\subset G\setminus A=U\cup(G\setminus U^2)$. Since $US\subset U^2$, we conclude that $US\subset U$. Therefore
\[US^n\subset U,\quad\forall n\in\mathbb{Z}_{\geq 0},\]and
\[H:=\bigcup_{n\in\mathbb{Z}_{\geq 0}}\left(S\cap S^{-1}\right)^n\]is a subgroup and a neighborhood of $1$, hence an open subgroup. Then the core, i.e., the largest normal subgroup in $H$, is open as well:
\[\operatorname{Core}(H)=\bigcap_{g\in G}g^{-1}Hg=\bigcap_{g\in G/N_G(H)}g^{-1}Hg.\]Putting these together, we have proved 3. implies 2.
Lemma 1. If $X$ is a compact Hausdorff space, then the connected component $A$ of $x\in X$ is the intersection of all clopen neighborhoods $N:=\mathcal{N}_{\text{clopen}}(x)$
\[T=\bigcap_{U\in N}U.\]Proof. Indeed, every $U\in N$ contains $A$, so $A\subset T$. Hence it suffices to prove $T$ is connected. Note that $T$ is closed. Suppose $T=U\cup V$ is a disjoint union of closed, hence compact, sets. Since $X$ is Hausdorff, there exists open neighborhoods $U’$, $V’$ separating $U$ and $V$.
Now $X-U’\cup V’$ is a closed set not intersecting $T$. By compactness, there exists a finite $N_0\subset N$, such that
\[T_0:=\left(\bigcap_{U\in N_0}U\right)\subset U'\cup V'.\]It is clear that $(T_0\cap U’$, $T_0\cap V’)$ are clopen and disjoint, and covers $T_0$. One of them contains $x$ and hence $T$. Then the other one cannot intersect $T$, proving that one of $U$, $V$ is empty.
With this understanding, it is now clear that all galois groups are either finite or uncountable:
Corollary. A locally compact Hausdorff group is either finite or uncountable.
Proof. The Baire category theorem applies. If the group is infinite, points cannot be open. Therefore a countable union of single points is nowhere dense.
Another quick proof with measure theory:
Proof’. We have a left Haar measure $\mu$ on $G$. In particular, $\mu(G)=1$. But if $G$ is countably infinite, $\mu(G)$ has to be $0$ or $\infty$, a contradiction.
Post date: 2024/11/21
Home page