Let $L/F$ be a Galois extension, not necessarily finite dimensional.

Topology on the Galois group

Recall that the Galois group $G := \Gal(L/F)$ of $L/F$ is the group of all field automorphisms $\sigma: L \lra L$ that restrict to the identity map on $F$ under the group operation of composition. In the case where the extension $L/F$ is infinite dimensional, the group $G$ comes equipped with a natural topology, which plays a key role in the statement of the Galois correspondence.

We define a subset $U$ of $G$ to be open if, for each $\sigma \in U$ there exists an intermediate field $K \subset L$ such that

• The degree $[K:F]$ is finite,
• If $\sigma'$ is another element of $G$ and the restrictions $\sigma|_K$ and $\sigma'|_K$ are equal, then $\sigma' \in U$

The resulting collection of open sets forms a topology on $G$ called the Krull topology, and $G$ is a topological group under the Krull topology. Another way to define the topology is to state that the subgroups $\Gal(L/K)$ for finite extensions $K/F$ form a neighborhood basis for $\Gal(L/F)$ at the identity.

Inverse limit structure

In this section we exhibit the group $G$ as a projective limit of an inverse system of finite groups. This construction shows that the Galois group $G$ is actually a profinite group.

Let $\A$ denote the set of finite normal extensions $K$ of $F$ which are contained in $L$ The set $\A$ is a partially ordered set under the inclusion relation. Form the inverse limit $$ \Gamma := \ilim \Gal(K/F) \subset \prod_{K \in \A} \Gal(K/F) $$ consisting, as usual, of the set of all $(\sigma_K) \in \prod_K \Gal(K/F)$ such that $\sigma_{K'}|_K = \sigma_K$ for all $K,K' \in \A$ with $K \subset K'$ We make $\Gamma$ into a topological space by putting the discrete topology on each finite set $\Gal(K/F)$ and giving $\Gamma$ the subspace topology induced by the product topology on $\prod_K \Gal(K/F)$ The group $\Gamma$ is a closed subset of the compact group $\prod_K \Gal(K/F)$ and is therefore compact.

Let $$ \phi: G \lra \prod_{K \in \A} \Gal(K/F) $$ be the group homomorphism which sends an element $\sigma \in G$ to the element $(\sigma_K)$ of $\prod_K \Gal(K/F)$ whose $K$ -th coordinate is the automorphism $\sigma|_K \in \Gal(K/F)$ Then the function $\phi$ has image equal to $\Gamma$ and in fact is a homeomorphism between $G$ and $\Gamma$ Since $\Gamma$ is profinite, it follows that $G$ is profinite as well.

The Galois correspondence