This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
We previously defined Galois extensions and Galois groups. The central result of Galois theory is that for a field extension \(\mathbb{L}/\mathbb{K}\), there exists an order-preserving bijection between the lattice of subgroups of the Galois group \(\Gal(\mathbb{L}/\mathbb{K})\) and the lattice of Galois subextensions of \(\mathbb{L}/\mathbb{K}\). In many cases this result treats only the case where the Galois group \(\Gal(\mathbb{L}/\mathbb{K})\) is finite, but since we will also treat the case where \(\Gal(\mathbb{L}/\mathbb{K})\) is infinite, we must endow \(\Gal(\mathbb{L}/\mathbb{K})\) with an appropriate topological structure.
The Topological Structure of Galois Groups
Let a Galois extension \(\mathbb{L}/\mathbb{K}\) be given, and let \(\Gal(\mathbb{L}/\mathbb{K})\) be its Galois group. Since the Galois group is in any case a collection of functions from the set \(\mathbb{L}\) to \(\mathbb{L}\), if we endow the collection of functions from \(\mathbb{L}\) to \(\mathbb{L}\), denoted \(\Fun(\mathbb{L},\mathbb{L})=\mathbb{L}^\mathbb{L}\), with a topological structure, we can give \(\Gal(\mathbb{L}/\mathbb{K})\) a topological structure as a subset of this set. ([Topology] §Subspaces, ⁋Definition 1)
To this end, let us endow \(\mathbb{L}\) with the discrete topology. ([Topology] §Open Sets, ⁋Example 2) From the results of [Topology] §Product Spaces, we know that a subbase for this set consists of sets of the form
\[U_{x,y}=\left\{\sigma\mid\sigma(x)=y \right\}\]and so, if we give \(\Gal(\mathbb{L}/\mathbb{K})\) the subspace topology, we know that for any \(\sigma\in\Gal(\mathbb{L}/\mathbb{K})\) the collection of sets of the form
\[U_{x_1,\ldots,x_n}=\left\{\tau\in\Gal(\mathbb{L}/\mathbb{K})\mid \text{$\tau(x_i)=\sigma(x_i)$ for all $i$}\right\}\]constitutes a local base at \(\sigma\). ([Topology] §Bases of Topological Spaces, ⁋Definition 4)
Meanwhile, the functions satisfying the above condition are those that agree with \(\sigma\) when restricted to a finite subextension \(\mathbb{M}=\mathbb{L}(x_1,\ldots,x_n )\) of \(\mathbb{L}\), and conversely any finite subextension \(\mathbb{M}/\mathbb{K}\) defines an element of the local base at \(\sigma\) in this manner. That is, letting \(\Lambda\) be the collection of finite subextensions of \(\mathbb{L}/\mathbb{K}\), and for any \(\mathbb{M}/\mathbb{K}\in \Lambda\) and any \(\sigma\in \Gal(\mathbb{L}/\mathbb{K})\), defining the subset \(U_\mathbb{M}(\sigma)\) of \(\Gal(\mathbb{L}/\mathbb{K})\) by the formula
\[U_\mathbb{M}(\sigma)=\left\{\tau\in \Gal(\mathbb{L}/\mathbb{K})\mid \sigma\vert_\mathbb{M}=\tau\vert_\mathbb{M}\right\}\]this set becomes an element of the local base at \(\sigma\), and the collection $(U_\mathbb{M}(\sigma))_{\sigma\in\Lambda}$ is exactly the local base at \(\sigma\).
Example 1 In particular, consider the case where \(\mathbb{L}/\mathbb{K}\) is a finite degree Galois extension. Then from the discussion following §Galois Extensions, ⁋Definition 12, we know that \(\Gal(\mathbb{L}/\mathbb{K})\) is a finite set. Meanwhile, from the above local base, since \(\mathbb{L}/\mathbb{K}\) is of finite degree, \(\mathbb{L}/\mathbb{K}\) is already an element of \(\Lambda\) and therefore
\[U_\mathbb{L}(\sigma)=\left\{\tau\in\Gal(\mathbb{L}/\mathbb{K})\mid \sigma\vert_\mathbb{L}=\tau\vert_\mathbb{L}\right\}=\left\{\sigma\right\}\]so in this case \(\Gal(\mathbb{L}/\mathbb{K})\) becomes a set equipped with the discrete topology.
Meanwhile, the topological space \(\Gal(\mathbb{L}/\mathbb{K})\) defined as above is originally a group with the composition of \(\mathbb{K}\)-automorphisms as its operation, and it is not difficult to show that the composition of functions is compatible with this topological structure.
Proposition 2 \(\Gal(\mathbb{L}/\mathbb{K})\) defined above is a topological group.
Proof
That is, we must show that the two homomorphisms
\[\Gal(\mathbb{L}/\mathbb{K})\times\Gal(\mathbb{L}/\mathbb{K})\rightarrow\Gal(\mathbb{L}/\mathbb{K});\quad (\sigma,\sigma')\mapsto \sigma\sigma',\qquad \Gal(\mathbb{L}/\mathbb{K})\rightarrow\Gal(\mathbb{L}/\mathbb{K});\quad \sigma\mapsto \sigma^{-1}\]are continuous. First, considering any element \(U_\mathbb{M}(\sigma\sigma')\) of the local base at \(\sigma\sigma'\), by definition
\[U_\mathbb{M}(\sigma\sigma')=\left\{\tau\in\Gal(\mathbb{L}/\mathbb{K})\mid \tau\vert_\mathbb{M}=\sigma\sigma'\vert_\mathbb{M}\right\}\]and therefore for obvious reasons the open set \(U_\mathbb{M}(\sigma)\times U_\mathbb{M}(\sigma')\) of \(\Gal(\mathbb{L}/\mathbb{K})\times\Gal(\mathbb{L}/\mathbb{K})\) belongs to the preimage of the above set, and thus the multiplication map is continuous.
In a similar manner, the local base \(U_\mathbb{M}(\sigma^{-1})\) at \(\sigma^{-1}\) is given by the formula
\[U_\mathbb{M}(\sigma^{-1})=\left\{\tau\in\Gal(\mathbb{L}/\mathbb{K})\mid \tau\vert_\mathbb{M}=\sigma^{-1}\vert_\mathbb{M}\right\}\]and considering the local base \(U_\mathbb{M}(\sigma)\) at \(\sigma\), this set belongs to the preimage of the above set.
In particular, the local base at any \(\sigma\) is obtained by translating the local base at the identity \(\id_\mathbb{L}\) along the translation map. That is, for any \(\sigma\in \Gal(\mathbb{L}/\mathbb{K})\) the formula
\[U_\mathbb{M}(\sigma)=U_\mathbb{M}(\id_\mathbb{L})\sigma=\sigma U_\mathbb{M}(\id_\mathbb{L})\]holds. From this we know that instead of the above sets, we may consider only the set
\[U_\mathbb{M}(\id_\mathbb{L})=\left\{\tau\in \Gal(\mathbb{L}/\mathbb{K})\mid \tau\vert_\mathbb{M}=\id_\mathbb{M}\right\}\]alone. Then
\[U_\mathbb{M}(\id_\mathbb{L})\subseteq U_\mathbb{N}(\id_\mathbb{L})\iff \mathbb{M}\supseteq \mathbb{N}\]and from the above notation, as a set we know that
\[U_\mathbb{M}(\id_\mathbb{L})=\Gal(\mathbb{L}/\mathbb{M})\]The inclusion on the right is simply obtained by viewing an \(\mathbb{M}\)-automorphism as a \(\mathbb{K}\)-automorphism, and moreover we know that the topological structure defined on \(\Gal(\mathbb{L}/\mathbb{M})\) is exactly the same as that of \(U_\mathbb{M}(\id_\mathbb{L})\).
Now, considering the collection \(\Lambda'\) of finite degree Galois extensions, by §Galois Extensions, ⁋Proposition 11 we know that this is a cofinal subset of \(\Lambda\). That is, $(U_\mathbb{M}(\id_\mathbb{L}))_{\mathbb{M}\in\Lambda}$ is also a local base at \(\id_\mathbb{L}\). Then for any \(\mathbb{M}\in \Lambda'\), considering the restriction homomorphism \(\rho:\Gal(\mathbb{L}/\mathbb{K})\rightarrow\Gal(\mathbb{M}/\mathbb{K})\) examined in §Galois Extensions, ⁋Proposition 13, since any finite degree subextension of \(\mathbb{M}\) is also a finite degree extension of \(\mathbb{L}\), this restriction homomorphism is continuous with respect to the topological structure defined above. In this situation, since \(\rho\) is a continuous function from \(\Gal(\mathbb{L}/\mathbb{K})\) to the finite discrete space \(\Gal(\mathbb{M}/\mathbb{K})\) (Example 1), \(\ker\rho\) is a closed subgroup of \(\Gal(\mathbb{L}/\mathbb{K})\). However, by definition
\[\sigma\in\ker\rho\iff \sigma\vert_\mathbb{M}=\id\vert_\mathbb{M}\iff\sigma\in U_\mathbb{M}(\id_\mathbb{L})\]so each \(U_\mathbb{M}(\id_\mathbb{L})\) is clopen. Meanwhile, any clopen set can always be written as a union of connected components, and therefore any nonempty intersection of clopen sets must contain a connected component. However, the following holds.
Proposition 3 In the above situation the formula
\[\{\id_\mathbb{L}\}=\bigcap_{\mathbb{M}\in \Lambda'}U_\mathbb{M}(\id_\mathbb{L})\]holds.
Proof
Let any \(\sigma\in \Gal(\mathbb{L}/\mathbb{K})\) be given. If \(\sigma\neq\id_\mathbb{L}\), then there exists \(x\in \mathbb{L}\) such that \(\sigma(x)\neq x\). Then taking \(\mathbb{M}\mathbb{K}(x)\), we have \(\sigma\not\in U_\mathbb{M}(\id_\mathbb{L})\). Now, as we saw earlier, since \(\Lambda'\) is a cofinal subset of \(\Lambda\), we obtain the desired result.
Therefore, by the result of this proposition the connected component containing \(\id_\mathbb{L}\) is \(\left\{\id_\mathbb{L}\right\}\), and from this we know that \(\Gal(\mathbb{L}/\mathbb{K})\) is a totally disconnected space. ([Topology] §Connected Spaces, ⁋Definition 7) Moreover, the following holds.
Proposition 4 \(\Gal(\mathbb{L}/\mathbb{K})\) is compact.
Proof
First, for each \(x\in \mathbb{L}\), since \(\mathbb{L}/\mathbb{K}\) is an algebraic extension, \(x\) is algebraic, and therefore there are only finitely many elements conjugate to \(x\). (§Galois Extensions, ⁋Proposition 3) In other words,
\[\Gal(\mathbb{L}/\mathbb{K})\hookrightarrow \prod_{x\in \mathbb{L}}\mathbb{L}\overset{\pr_x}{\longrightarrow}\mathbb{L};\qquad \sigma\mapsto \sigma(x)\]considering this, the image of this function is a finite set. Therefore \(\Gal(\mathbb{L}/\mathbb{K})\) is a subset of a product of finite sets, and since finite sets are compact, this product is also compact. ([Topology] §Compactness, ⁋Theorem 2) Thus proving the given proposition is the same as showing that \(\Gal(\mathbb{L}/\mathbb{K})\) is closed in \(\mathbb{L}^\mathbb{L}\).
Suppose a function \(u\) belongs to the closure of \(\Gal(\mathbb{L}/\mathbb{K})\) in \(\mathbb{L}^\mathbb{L}\). If \(u\) is not an element of \(\Gal(\mathbb{L}/\mathbb{K})\), then either \(u\) is not a field homomorphism or \(u\) does not fix \(\mathbb{K}\). Adopting the first assumption, suppose there exist \(x,y\in\mathbb{L}\) such that \(u(x+y)\neq u(x)+u(y)\), for example. Then the set
\[\left\{f\in \mathbb{L}^\mathbb{L}\mid f(x)=u(x),f(y)=u(y),f(x+y)=u(x+y)\right\}\]is an element of the basis form of \(\mathbb{L}^\mathbb{L}\), so it is an open set and moreover contains \(u\). That is, this set is an open neighborhood of \(u\). However, from the assumption
\[f(x+y)=u(x+y)\neq u(x)+u(y)=f(x)+f(y)\]the \(f\)’s also do not become field homomorphisms. That is, the above open neighborhood does not meet \(\Gal(\mathbb{L}/\mathbb{K})\), which contradicts the assumption that \(u\) belongs to the closure of \(\Gal(\mathbb{L}/\mathbb{K})\). By similar logic all other cases can also be excluded, and from this we can prove that \(\Gal(\mathbb{L}/\mathbb{K})\) is closed in \(\mathbb{L}^\mathbb{L}\).
Meanwhile, let \(\mathbb{L}/\mathbb{K}\) be a Galois extension, and let \(\mathbb{L}_i/\mathbb{K}\) be Galois subextensions of this extension satisfying \(\mathbb{L}=\bigcup_{i\in I}\mathbb{L}_i\). Then we give this a partial order
\[i\leq j \iff \mathbb{L}_i\subset \mathbb{L}_j\]and under this partial order we can define the following restriction maps
\[\rho_{ij}:\Gal(\mathbb{L}_j/\mathbb{K}) \rightarrow \Gal(\mathbb{L}_i/\mathbb{K})\qquad \text{whenever $i\leq j$}\]These are continuous homomorphisms, and therefore their inverse limit
\[\varprojlim_{i\in I}\Gal(\mathbb{L}_i/\mathbb{K})\]and the canonical morphisms \(\rho_i:\varprojlim \Gal(\mathbb{L}_i/\mathbb{K})\rightarrow\Gal(\mathbb{L}_i/\mathbb{K})\) exist.
Meanwhile, considering the restriction maps
\[\lambda_i:\Gal(\mathbb{L}/\mathbb{K})\rightarrow\Gal(\mathbb{L}_i/\mathbb{K})\]these satisfy \(\lambda_i=\rho_{ij}\circ\lambda_j\), so there exists a continuous homomorphism \(\lambda:\Gal(\mathbb{L}/\mathbb{K})\rightarrow\varprojlim\Gal(\mathbb{L}_i/\mathbb{K})\) induced by them.
Proposition 5 The \(\lambda\) defined above defines an isomorphism between topological groups.
Proof
By Proposition 3, \(\Gal(\mathbb{L}_i/\mathbb{K})\) is Hausdorff, and since products and subspaces of Hausdorff spaces are again Hausdorff, their inverse limit \(\varprojlim \Gal9\mathbb{L}_i/\mathbb{K})\) is also Hausdorff. Meanwhile, since \(\Gal(\mathbb{L}/\mathbb{K})\) is compact by Proposition 4, by [Topology] §Compact Spaces, ⁋Proposition 9 it suffices to show that \(\lambda\) is bijective, which is almost obvious from \(\mathbb{L}= \bigcup \mathbb{L}_i\).
In particular, this proposition applies well to the family \(\Lambda'\).
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