[proofplan] We prove that $\operatorname{Gal}(M/K)$ is compact and Hausdorff by realising it as a closed subgroup of a product of finite discrete groups (via the Krull topology) and applying Tychonoff's theorem. The Hausdorff property follows from the fact that any two distinct automorphisms differ on some finite subextension. The existence of an open normal subgroup inside any open neighbourhood of the identity follows from the definition of the Krull topology (basic open sets are cosets of $\operatorname{Gal}(M/L)$ for finite $L/K$) and the passage to the Galois closure. [/proofplan] [step:Realise $\operatorname{Gal}(M/K)$ as a closed subspace of a product of finite groups] Let $\{L_i/K\}_{i \in I}$ be the family of all finite Galois subextensions of $M/K$. Each $\operatorname{Gal}(L_i/K)$ is a finite group, hence compact and Hausdorff in the discrete topology. Consider the restriction map: \begin{align*} \Phi: \operatorname{Gal}(M/K) &\to \prod_{i \in I} \operatorname{Gal}(L_i/K) \\ \sigma &\mapsto (\sigma|_{L_i})_{i \in I}. \end{align*} This map is injective: if $\sigma|_{L_i} = \tau|_{L_i}$ for all $i$, then $\sigma$ and $\tau$ agree on every finite Galois subextension. Since $M = \bigcup_i L_i$ (every element of $M$ lies in some finite Galois extension of $K$ contained in $M$), we have $\sigma = \tau$. The image of $\Phi$ is the set of compatible families: $(\sigma_i)_{i \in I} \in \prod_i \operatorname{Gal}(L_i/K)$ such that $\sigma_j|_{L_i} = \sigma_i$ whenever $L_i \subseteq L_j$. This compatibility condition defines a closed subset of $\prod_i \operatorname{Gal}(L_i/K)$: for each pair $i, j$ with $L_i \subseteq L_j$, the set $\{(\sigma_k) : \sigma_j|_{L_i} = \sigma_i\}$ is closed (it is the preimage of the diagonal under the continuous map $(\sigma_k) \mapsto (\sigma_j|_{L_i}, \sigma_i)$ into the discrete space $\operatorname{Gal}(L_i/K) \times \operatorname{Gal}(L_i/K)$). The image is the intersection of all such closed sets, hence closed. The Krull topology on $\operatorname{Gal}(M/K)$ coincides with the subspace topology inherited from $\prod_i \operatorname{Gal}(L_i/K)$ via $\Phi$: a basic open set in the Krull topology is a coset $\sigma \cdot \operatorname{Gal}(M/L_i)$, which corresponds to $\Phi^{-1}(\{\sigma_i\} \times \prod_{j \neq i} \operatorname{Gal}(L_j/K))$, a basic open set in the product topology. [guided] The realisation as an inverse limit is the conceptual backbone: $\operatorname{Gal}(M/K) = \varprojlim_i \operatorname{Gal}(L_i/K)$, where the inverse system is ordered by inclusion of fields, with transition maps given by restriction $\operatorname{Gal}(L_j/K) \to \operatorname{Gal}(L_i/K)$ for $L_i \subseteq L_j$. The inverse limit is a closed subspace of the product $\prod_i \operatorname{Gal}(L_i/K)$ (cut out by the compatibility conditions), and the product topology restricts to the profinite topology, which is the Krull topology. The injectivity of $\Phi$ is the content of the statement that $M/K$ is Galois: every element of $M$ is separable and normal over $K$, hence lies in some finite Galois subextension, and an automorphism is determined by its action on all elements. [/guided] [/step] [step:Conclude compactness and the Hausdorff property] **Compactness:** By Tychonoff's theorem, the product $\prod_{i \in I} \operatorname{Gal}(L_i/K)$ is compact (each factor is a finite discrete space, hence compact). Since $\operatorname{Gal}(M/K)$ is homeomorphic to a closed subspace of a compact space (via $\Phi$), it is compact. **Hausdorff:** The product of Hausdorff spaces is Hausdorff (in the product topology), and a subspace of a Hausdorff space is Hausdorff. Since each $\operatorname{Gal}(L_i/K)$ is discrete (hence Hausdorff), the product is Hausdorff, and so is $\operatorname{Gal}(M/K)$. Alternatively: given $\sigma \neq \tau$ in $\operatorname{Gal}(M/K)$, there exists $\alpha \in M$ with $\sigma(\alpha) \neq \tau(\alpha)$. Let $L$ be a finite Galois subextension containing $\alpha$. Then $\sigma|_L \neq \tau|_L$, so $\sigma$ and $\tau$ lie in different cosets of $\operatorname{Gal}(M/L)$, which are disjoint open sets separating them. [/step] [step:Find an open normal subgroup inside any open neighbourhood of the identity] Let $U$ be an open subset of $\operatorname{Gal}(M/K)$ containing the identity $\operatorname{id}$. By definition of the Krull topology, there exists a finite Galois subextension $L/K$ such that $\operatorname{Gal}(M/L) \subseteq U$. If $L/K$ is already Galois, set $N = \operatorname{Gal}(M/L)$. Then $N$ is open (by definition of the Krull topology, it is a basic open set), and $N$ is normal in $\operatorname{Gal}(M/K)$: for any $\sigma \in \operatorname{Gal}(M/K)$ and $\tau \in \operatorname{Gal}(M/L)$, the conjugate $\sigma\tau\sigma^{-1}$ fixes $L$ pointwise because $L/K$ is Galois (i.e., $\sigma(L) = L$), so $\sigma\tau\sigma^{-1} \in \operatorname{Gal}(M/L)$. If $L/K$ is not necessarily Galois (the Krull topology basis may involve non-Galois finite subextensions in some formulations), let $L'$ be the Galois closure of $L$ over $K$ inside $M$. Since $L'/K$ is a finite Galois extension and $L \subseteq L'$, we have $\operatorname{Gal}(M/L') \subseteq \operatorname{Gal}(M/L) \subseteq U$. Set $N = \operatorname{Gal}(M/L')$. This is open (as above) and normal (since $L'/K$ is Galois). [guided] Why is $\operatorname{Gal}(M/L)$ normal when $L/K$ is Galois? Normality means: for every $\sigma \in \operatorname{Gal}(M/K)$ and $\tau \in \operatorname{Gal}(M/L)$, we have $\sigma\tau\sigma^{-1} \in \operatorname{Gal}(M/L)$. Take any $\ell \in L$. Since $L/K$ is Galois, $\sigma^{-1}(\ell) \in L$ (because $\sigma^{-1}$ permutes the roots of any polynomial over $K$ that splits in $L$, and $L/K$ being Galois means $\sigma^{-1}(L) = L$). Therefore $\tau(\sigma^{-1}(\ell)) = \sigma^{-1}(\ell)$ (since $\tau$ fixes $L$), and $\sigma(\tau(\sigma^{-1}(\ell))) = \sigma(\sigma^{-1}(\ell)) = \ell$. So $\sigma\tau\sigma^{-1}$ fixes $\ell$, and since $\ell \in L$ was arbitrary, $\sigma\tau\sigma^{-1} \in \operatorname{Gal}(M/L)$. The passage to the Galois closure is necessary because the Krull topology might initially provide only a finite (not necessarily Galois) subextension $L/K$ with $\operatorname{Gal}(M/L) \subseteq U$. The Galois closure $L'$ of $L$ over $K$ is still finite over $K$ (since $L$ is finite over $K$), and $L' \subseteq M$ (since $M/K$ is Galois, all conjugates of elements of $L$ lie in $M$). Then $\operatorname{Gal}(M/L')$ is a smaller open subgroup, contained in $\operatorname{Gal}(M/L) \subseteq U$, and it is normal since $L'/K$ is Galois. [/guided] [/step]