[proofplan] We prove the three properties of the Weil group $W(M/K)$ in sequence. For density (Part 1), we reduce to showing that the restriction $W(M/K) \to \operatorname{Gal}(L/K)$ is surjective for every finite Galois subextension $L/K$, which we establish by a diagram chase using the exact sequence $\operatorname{Gal}(M/T) \hookrightarrow W(M/K) \twoheadrightarrow \operatorname{Frob}_{T/K}^{\mathbb{Z}}$ (where $T = T_{M/K}$ is the maximal unramified subextension) and the surjectivity of the Galois-theoretic restriction maps on both the inertia and unramified components. For Part 2, we use the inclusion $L \cdot T_{M/K} \subseteq T_{M/L}$ to show that $W(M/L) = W(M/K) \cap \operatorname{Gal}(M/L)$ by unravelling the Frobenius characterisation. For Part 3, we combine Part 2 with density and the second isomorphism theorem. [/proofplan] [step:Prove density by showing surjectivity of $W(M/K) \to \operatorname{Gal}(L/K)$ for each finite Galois $L/K$] Let $T = T_{M/K}$ denote the maximal unramified subextension of $M/K$. By definition, $W(M/K)$ consists of all $\sigma \in \operatorname{Gal}(M/K)$ whose restriction $\sigma|_T$ is an integer power of the Frobenius $\operatorname{Frob}_{T/K}$. This gives the exact sequence \begin{align*} 1 \to \operatorname{Gal}(M/T) \to W(M/K) \xrightarrow{\operatorname{res}_T} \operatorname{Frob}_{T/K}^{\mathbb{Z}} \to 1. \end{align*} Density of $W(M/K)$ in $\operatorname{Gal}(M/K)$ is equivalent (by the definition of the Krull topology) to the surjectivity of the restriction $\operatorname{res}_L \colon W(M/K) \to \operatorname{Gal}(L/K)$ for every finite Galois subextension $L/K$ of $M/K$. Fix such an $L/K$. We perform a diagram chase. Consider the commutative diagram with exact rows: \begin{align*} 1 \to \operatorname{Gal}(M/T) &\to W(M/K) \to \operatorname{Frob}_{T/K}^{\mathbb{Z}} \to 1 \\ \downarrow^{\operatorname{res}_{T \cap L}} &\quad \downarrow^{\operatorname{res}_L} \quad\quad \downarrow^{\operatorname{res}_{T \cap L}} \\ 1 \to \operatorname{Gal}(L/T \cap L) &\to \operatorname{Gal}(L/K) \to \operatorname{Gal}(T \cap L/K) \to 1 \end{align*} The bottom row is exact by the fundamental theorem of Galois theory: $T \cap L$ is the fixed field of $\operatorname{Gal}(L/T \cap L)$ in $L$, and $\operatorname{Gal}(T \cap L / K)$ is the quotient $\operatorname{Gal}(L/K) / \operatorname{Gal}(L / T \cap L)$. **Left vertical map is surjective.** The extension $T/K$ is unramified, so $T \cap L / K$ is unramified. Therefore $L / T \cap L$ is a finite Galois extension, and $T \supseteq T \cap L$. Since $M \supseteq T$, the restriction $\operatorname{Gal}(M/T) \to \operatorname{Gal}(L / T \cap L)$ is surjective: every $K$-automorphism of $L$ fixing $T \cap L$ extends to an automorphism of $M$ fixing $T$ (because $M/T$ is Galois, and $\operatorname{Gal}(M/T) \to \operatorname{Gal}(L / T \cap L)$ is the restriction map for the extension $M/T$ and its subextension $L / T \cap L$, which is surjective by the surjectivity of restriction in Galois theory). **Right vertical map is surjective.** The extension $T \cap L / K$ is a finite unramified extension of $K$. Its Galois group $\operatorname{Gal}(T \cap L / K)$ is generated by the Frobenius $\operatorname{Frob}_{(T \cap L)/K}$, which is the image of $\operatorname{Frob}_{T/K}$ under restriction. Therefore the image of $\operatorname{Frob}_{T/K}^{\mathbb{Z}}$ under the right vertical map is $\operatorname{Frob}_{(T \cap L)/K}^{\mathbb{Z}} = \operatorname{Gal}(T \cap L / K)$. **Diagram chase.** Let $\sigma \in \operatorname{Gal}(L/K)$. By surjectivity of the right vertical map, there exists $w \in W(M/K)$ with $\operatorname{res}_{T \cap L}(w) = \operatorname{res}_{T \cap L}(\sigma)$, i.e., $w|_{T \cap L} = \sigma|_{T \cap L}$. Then $\sigma \cdot (w|_L)^{-1} \in \operatorname{Gal}(L / T \cap L)$. By surjectivity of the left vertical map, there exists $g \in \operatorname{Gal}(M/T) \subset W(M/K)$ with $g|_L = \sigma \cdot (w|_L)^{-1}$. Then $(gw)|_L = g|_L \cdot w|_L = \sigma$, and $gw \in W(M/K)$ (since both $g$ and $w$ lie in $W(M/K)$, and $W(M/K)$ is a group). Therefore $\operatorname{res}_L$ is surjective. [guided] The density of the Weil group is not formal — it requires the surjectivity of the Frobenius restriction and the inertia restriction, and the proof is a genuine diagram chase. The key structures are: - The Weil group $W(M/K)$ sits between the inertia group $\operatorname{Gal}(M/T)$ (a profinite group) and the discrete Frobenius powers $\operatorname{Frob}_{T/K}^{\mathbb{Z}}$. - The target $\operatorname{Gal}(L/K)$ decomposes analogously via its unramified part $T \cap L$. The diagram chase is a standard "lift from both sides" argument. Given $\sigma \in \operatorname{Gal}(L/K)$, we first lift its unramified component (the image in $\operatorname{Gal}(T \cap L / K)$) to a Weil group element $w$. The "error" $\sigma (w|_L)^{-1}$ lies in the inertia part $\operatorname{Gal}(L / T \cap L)$, which we then lift to an element $g$ of $\operatorname{Gal}(M/T)$. The product $gw$ lies in $W(M/K)$ and restricts to $\sigma$. Why is $W(M/K)$ dense but not equal to $\operatorname{Gal}(M/K)$? Because $\operatorname{Frob}_{T/K}^{\mathbb{Z}}$ is a proper dense subgroup of $\operatorname{Gal}(T/K) \cong \hat{\mathbb{Z}}$: the integers $\mathbb{Z}$ are dense in $\hat{\mathbb{Z}}$ but not equal to it. The Weil group "thins out" the Galois group by replacing the profinite Frobenius powers with the discrete ones. [/guided] [/step] [step:Prove $W(M/L) = W(M/K) \cap \operatorname{Gal}(M/L)$ for any finite subextension $L/K$] Let $T_K = T_{M/K}$ and $T_L = T_{M/L}$ denote the maximal unramified subextensions of $M/K$ and $M/L$ respectively. Since every unramified extension of $K$ contained in $M$ remains unramified over $L$ (the composite of an unramified extension with any extension is unramified over the larger base), we have $T_K \subseteq T_L$. Moreover, $L \cdot T_K \subseteq T_L$ since $L \cdot T_K / L$ is unramified ($T_K / K$ is unramified and unramified extensions are preserved under base change). For $\sigma \in \operatorname{Gal}(M/L)$, the following chain of equivalences holds: \begin{align*} \sigma \in W(M/L) &\iff \sigma|_{T_L} \in \operatorname{Frob}_{T_L/L}^{\mathbb{Z}} \\ &\iff \sigma|_{T_K} \in \operatorname{Frob}_{T_K/K}^{\mathbb{Z}} \\ &\iff \sigma \in W(M/K). \end{align*} **First equivalence.** This is the definition of $W(M/L)$. **Second equivalence ($\Rightarrow$).** If $\sigma|_{T_L} = \operatorname{Frob}_{T_L/L}^n$ for some $n \in \mathbb{Z}$, then restricting to $T_K \subseteq T_L$ gives $\sigma|_{T_K} = \operatorname{Frob}_{T_L/L}^n|_{T_K}$. Since $T_K/K$ is unramified and $\operatorname{Frob}_{T_L/L}|_{T_K}$ acts as the Frobenius of the residue field extension (which coincides with $\operatorname{Frob}_{T_K/K}$, since the Frobenius is determined by its action on the residue field), we get $\sigma|_{T_K} = \operatorname{Frob}_{T_K/K}^n$. **Second equivalence ($\Leftarrow$).** Suppose $\sigma|_{T_K} = \operatorname{Frob}_{T_K/K}^n$. Since $\sigma \in \operatorname{Gal}(M/L)$, $\sigma$ fixes $L$ pointwise. On $L \cdot T_K \subseteq T_L$, $\sigma$ acts as the identity on $L$ and as $\operatorname{Frob}_{T_K/K}^n$ on $T_K$. We need to show $\sigma|_{T_L} \in \operatorname{Frob}_{T_L/L}^{\mathbb{Z}}$. The Frobenius $\operatorname{Frob}_{T_L/L}$ is characterised by its action on the residue field: it acts as $x \mapsto x^{|k_L|}$. On $T_K$, $\operatorname{Frob}_{T_K/K}$ acts as $x \mapsto x^{|k_K|}$ on the residue field. Since $|k_L| = |k_K|^{f_{L/K}}$ where $f_{L/K} = [k_L : k_K]$, we have $\operatorname{Frob}_{T_L/L}|_{T_K} = \operatorname{Frob}_{T_K/K}^{f_{L/K}}$. The element $\sigma|_{T_L}$ restricts to an integer power of $\operatorname{Frob}_{T_K/K}$ on $T_K$ and fixes $L$, so by the structure of $\operatorname{Gal}(T_L/L) \cong \hat{\mathbb{Z}}$ (generated topologically by $\operatorname{Frob}_{T_L/L}$), $\sigma|_{T_L}$ is an integer power of $\operatorname{Frob}_{T_L/L}$. The combined equivalences give $W(M/L) = W(M/K) \cap \operatorname{Gal}(M/L)$. [/step] [step:Deduce $W(M/K)/W(M/L) \cong \operatorname{Gal}(L/K)$ for finite Galois $L/K$] Assume $L/K$ is Galois, so $\operatorname{Gal}(M/L)$ is a normal subgroup of $\operatorname{Gal}(M/K)$. By Part 2, $W(M/L) = W(M/K) \cap \operatorname{Gal}(M/L)$. Since $\operatorname{Gal}(M/L)$ is normal in $\operatorname{Gal}(M/K)$ and $W(M/K) \leq \operatorname{Gal}(M/K)$, the intersection $W(M/L) = W(M/K) \cap \operatorname{Gal}(M/L)$ is normal in $W(M/K)$. By the second isomorphism theorem for groups: \begin{align*} \frac{W(M/K)}{W(M/L)} = \frac{W(M/K)}{W(M/K) \cap \operatorname{Gal}(M/L)} \cong \frac{W(M/K) \cdot \operatorname{Gal}(M/L)}{\operatorname{Gal}(M/L)}. \end{align*} We claim $W(M/K) \cdot \operatorname{Gal}(M/L) = \operatorname{Gal}(M/K)$. By Part 1, $W(M/K)$ is dense in $\operatorname{Gal}(M/K)$, which means $W(M/K)$ meets every coset of $\operatorname{Gal}(M/L)$: for any $\sigma \in \operatorname{Gal}(M/K)$, the coset $\sigma \operatorname{Gal}(M/L)$ is open (since $\operatorname{Gal}(M/L)$ is open in the Krull topology), and a dense subgroup meets every non-empty open set. Therefore $\sigma \in W(M/K) \cdot \operatorname{Gal}(M/L)$, and since $\sigma$ was arbitrary, $W(M/K) \cdot \operatorname{Gal}(M/L) = \operatorname{Gal}(M/K)$. Substituting: \begin{align*} \frac{W(M/K)}{W(M/L)} \cong \frac{\operatorname{Gal}(M/K)}{\operatorname{Gal}(M/L)} \cong \operatorname{Gal}(L/K), \end{align*} where the last isomorphism is the fundamental theorem of Galois theory. [guided] The argument for Part 3 combines the previous two parts in a clean way. The second isomorphism theorem reduces the quotient $W(M/K) / W(M/L)$ to $W(M/K) \cdot \operatorname{Gal}(M/L) / \operatorname{Gal}(M/L)$, and density upgrades the numerator from $W(M/K) \cdot \operatorname{Gal}(M/L)$ to all of $\operatorname{Gal}(M/K)$. Why does density imply $W(M/K) \cdot \operatorname{Gal}(M/L) = \operatorname{Gal}(M/K)$? In a topological group $G$, a dense subgroup $D$ meets every non-empty open set (by definition of density). Since $\operatorname{Gal}(M/L)$ is an open subgroup of $\operatorname{Gal}(M/K)$ (by the Krull topology), every coset $\sigma \operatorname{Gal}(M/L)$ is also open. Therefore $D \cap \sigma \operatorname{Gal}(M/L) \neq \varnothing$, i.e., $\sigma \in D \cdot \operatorname{Gal}(M/L)$. Hence $G = D \cdot \operatorname{Gal}(M/L)$. The result says that, despite $W(M/K)$ being a proper (non-closed) subgroup of $\operatorname{Gal}(M/K)$, it "sees" the same finite Galois quotients. This is the key property that makes the Weil group useful in local class field theory: it carries the same information as the full Galois group at the level of finite extensions, while having the advantage of admitting a topology (the Weil topology) under which the Artin map is a topological isomorphism. [/guided] [/step]