[proofplan] We establish the inclusion-reversing bijection between open finite-index subgroups of $K^\times$ and finite abelian extensions of $K$ using the [Local Artin Reciprocity](/theorems/???) map. The direction $L/K \mapsto N(L/K)$ is well-defined by Artin reciprocity, which gives $K^\times / N(L/K) \cong \operatorname{Gal}(L/K)$, so $N(L/K)$ is open and of finite index. For the direction $H \mapsto (K^{ab})^{\operatorname{Art}_K(H)}$, we use the density of the Weil group to show the fixed field is a finite abelian extension. The bijection is verified by showing $N((K^{ab})^{\operatorname{Art}_K(H)}/K) = H$ and $(K^{ab})^{\operatorname{Art}_K(N(L/K))} = L$, using the density of the Weil group and the second isomorphism theorem. The lattice identities for $N(LM/K)$ and $N(L \cap M/K)$ follow from the inclusion-reversing nature and standard norm identities. [/proofplan] [step:Show that $N(L/K)$ is an open finite-index subgroup of $K^\times$ for each finite abelian $L/K$] Let $L/K$ be a finite abelian extension. By [Local Artin Reciprocity](/theorems/???), the Artin map induces an isomorphism \begin{align*} \frac{K^\times}{N(L/K)} \xrightarrow{\;\sim\;} \operatorname{Gal}(L/K), \end{align*} where $N(L/K) = N_{L/K}(L^\times)$ is the kernel of $\operatorname{Art}_K|_L \colon K^\times \to \operatorname{Gal}(L/K)$. Since $\operatorname{Gal}(L/K)$ is a finite group, $N(L/K)$ has finite index $[K^\times : N(L/K)] = [L : K]$ in $K^\times$. To show $N(L/K)$ is open: the norm map $N_{L/K} \colon L^\times \to K^\times$ is a continuous map of topological groups (with respect to the $\mathfrak{m}$-adic topologies). The group $1 + \mathfrak{m}_L^n$ is open in $L^\times$ for each $n \geq 1$, and $N_{L/K}(1 + \mathfrak{m}_L^n) \subseteq 1 + \mathfrak{m}_K^{\lceil n/e_{L/K} \rceil}$. In fact, $N_{L/K}$ maps a sufficiently small open subgroup of $\mathcal{O}_L^\times$ onto an open subgroup of $\mathcal{O}_K^\times$ (since $L/K$ is a finite extension of local fields, the norm map is locally a polynomial in the coefficients and is an open map near the identity). Therefore $N(L/K) \supseteq N_{L/K}(1 + \mathfrak{m}_L^n)$ contains an open subgroup of $K^\times$, so $N(L/K)$ is open. [/step] [step:Construct the inverse map $H \mapsto (K^{ab})^{\operatorname{Art}_K(H)}$ and verify it produces a finite abelian extension] Let $H$ be an open finite-index subgroup of $K^\times$. The Artin map $\operatorname{Art}_K \colon K^\times \xrightarrow{\sim} W(K^{ab}/K)$ sends $H$ to the subgroup $\operatorname{Art}_K(H) \leq W(K^{ab}/K)$. Since $H$ is open and of finite index, $\operatorname{Art}_K(H)$ is an open finite-index subgroup of $W(K^{ab}/K)$ (with the Weil topology). Define $L_H = (K^{ab})^{\operatorname{Art}_K(H)}$, the fixed field of $\operatorname{Art}_K(H)$ acting on $K^{ab}$. Since $\operatorname{Art}_K(H)$ has finite index in $W(K^{ab}/K)$, and $W(K^{ab}/K)$ is dense in $\operatorname{Gal}(K^{ab}/K)$, the closure $\overline{\operatorname{Art}_K(H)}$ in $\operatorname{Gal}(K^{ab}/K)$ is an open finite-index subgroup of $\operatorname{Gal}(K^{ab}/K)$. By the fundamental theorem of infinite Galois theory, the fixed field of a closed subgroup of $\operatorname{Gal}(K^{ab}/K)$ is a well-defined subextension of $K^{ab}/K$. Since $\operatorname{Art}_K(H)$ and $\overline{\operatorname{Art}_K(H)}$ have the same fixed field (an element fixed by $\operatorname{Art}_K(H)$ is fixed by every limit point, hence by the closure), $L_H = (K^{ab})^{\overline{\operatorname{Art}_K(H)}}$. The index $[\operatorname{Gal}(K^{ab}/K) : \overline{\operatorname{Art}_K(H)}]$ is finite (and equals $[K^\times : H]$), so $L_H / K$ is a finite extension. Since $L_H \subseteq K^{ab}$, the extension $L_H/K$ is abelian. [/step] [step:Verify the bijection: $N(L_H / K) = H$ and $L_{N(L/K)} = L$] **$N(L_H / K) = H$.** The Artin map gives $\operatorname{Art}_K|_{L_H} \colon K^\times \to \operatorname{Gal}(L_H/K)$ with kernel $N(L_H/K)$. An element $x \in K^\times$ lies in $\ker(\operatorname{Art}_K|_{L_H})$ if and only if $\operatorname{Art}_K(x)|_{L_H} = \operatorname{id}$, i.e., $\operatorname{Art}_K(x)$ fixes $L_H$. By definition of $L_H = (K^{ab})^{\operatorname{Art}_K(H)}$, the group $\operatorname{Art}_K(H)$ fixes $L_H$. The group of all elements of $W(K^{ab}/K)$ fixing $L_H$ is $\overline{\operatorname{Art}_K(H)} \cap W(K^{ab}/K) = \operatorname{Art}_K(H)$ (since $\operatorname{Art}_K(H)$ is already closed in the Weil topology, and the Weil topology refines the restriction of the Krull topology to the Weil group). Therefore $\operatorname{Art}_K(x)$ fixes $L_H$ if and only if $\operatorname{Art}_K(x) \in \operatorname{Art}_K(H)$, if and only if $x \in H$ (since $\operatorname{Art}_K$ is injective). Hence $N(L_H/K) = H$. **$L_{N(L/K)} = L$.** Let $L/K$ be a finite abelian extension and set $H = N(L/K)$. By definition, $L_H = (K^{ab})^{\operatorname{Art}_K(N(L/K))}$. The kernel of $\operatorname{Art}_K|_L$ is $N(L/K)$, so $\operatorname{Art}_K(N(L/K)) = \ker(\operatorname{Art}_K|_L \to \operatorname{Gal}(L/K))$ when viewed as a subgroup of $W(K^{ab}/K)$, which equals $W(K^{ab}/K) \cap \operatorname{Gal}(K^{ab}/L) = W(K^{ab}/L)$ (by the [Density of the Weil Group](/theorems/???), Part 2). The fixed field of $W(K^{ab}/L)$ in $K^{ab}$ is $L$ (since $W(K^{ab}/L)$ is dense in $\operatorname{Gal}(K^{ab}/L)$, and the fixed field of a dense subgroup equals the fixed field of its closure). Therefore $L_H = L$. [guided] The two verifications constitute the proof that the maps $L \mapsto N(L/K)$ and $H \mapsto L_H$ are mutually inverse. For $N(L_H/K) = H$: the crucial point is that the Artin map is a *topological* isomorphism (with the Weil topology on the target), so $\operatorname{Art}_K$ maps open subgroups to open subgroups, and the fixed-field/kernel correspondence is exact. For $L_{N(L/K)} = L$: we use the fact that $\operatorname{Art}_K(N(L/K)) = W(K^{ab}/L)$. This is because $N(L/K)$ is exactly the set of $x \in K^\times$ with $\operatorname{Art}_K(x)|_L = \operatorname{id}$, and the image of this set under $\operatorname{Art}_K$ is the Weil group of $K^{ab}/L$. The density of $W(K^{ab}/L)$ in $\operatorname{Gal}(K^{ab}/L)$ ensures the fixed field is exactly $L$. The inclusion-reversing property is immediate: if $M \subseteq L$ then $N_{L/K} = N_{M/K} \circ N_{L/M}$, so $N(L/K) = N_{M/K}(N_{L/M}(L^\times)) \subseteq N_{M/K}(M^\times) = N(M/K)$. Larger extensions have smaller norm groups. [/guided] [/step] [step:Prove the lattice identities $N(LM/K) = N(L/K) \cap N(M/K)$ and $N(L \cap M/K) = N(L/K) \cdot N(M/K)$] Let $L/K$ and $M/K$ be finite abelian extensions. **$N(LM/K) = N(L/K) \cap N(M/K)$.** Under the bijection, $LM$ corresponds to $N(LM/K)$ and is the smallest extension containing both $L$ and $M$. By the inclusion-reversing property, $L \subseteq LM$ gives $N(LM/K) \subseteq N(L/K)$, and $M \subseteq LM$ gives $N(LM/K) \subseteq N(M/K)$. Therefore $N(LM/K) \subseteq N(L/K) \cap N(M/K)$. For the reverse inclusion: $N(L/K) \cap N(M/K)$ is an open finite-index subgroup of $K^\times$ (an intersection of two open finite-index subgroups is open and of finite index). Under the bijection, it corresponds to some finite abelian extension $E/K$ with $N(E/K) = N(L/K) \cap N(M/K)$. Since $N(E/K) \subseteq N(L/K)$, the inclusion-reversing bijection gives $L \subseteq E$. Similarly $M \subseteq E$. Therefore $LM \subseteq E$, so $N(E/K) \subseteq N(LM/K)$. Hence $N(L/K) \cap N(M/K) = N(E/K) \subseteq N(LM/K)$. **$N(L \cap M/K) = N(L/K) \cdot N(M/K)$.** The extension $L \cap M$ is the largest extension contained in both $L$ and $M$. By inclusion-reversal, $L \cap M \subseteq L$ gives $N(L/K) \subseteq N(L \cap M / K)$, and similarly $N(M/K) \subseteq N(L \cap M / K)$. Therefore $N(L/K) \cdot N(M/K) \subseteq N(L \cap M / K)$. For the reverse: $N(L/K) \cdot N(M/K)$ is a subgroup of $K^\times$ containing both $N(L/K)$ and $N(M/K)$. It is open (since $N(L/K)$ is open and $N(L/K) \subseteq N(L/K) \cdot N(M/K)$) and of finite index (since $[K^\times : N(L/K) \cdot N(M/K)] \leq [K^\times : N(L/K)]$). Under the bijection, it corresponds to a finite abelian extension $F/K$ with $N(F/K) = N(L/K) \cdot N(M/K)$. Since $N(L/K) \subseteq N(F/K)$, we get $F \subseteq L$. Similarly $F \subseteq M$. Therefore $F \subseteq L \cap M$, so $N(L \cap M / K) \subseteq N(F/K) = N(L/K) \cdot N(M/K)$. [/step]