[proofplan]
We reduce to the case where $L/K$ is abelian using the index bound theorem, then prove both directions. The forward direction follows from transitivity of norms. The converse uses the density of the Weil group in the Galois group: if $N(L/K) \subseteq N(M/K)$, then the Artin map sends every element of $W(K^{\mathrm{ab}}/L)$ to an automorphism that acts directly on $M$, forcing $M \subseteq L$ by Galois theory.
[/proofplan]
[step:Reduce to the case where $L/K$ is abelian]
By the [Index Bound for Norm Groups](/theorems/2388), the norm group satisfies $N(L/K) = N((L \cap K^{\mathrm{ab}})/K)$. Replacing $L$ by $L \cap K^{\mathrm{ab}}$ does not change $N(L/K)$, and $M \subseteq L$ if and only if $M \subseteq L \cap K^{\mathrm{ab}}$ (since $M/K$ is abelian, $M \subseteq K^{\mathrm{ab}}$). We may therefore assume $L/K$ is abelian.
[guided]
The containment $N(L/K) \subseteq N(M/K)$ involves only the norm group of $L$, and the [Index Bound for Norm Groups](/theorems/2388) tells us that $N(L/K) = N((L \cap K^{\mathrm{ab}})/K)$. So the norm group depends only on the abelian part of $L$. Similarly, the condition $M \subseteq L$ reduces to $M \subseteq L \cap K^{\mathrm{ab}}$ because $M/K$ is abelian and hence $M \subseteq K^{\mathrm{ab}}$ automatically. After replacing $L$ by $L \cap K^{\mathrm{ab}}$, we may assume $L/K$ is itself abelian. This is a standard reduction in local class field theory: norm group questions about arbitrary extensions can always be reduced to abelian extensions.
[/guided]
[/step]
[step:Prove the forward direction: $M \subseteq L$ implies $N(L/K) \subseteq N(M/K)$]
Assume $M \subseteq L$. By the transitivity of field norms, $N_{L/K} = N_{M/K} \circ N_{L/M}$. Therefore
\begin{align*}
N(L/K) = N_{L/K}(L^\times) = N_{M/K}(N_{L/M}(L^\times)) \subseteq N_{M/K}(M^\times) = N(M/K).
\end{align*}
[/step]
[step:Prove the converse: $N(L/K) \subseteq N(M/K)$ implies $M \subseteq L$]
Assume $N(L/K) \subseteq N(M/K)$. We must show $M \subseteq L$, which by Galois theory is equivalent to $\operatorname{Gal}(K^{\mathrm{ab}}/L) \subseteq \operatorname{Gal}(K^{\mathrm{ab}}/M)$, i.e., every $\sigma \in \operatorname{Gal}(K^{\mathrm{ab}}/L)$ acts directly on $M$.
The Weil group $W(K^{\mathrm{ab}}/L)$ is dense in $\operatorname{Gal}(K^{\mathrm{ab}}/L)$ in the Krull topology. Since the restriction map $\sigma \mapsto \sigma|_M$ is continuous and $\operatorname{Gal}(M/K)$ is finite (hence discrete), it suffices to show $\sigma|_M = \operatorname{id}_M$ for every $\sigma \in W(K^{\mathrm{ab}}/L)$.
By the Artin reciprocity map, $W(K^{\mathrm{ab}}/L) = \operatorname{Art}_K(N(L/K))$. Since $N(L/K) \subseteq N(M/K)$ by hypothesis, every $\sigma \in W(K^{\mathrm{ab}}/L)$ has the form $\sigma = \operatorname{Art}_K(a)$ for some $a \in N(M/K)$. Write $a = N_{M/K}(b)$ for some $b \in M^\times$. The fundamental property of the Artin map gives
\begin{align*}
\operatorname{Art}_K(N_{M/K}(b))\big|_M = \operatorname{id}_M.
\end{align*}
Hence $\sigma|_M = \operatorname{id}_M$, as required.
[guided]
We assume $N(L/K) \subseteq N(M/K)$ and aim to show $M \subseteq L$. The key tool is the Artin reciprocity map and the density of the Weil group.
By Galois theory for $K^{\mathrm{ab}}/K$, the inclusion $M \subseteq L$ holds if and only if $\operatorname{Gal}(K^{\mathrm{ab}}/L) \subseteq \operatorname{Gal}(K^{\mathrm{ab}}/M)$, i.e., every automorphism fixing $L$ also fixes $M$. We need to check this for every $\sigma \in \operatorname{Gal}(K^{\mathrm{ab}}/L)$.
Why is the Weil group sufficient? The Weil group $W(K^{\mathrm{ab}}/L)$ is dense in $\operatorname{Gal}(K^{\mathrm{ab}}/L)$ in the Krull topology. The restriction map $\operatorname{res}: \operatorname{Gal}(K^{\mathrm{ab}}/K) \to \operatorname{Gal}(M/K)$ is continuous, and $\operatorname{Gal}(M/K)$ is a finite group equipped with the discrete topology. A continuous map from a topological space to a discrete space is constant on dense subsets: if $\operatorname{res}(\sigma) = \operatorname{id}_M$ for all $\sigma$ in the dense subset $W(K^{\mathrm{ab}}/L)$, then $\operatorname{res}(\sigma) = \operatorname{id}_M$ for all $\sigma \in \operatorname{Gal}(K^{\mathrm{ab}}/L)$.
Now take any $\sigma \in W(K^{\mathrm{ab}}/L)$. The Artin map provides a bijection $\operatorname{Art}_K: K^\times / N(L/K) \xrightarrow{\sim} \operatorname{Gal}(L/K)$, and under this correspondence $W(K^{\mathrm{ab}}/L) = \operatorname{Art}_K(N(L/K))$. So $\sigma = \operatorname{Art}_K(a)$ for some $a \in N(L/K)$.
By hypothesis, $a \in N(L/K) \subseteq N(M/K)$, so $a = N_{M/K}(b)$ for some $b \in M^\times$. The fundamental property of the Artin map states that $\operatorname{Art}_K(N_{M/K}(b))$ acts directly on $M$:
\begin{align*}
\operatorname{Art}_K(N_{M/K}(b))\big|_M = \operatorname{id}_M.
\end{align*}
This is the content of the norm-fixing property: elements in the image of the norm map from $M$ correspond to automorphisms that fix $M$. Therefore $\sigma|_M = \operatorname{id}_M$, and since this holds for every $\sigma$ in the dense subset $W(K^{\mathrm{ab}}/L)$, it holds for all of $\operatorname{Gal}(K^{\mathrm{ab}}/L)$. Hence $M \subseteq L$.
[/guided]
[/step]
[step:Combine both directions to conclude the equivalence]
We have shown:
- If $M \subseteq L$, then $N(L/K) \subseteq N(M/K)$ (by transitivity of norms).
- If $N(L/K) \subseteq N(M/K)$, then $M \subseteq L$ (by the Artin map and Weil group density).
This establishes the biconditional: $N(L/K) \subseteq N(M/K)$ if and only if $M \subseteq L$.
[/step]
