Notes on class field theory

Let us first check that $H_T^i(G,M)=0$ for all $i\ge 0\text{.}$ For $G$ cyclic, this follows by periodicity (Theorem 3.4.1) from the hypothesis that $H^1(G,M)=H^2(G,M)=0\text{.}$ In fact this even yields vanishing for $i<0\text{,}$ but we will not be able to carry that through the following steps.

Since $\mathrm{Cor}\circ \mathrm{Res}=[G:H]$ (Example 3.2.24), $\mathrm{Res}(\gamma )$ generates $H^2(H,{\mathrm{Res}}_H^{G}M)$ for any $H\text{.}$ We start out by explicitly constructing a $G$-module containing $M$ in which $\gamma$ becomes a coboundary.

This amounts to carefully tracing through the dimension shifts we used in Proposition 4.3.1 to construct the cup product. For the details, see [48], XI.3, Proposition 2.

Let $M/K$ be an abelian extension of local fields. Then ${\mathrm{Norm}}_{L/K}$ carries ${\mathrm{Norm}}_{ML/L}(ML{)}^{\ast }$ to ${\mathrm{Norm}}_{ML/K}(ML{)}^{\ast }\subseteq {\mathrm{Norm}}_{M/K}{M}^{\ast }$ and thus induces a map $L^{\ast }/{\mathrm{Norm}}_{ML/L}(ML{)}^{\ast }\to K^{\ast }/{\mathrm{Norm}}_{M/K}{K}^{\ast }\text{;}$ the claim is that this is compatible with the natural map $\mathrm{Gal}(ML/L)\to \mathrm{Gal}(M/K)\text{.}$

For this, it is convenient to replace $L^{\ast }$ with the group $(L{\otimes }_{K}M{)}^{\ast }$ and identify the latter with ${\mathrm{Ind}}_H^G{L}^{\ast }$ for $G=\mathrm{Gal}(M/K),H=\mathrm{Gal}(ML/L)\text{.}$ We can then build a diagram that looks like the one we want by applying ordinary functoriality for the map ${\mathrm{Norm}}_{L/K}:(L{\otimes }_{K}M{)}^{\ast }\to M^{\ast }$ to the construction of Remark 4.3.7, translating back into the terms we want using Shapiro’s lemma (Lemma 3.2.3), and observing the relationship between the local invariant maps for $K$ and $L$ (as in the proof of Proposition 4.2.17).

Let $M$ be the compositum of all cyclic $\ell$-extensions of $K\text{.}$ The group $K^{\ast }/(K^{\ast }{)}^{\ell }$ is finite (Exercise 2), and hence is isomorphic to $(\mathbb{Z}/\ell \mathbb{Z}{)}^n$ for some positive integer $n\text{.}$ By Kummer theory (Theorem 1.2.9), we also have $\mathrm{Gal}(M/K)\cong (\mathbb{Z}/\ell \mathbb{Z}{)}^n\text{.}$ By the local reciprocity isomorphism (in the form obtained in Remark 4.3.7), $K^{\ast }/{\mathrm{Norm}}_{M/K}{M}^{\ast }\cong (\mathbb{Z}/\ell \mathbb{Z}{)}^n\text{;}$ consequently, on one hand $(K^{\ast }{)}^{\ell }\subseteq {\mathrm{Norm}}_{M/K}{M}^{\ast }\text{,}$ and on other hand these subgroups of $K^{\ast }$ have the same index ${\ell }^n\text{.}$ They are thus equal, proving the claim.

Apply Lemma 4.3.13 to the field $K({\zeta }_{\ell })$ obtained by adjoining a primitive $\ell$-th root of unity to $K\text{:}$ this tells us that $x$ becomes an $\ell$-th power in $K({\zeta }_{\ell }{)}^{\ast }\text{.}$ By taking norms back to $K\text{,}$ we deduce that $x^{[K({\zeta }_{\ell }):K]}$ is an $\ell$-th power in $K^{\ast }\text{;}$ since $[K({\zeta }_{\ell }):K]$ divides $\ell -1\text{,}$ it is coprime to $\ell \text{,}$ and so we deduce that $x\in (K^{\ast }{)}^{\ell }\text{.}$

Let $D_K$ be the intersection in question; note that $D_K\subseteq {\mathfrak{o}}_K^{\ast }$ by considering unramified extensions of $K\text{,}$ so $D_K$ is in particular a compact topological group. By Lemma 4.3.15, for every prime $\ell \text{,}$ every element of $D_K$ is the $\ell$-th power of an element of $D_K$ we will show that in fact every element of $D_K$ is the $\ell$-th power of an element of $D_K$ itself. This will show that $D_K$ is a divisible abelian group, and in particular every element is an $n$-th power for every positive integer $n\text{;}$ it will then follow from Exercise 3 that $D_K$ is the trivial group. (Alternatively, one can follow the suggestion of Remark 4.3.14 and prove that the conclusion of Lemma 4.3.13 retains true when $\ell$ is replaced by an arbitrary positive integer $n\text{,}$ and then apply Exercise 3 directly.)

Now let $\ell$ be a prime and choose $x\in D_K\text{.}$ For each finite extension $L$ of $K$ containing a primitive $\ell$-th root of unity, let $E(L)$ be the set of $\ell$-th roots of $x$ in $K$ which belong to ${\mathrm{Norm}}_{L/K}{L}^{\ast }\text{.}$ This set is finite (of cardinality at most $\ell$) and nonempty: we have $x={\mathrm{Norm}}_{L/K}(y)$ for some $y\in D_L$ by the previous paragraph, so $y$ has an $\ell$-th root $z$ in $L$ and ${\mathrm{Norm}}_{L/K}(z)\in E(L)\text{.}$ By the previous paragraph, $E(M)\subseteq E(L)$ whenever $L\subseteq M\text{,}$ so we may again conclude using the finite intersection property. Alternatively, just note that if each of the (finitely many!) elements of $E(K)$ fails to survive to some larger field, we can take a compositum to get a single field $L$ such that no element of $E(K)$ belongs to $E(L)\text{,}$ which is absurd since $E(L)\ne \mathrm{\varnothing }\text{.}$

By the norm limitation theorem (Theorem 4.1.7), it suffices to produce a finite extension $L/K\text{,}$ not necessarily abelian, such that ${\mathrm{Norm}}_{L/K}{L}^{\ast }\subseteq U\text{.}$ This will give us a useful amount of flexibility in what follows.

By Lemma 4.3.17, we can find a finite abelian extension $M$ of $K$ such that ${\mathrm{Norm}}_{M/K}{M}^{\ast }\subseteq U\text{.}$ By the local reciprocity isomorphism (as obtained in Remark 4.3.7), we then have $\mathrm{Gal}(M/K)\cong K^{\ast }/{\mathrm{Norm}}_{M/K}{M}^{\ast }\text{.}$ Now take $L$ to be the fixed field of the subgroup of $\mathrm{Gal}(M/K)$ corresponding to $U/{\mathrm{Norm}}_{M/K}{M}^{\ast }\text{;}$ using Remark 4.3.9, we see that this has the desired effect.