Notes on class field theory

Let $g\in \mathrm{Gal}(L/K)$ be the Frobenius and choose $h\in \mathrm{Gal}(L^{\mathrm{unr}}/K)$ lifting $g\text{.}$ Then the fixed field of $h$ is $K$ itself, and from the definition of $r^{\prime }\text{,}$ $r(g)=r^{\prime }(h)$ is a uniformizer of $K\text{.}$ By Proposition 5.1.14, $r(g)$ generates $H^0(\mathrm{Gal}(L/K),A_L)\text{.}$

By the class field axiom, $r_{L/K}$ maps between two groups of the same order, and the previous paragraph implies that the map is surjective. It is thus an isomorphism.

Put $n=[L:K]\text{.}$ The extension $L^{\mathrm{unr}}/K$ is the compositum of two linearly disjoint extensions $L/K$ and $K^{\mathrm{unr}}/K\text{,}$ so its Galois group is canonically a product $\mathrm{Gal}(L/K)\times \mathrm{Gal}(K^{\mathrm{unr}}/K)\text{.}$ Let $g$ be a generator of the first factor (which we can also identify with $\mathrm{Gal}(L^{\mathrm{unr}}/K^{\mathrm{unr}})$) and let $\varphi$ be a generator of the second factor with $d_K(\varphi )=1\text{.}$ Put $\tau =g\varphi$ and let $M$ be the fixed field of $\tau \text{.}$ Let $N$ be the compositum of $L$ and $M$ and put $N_0=N\cap K^{\mathrm{unr}}\text{.}$ We now have the field diagram Figure 5.3.3 in which each line denotes a $\mathbb{Z}/n\mathbb{Z}$-extension, the dashed lines represent unramified extensions, and each label indicates one or more generators of the Galois group $g\text{.}$

Pick uniformizers ${\pi }_L$ and ${\pi }_M$ of $L$ and $M\text{,}$ respectively. Since $d_K(\tau )=1\text{,}$ by Proposition 5.2.3 we have $r_{L/K}(g)={\mathrm{Norm}}_{L^{\mathrm{unr}}/K^{\mathrm{unr}}}({\pi }_M)\text{.}$

By the class field axiom, $r_{L/K}$ maps between two groups of the same order $n\text{,}$ and the previous paragraph implies that the map has image of size at least $n\text{.}$ It is thus an isomorphism.

If $L/K$ is cyclic of prime order, We induct on $[L:K]\text{.}$ then either it is unramified or totally ramified, and we already know $r_{L/K}$ is an isomorphism in those cases (by Lemma 5.3.1 or Lemma 5.3.2, respectively). Otherwise, let $M$ be a subextension of $L/K\text{.}$ Then diagram chasing through Figure 5.3.5 gives that $r_{L/K}$ is surjective. If $L/K$ is cyclic, then the class field axiom implies that $r_{L/K}$ is a map between two groups of the same order, and hence must be an isomorphism. Otherwise, we see from Figure 5.3.5 again that the kernel of $r_{L/K}$ lies in the kernel of $\mathrm{Gal}(L/K)\to \mathrm{Gal}(N/K)$ for every cyclic subextension $N$ of $L/K\text{.}$ Since $L/K$ is abelian and not cyclic, the intersection of these kernels is trivial. Thus $r_{L/K}$ is also injective, so is an isomorphism.

Let $M$ be the maximal abelian subextension of $L/K\text{.}$ We have the following commutative diagram:

in which the left vertical arrow and bottom horizontal arrows are isomorphisms (the latter by Lemma 5.3.6). Thus the composite $\mathrm{Gal}(L/K{)}^{\mathrm{ab}}\to A_K/{\mathrm{Norm}}_{M/K}{A}_M$ is an isomorphism, so $r_{L/K}$ must be injective.

The map in question is injective by Lemma 5.3.7, so it only remains to check that $r_{L/K}$ itself is surjective. If $L/K$ is solvable, we may deduce surjectivity from Lemma 5.3.6 by induction on $[L:K]$ again by a diagram chase on Figure 5.3.5.

Here some caution is required because $M/K$ need not be Galois, so we cannot draw the full diagram Figure 5.3.5. However, the left square in that diagram still makes sense and commutes. Meanwhile, we may apply the previous paragraph to see that the left vertical arrow $r_{L/M}$ is an isomorphism. Now note that the composition $A_K\subseteq A_M\stackrel{{\mathrm{Norm}}_{M/K}}{\to }{A}_K$ is multiplication by $[M:K]\text{,}$ which is coprime to $p\text{;}$ it follows that the bottom horizontal arrow induces a surjection of Sylow $p$-subgroups. (One can also apply Proposition 5.2.10 here.)

The only issue is the inclusion ${\mathrm{Norm}}_{M/K}{A}_M\subseteq {\mathrm{Norm}}_{L/K}{A}_L\text{,}$ which we are free to check after enlarging $L$ (as long as we do not change $M$). We may thus assume that $L/K$ is Galois.

By Proposition 5.2.7 and Theorem 5.3.9, we have a commutative diagram

in which the horizontal arrows are isomorphisms. This implies the claim.

The compositum $L=L_1{L}_2$ is also a finite abelian extension of $K\text{.}$ By Proposition 5.2.7, $\mathrm{Gal}(L_1/K)\cong A_K/{\mathrm{Norm}}_{L_1/K}{A}_{L_1}$ and $\mathrm{Gal}(L_2/K)\cong {\mathrm{Norm}}_{L_2/K}{A}_{L_2}$ must be the same quotient of $\mathrm{Gal}(L/K)\cong A_K/{\mathrm{Norm}}_{L/K}{A}_L\text{,}$ which forces $L_1=L_2\text{.}$