Notes on class field theory

For the first assertion, see Proposition 4.2.16 and Proposition 4.2.17. For the second assertion, see Lemma 4.2.20.

One can certainly give an elementary proof of this using the fact that $L^{\ast }$ is cyclic (see Exercise 1). But one can also see it using the machinery we have at hand. Because $L^{\ast }$ is a finite module, its Herbrand quotient is 1. Also, we know $H_T^1(L/K)$ is trivial by Lemma 1.2.3. Thus $H_T^0(L/K)$ is trivial too, that is, ${\mathrm{Norm}}_{L/K}:L^{\ast }\to K^{\ast }$ is surjective.

Say $u\in {\mathfrak{o}}_K^{\ast }$ is a unit. By Proposition 4.2.4, we may pick $v_0\in {\mathfrak{o}}_L^{\ast }$ such that in the residue fields, the norm of $v_0$ coincides with $u\text{.}$ Thus $u/\mathrm{Norm}(v_0)\equiv 1(\mathrm{mod}\pi )\text{,}$ where $\pi$ is a uniformizer of $K\text{.}$ Now we construct units $v_i\equiv 1(\mathrm{mod}{\pi }^i)$ such that $u_i=u/\mathrm{Norm}(v_0\cdots v_i)\equiv 1(\mathrm{mod}{\pi }^{i+1})\text{:}$ simply take $v_i$ so that $\mathrm{Trace}((1-v_i)/{\pi }^i)\equiv (1-u_{i-1})/{\pi }^i(\mathrm{mod}\pi )\text{.}$ (That’s possible because the trace map on residue fields is surjective by the normal basis theorem.) Then the product $v_0{v}_1\cdots$ converges to a unit $v$ with norm $u\text{.}$

Again, $\mathrm{Gal}(L/K)$ is cyclic, so by Theorem 3.4.1 we need only check this for $i=0,1\text{.}$ For $i=0\text{,}$ it is Proposition 4.2.5. For $i=1\text{,}$ note that because $L/K$ is unramified, we can split the surjection $L^{\ast }\to L^{\ast }/{\mathfrak{o}}_L^{\ast }$ by choosing a uniformizer ${\pi }_K$ of $K$ and writing $L^{\ast }={\mathfrak{o}}_L^{\ast }{\pi }_K^{\mathbb{Z}}\text{.}$ Hence $H_T^1(\mathrm{Gal}(L/K),{\mathfrak{o}}_L^{\ast })$ is a direct summand of $H_T^1(\mathrm{Gal}(L/K),L^{\ast })\text{,}$ and the latter vanishes by Lemma 1.2.3.

By the normal basis theorem, there exists $\alpha \in L$ such that $\{{\alpha }^g:g\in \mathrm{Gal}(L/K)\}$ is a basis for $L$ over $K\text{.}$ Without loss of generality, we may rescale to get $\alpha \in {\mathfrak{o}}_L\text{;}$ then put $V:=\sum {\mathfrak{o}}_K{\alpha }^g\text{.}$ As in the proof of Theorem 3.2.8, $V={\mathrm{Ind}}_1^G{\mathfrak{o}}_K$ is induced and hence acyclic.

Take $W$ as in Lemma 4.2.10. Since $W$ has finite index in ${\mathfrak{o}}_L^{\ast }\text{,}$ we have $h({\mathfrak{o}}_L^{\ast }/W)=1$ (Remark 3.4.8) and hence $h({\mathfrak{o}}_L^{\ast })=h(W)=1$ by Lemma 4.2.10. So again we may conclude that $h(L^{\ast })=h({\mathfrak{o}}_L^{\ast })h(\mathbb{Z})=[L:K]\text{,}$ and so $H_T^0(\mathrm{Gal}(L/K),L^{\ast })=[L:K]\text{.}$

In Figure 4.2.14, the second vertical arrow arises from the long exact sequence for $G$-cohomology; since ${\mathrm{Ind}}_1^G{\mathrm{Res}}_1^{G}M$ is an induced $G$-module, this arrow is an isomorphism. Similarly, the third vertical arrow arises from the long exact sequence for $H$-cohomology, and it is an isomorphism because ${\mathrm{Ind}}_1^G{\mathrm{Res}}_1^{G}M$ is also an induced $H$-module; moreover, $H^i(H,N)=0$ for $i=1,\dots ,r-2\text{.}$

This follows from Proposition 4.2.13 using Lemma 1.2.3.

Consider the diagram

in which the bottom row is exact and the vertical arrow is injective, both by Corollary 4.2.15. It suffices to show that the diagonal arrow $H^2(M/K)\to H^2(ML/L)$ is the zero map, as this will imply an inclusion $H^2(M/K)\subseteq H^2(L/K)$ within $H^2(ML/K)$ and we already know that $H^2(M/K)$ is cyclic of order $[M:K]=[L:K]$ by Proposition 4.2.7.

Let $U/K$ be the maximal unramified subextension of $L/K\text{;}$ then $U/K$ is also a subextension of $M/K$ (because the latter is cyclic and there is a unique unramified extension of $K$ of any given degree). We can thus enlarge our previous diagram to

where the maps out of $H^2(U/K)$ are again injective by Corollary 4.2.15; moreover, the cokernel of $H^2(U/K)\stackrel{\mathrm{Inf}}{\to }{H}^2(M/K)$ maps injectively via $\mathrm{Res}$ into $H^2(M/U)\text{.}$ This means that we can replace $K$ by $U$ before checking that the diagonal arrow is injective; in other words, we may safely assume that $L/K$ is totally ramified.

By Corollary 4.2.15 we have an injection $H^2(K^{\mathrm{unr}}/K)\to H^2(\bar{K}/K)\text{,}$ and the former is canonically isomorphic to $\mathbb{Q}/\mathbb{Z}\text{;}$ so we have to prove that this injection is actually also surjective. Remember that $H^2(\bar{K}/K)$ is the direct limit of $H^2(L/K)$ running over all finite extensions $L$ of $K\text{.}$ By Proposition 4.2.17, if $L/K$ is a finite extension and $M$ is the unramified extension of $K$ of degree $[L:K]\text{,}$ then the images of $H^2(L/K)$ and $H^2(M/K)$ in $H^2(ML/K)$ are the same. In particular, $H^2(L/K)$ is in the image of the map $H^2(K^{\mathrm{unr}}/K)\to H^2(\bar{K}/K)\text{.}$ Since that’s true for any $L\text{,}$ we get that the map is indeed surjective, hence an isomorphism.

We compute the map following [37], Proposition III.1.8. Put $e=e_{L/K},f=f_{L/K}\text{.}$ We form a commutative diagram

as follows. The left square comes from the valuation maps. The middle square comes from the connecting homomorphisms for the sequence $0\to \mathbb{Z}\to \mathbb{Q}\to \mathbb{Q}/\mathbb{Z}\to 0$ with the trivial actions; note that these connecting homomorphisms are isomorphisms by Exercise 2. The right square comes from evaluating crossed homomorphisms at Frobenius. Since $ef=n\text{,}$ this yields the claim.