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Notes on class field theory

Kiran S. Kedlaya

Section 5.3 The theorems of abstract class field theory

We now establish that the reciprocity map is an isomorphism (Theorem 5.3.9). We also obtain an analogue of the norm limitation theorem (Corollary 5.3.11) and some tools which will help with the existence theorem (Remark 5.3.14).

Subsection Proof of the reciprocity law

Our goal is to prove that the homomorphism \(r_{L/K}\) from Definition 5.2.6 induces an isomorphism \(\Gal(L/K)^{\ab} \to A_K / \Norm_{L/K} A_L\text{.}\) Any resemblance with the proof of the local reciprocity law is not at all coincidental!

Proof.

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

Proof.

Put \(n = [L:K]\text{.}\) The extension \(L^{\unr}/K\) is the compositum of two linearly disjoint extensions \(L/K\) and \(K^{\unr}/K\text{,}\) so its Galois group is canonically a product \(\Gal(L/K) \times \Gal(K^{\unr}/K)\text{.}\) Let \(g\) be a generator of the first factor (which we can also identify with \(\Gal(L^{\unr}/K^{\unr})\)) and let \(\phi\) be a generator of the second factor with \(d_K(\phi) = 1\text{.}\) Put \(\tau = g\phi\) and let \(M\) be the fixed field of \(\tau\text{.}\) Let \(N\) be the compositum of \(L\) and \(M\) and put \(N_0 = N \cap K^{\unr}\text{.}\) We now have the field diagram Figure 5.3.3 in which each line denotes a \(\ZZ/n\ZZ\)-extension, the dashed lines represent unramified extensions, and each label indicates one or more generators of the Galois group \(g\text{.}\)
Figure 5.3.3.
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 \(r_{L/K}(g) = \Norm_{L^{\unr}/K^{\unr}}(\pi_M)\text{.}\)
Let \(j\) be the order of \(r_{L/K}(g)\) in \(A_K/\Norm_{L/K} A_L\) and put \(u = \pi_M^j/\pi_L^j \in U_N\text{.}\) Since both \(\Norm_{L^{\unr}/K^{\unr}}(\pi_M^j)\) and \(\Norm_{L^{\unr}/K^{\unr}}(\pi_L^j) = \Norm_{L/K}(\pi_L^j)\) belong to \(\Norm_{L/K} A_L\text{,}\) we can choose \(v \in A_L\) such that
\begin{equation*} \Norm_{L^{\unr}/K^{\unr}}(u) = \Norm_{L/K}(v) = \Norm_{L^{\unr}/K^{\unr}}(v). \end{equation*}
Since \(\Norm_{L/K}(v) \in A_K \cap U_N = U_K\text{,}\) we must have \(v \in U_L\text{.}\)
Applying the class field axiom to \(N/N_0\) yields
\begin{equation*} 0 = H^{-1}_T(\Gal(N/N_0), A_N) = \frac{\ker(\Norm_{N/N_0}: A_N \to A_{N_0})}{ \{ a^{g-1}: a \in A_N \}}. \end{equation*}
Since \(\Norm_{N/N_0}(u/v) = \Norm_{L^{\unr}/K^{\unr}}(u/v) = 1\text{,}\) we can write \(u/v\) in the form \(a^{g-1} = a^g/a\) for some \(a \in A_N\text{.}\) Then
\begin{equation*} (\pi_L^j v)^{g-1} = (\pi_L^j v)^{\tau-1} = (\pi_M^j v/u)^{\tau - 1} = (v/u)^{\tau-1} = (a/a^\tau)^{g-1}. \end{equation*}
If we put \(x = (\pi_L^j v)(a^\tau/a)\text{,}\) then \(x^g = x\) and so \(x \in A_{N_0}\text{.}\) Hence
\begin{equation*} j = v_N(x) = nv_{N_0}(x) \in n \widehat{\ZZ}. \end{equation*}
That is, the order of \(r_{L/K}(g)\) in \(A_K/\Norm_{L/K} A_L\) is divisible by \(n\) in \(\widehat{\ZZ}\text{,}\) and hence also in \(\ZZ\text{.}\)
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.
Before continuing, we record a key commutative diagram which will be the scene of a lot of diagram-chasing.

Remark 5.3.4.

For \(L/K\) a Galois extension of finite subextensions of \(\kbar/k\) and \(M/K\) a Galois subextension, the diagram
Figure 5.3.5.
commutes (thanks to Proposition 5.2.7) and the rows are exact.

Proof.

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 \(\Gal(L/K) \to \Gal(N/K)\) 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.

Proof.

Let \(M\) be the maximal abelian subextension of \(L/K\text{.}\) We have the following commutative diagram:
Figure 5.3.8.
in which the left vertical arrow and bottom horizontal arrows are isomorphisms (the latter by Lemma 5.3.6). Thus the composite \(\Gal(L/K)^{\ab} \to A_K / \Norm_{M/K} A_M\) is an isomorphism, so \(r_{L/K}\) must be injective.

Proof.

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.
For general \(L/K\text{,}\) we instead check that \(r_{L/K}\) becomes a surjection upon restriction to \(p\)-Sylow subgroups for each prime \(p\text{.}\) That is, for \(M\) the fixed field of a Sylow \(p\)-subgroup of \(\Gal(L/K)\) and \(S_p\) the Sylow \(p\)-subgroup of \(A_K/\Norm_{L/K}A_L\text{,}\) the composition
\begin{equation*} \Gal(L/M) \to \Gal(L/K) \stackrel{r_{L/K}}{\to} A_K/\Norm_{L/K} A_L \to S_p \end{equation*}
is surjective. (Compare the proof of Lemma 4.3.2.)
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 \(A_K \subseteq A_M \stackrel{\Norm_{M/K}}{\to} A_K\) 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.)

Remark 5.3.10.

Alternatively, one can derive Theorem 5.3.9 by an argument closer to what we did in local class field theory. In this approach, one first simulates the proofs of Proposition 4.2.17 and Proposition 4.2.18 to show that \(H^2(\Gal(L/K), A_L)\) is cyclic of order \([L:K]\text{;}\) the latter argument ends up being quite similar to the proof of Lemma 5.3.2, with the role of Theorem 90 (Lemma 1.2.3) being played by the \(H^{-1}_T\) aspect of the class field axiom. One must then check that the reciprocity map agrees with the map given by Tate’s theorem Theorem 4.3.1; we leave the details to the interested reader, but see Section 7.5 for a similar argument in the setting of global class field theory.
This directly implies a version of the norm limitation theorem.

Proof.

The only issue is the inclusion \(\Norm_{M/K} A_M \subseteq \Norm_{L/K} A_L\text{,}\) 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
Figure 5.3.12.
in which the horizontal arrows are isomorphisms. This implies the claim.
By similar logic, we also obtain a uniqueness result.

Proof.

The compositum \(L = L_1 L_2\) is also a finite abelian extension of \(K\text{.}\) By Proposition 5.2.7, \(\Gal(L_1/K) \cong A_K/\Norm_{L_1/K} A_{L_1}\) and \(\Gal(L_2/K) \cong \Norm_{L_2/K} A_{L_2}\) must be the same quotient of \(\Gal(L/K) \cong A_K/\Norm_{L/K} A_L\text{,}\) which forces \(L_1=L_2\text{.}\)

Remark 5.3.14.

In a similar vein, note that every subgroup of \(A_K\) containing a subgroup of the form \(\Norm_{M/K} A_M\) for some finite extension \(M/K\) must itself occur as \(\Norm_{L/K} A_L\) for some finite (and even abelian) extension \(L/K\text{.}\) Consequently, proving an analogue of the existence theorem in this setting amounts to computing the intersection of the groups \(\Norm_{M/K} A_M\text{.}\)
Following [37], one can view the groups \(\Norm_{M/K} A_M\) as the open subgroups for a certain topology on \(A_K\text{,}\) called the norm topology. One can then assert that \(\Gal(K^{\ab}/K)\) is isomorphic to the profinite completion of \(A_K\text{,}\) or equivalently its maximal Hausdorff quotient, for the family of quotients by open subgroups in the norm topology.
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