Notes on class field theory

Let \(\gamma\) be a generator of \(H^2(G, M)\text{.}\) Since \(\Cor \circ \Res = [G:H]\) (Example 3.2.22), \(\Res(\gamma)\) generates \(H^2(H,M)\) for any \(H\text{.}\) We start out by explicitly constructing a \(G\)-module containing \(M\) in which \(\gamma\) becomes a coboundary.

Choose a 2-cocycle \(\phi: G^3 \to M\) representing \(\gamma\text{;}\) by the definition of a cocycle,

Moreover, \(\phi\) is a coboundary if and only if it's of the form \(d(\rho)\text{,}\) that is, \(\phi(g_0, g_1, g_2) = \rho(g_1, g_2) - \rho(g_0, g_2) + \rho(g_0, g_1)\text{.}\) This \(\rho\) must itself be \(G\)-invariant: \(\rho(g_0, g_1)^g = \rho(g_0g, g_1g)\text{.}\) Thus \(\phi\) is a coboundary if \(\phi(e, g, hg) = \rho(e,h)^g - \rho(e,hg) + \rho(e,g)\text{.}\)

Let \(M[\phi]\) be the direct sum of \(M\) with the free abelian group with one generator \(x_g\) for each element \(g\) of \(G - \{e\}\text{,}\) with the \(G\)-action

(The symbol \(x_e\) should be interpreted as the element \(\phi(e,e,e)\) of \(M\text{.}\)) Using the cocycle property of \(\phi\text{,}\) one may verify that this is indeed a \(G\)-action; by construction, the cocycle \(\phi\) becomes zero in \(H^2(G, M[\phi])\) by setting \(\rho(e,g) = x_g\text{.}\) (Milne calls \(M[\phi]\) the splitting module of \(\phi\text{.}\)) Moreover, by the same token, for any \(H\text{,}\) the restriction of \(\phi\) to \(H\) also becomes zero in \(H^2(H, M).\)

The map \(\alpha: M[\phi] \to \ZZ[G]\) sending \(M\) to zero and \(x_g\) to \([g]-1\) is a homomorphism of \(G\)-modules. Actually it maps into the augmentation ideal \(I_G\text{,}\) and the sequence

is exact. (Note that this is backwards from the usual exact sequence featuring \(I_G\) as a submodule, which will appear again momentarily.) For any subgroup \(H\) of \(G\text{,}\) we can restrict to \(H\)-modules, then take the long exact sequence:

To make headway with this, view \(0 \to I_G \to \ZZ[G] \to \ZZ \to 0\) as an exact sequence of \(H\)-modules. Since \(\ZZ[G]\) is induced, its Tate groups all vanish. So we get a dimension shift:

Similarly, \(H^2(H, I_G) \cong H^1(H, \ZZ) = 0\text{.}\) Also, the map \(H^2(H, M) \to H^2(H, M[\phi])\) is zero because we constructed this map so as to kill off the generator \(\phi\text{.}\) Thus \(H^2(H, M[\phi]) = 0\) and \(H^1(H, I_G) \to H^2(H,M)\) is surjective. But these groups are both finite of the same order! So the map is also injective, and \(H^1(H, M[\phi])\) is also zero.

Now apply Lemma 4.3.2 below to conclude that \(H^i_T(G, M[\phi]) = 0\) for all \(i\text{.}\) This allows us to use the four-term exact sequence

(as in the proof of Theorem 3.4.1) to conclude that \(H^i_T(G, \ZZ) \cong H^{i+2}_T(G, M)\text{.}\)

Let us first check that \(H^i_T(G,M) = 0\) for all \(i \geq 0\text{.}\) For \(G\) cyclic, this follows by periodicity. For \(G\) solvable, we prove the general result by induction on \(\#G\text{.}\) Since \(G\) is solvable, it has a proper subgroup \(H\) for which \(G/H\) is cyclic. By the induction hypothesis, \(H^i_T(H,M) = 0\) for all \(i\text{.}\) Thus by the inflation-restriction exact sequence (Proposition 4.2.14),

is exact for all \(i>0\text{.}\) The term on the end being zero, we have \(H^i(G/H, M^H) \cong H^i(G,M) = 0\) for \(i=1, 2\text{.}\) By periodicity (Theorem 3.4.1), \(H^i_T(G/H, M^H) = 0\) for all \(i\text{,}\) so \(H^i(G/H, M^H) = 0\) for all \(i>0\text{,}\) and \(H^i(G,M) = 0\) for \(i>0\text{.}\) As for \(i=0\text{,}\) note that \(H^0_T(G/H, M^H) = H^2(G/H, M^H) = 0\text{,}\) so for any \(x \in M^G\) there exists \(y \in M^H\) such that \(\Norm_{G/H}(y) = x\text{.}\) Since \(H^0_T(H,M) = 0\text{,}\) there exists \(z \in M\) such that \(\Norm_{H}(z) = x\text{.}\) Now \(\Norm_G(z) = \Norm_{G/H} \circ \Norm_H(z) = x\text{.}\) Thus \(H^0_T(G,M) = 0\text{,}\) as desired.

We next extend the previous argument from \(G\) solvable to \(G\) general (this can be skipped if you only want the final result for solvable \(G\)). For \(i>0\text{,}\) we already know that the group \(H^i(G,M)\) is torsion (Example 3.2.22), so it suffices to show that its \(p\)-primary component vanishes for any prime \(p\text{.}\) To check this, let \(G_p\) be any Sylow \(p\)-subgroup of \(G\text{.}\) As per Example 3.2.22 again, the composition of \(\Res: H^i(G, M) \to H^i(H, M)\) with \(\Cor: H^i(H, M) \to H^i(G, M)\) is multiplication by \([G:G_p]\text{,}\) which is prime to \(p\text{.}\) Consequently, \(\Res\) induces an injective map on \(p\)-primary components. Since \(G_p\) is solvable, we already know that \(H^i(G_p, M) = 0\text{,}\) yielding the desired vanishing. For \(i=0\text{,}\) we argue as in Remark 3.3.13: we know that \(H^0_T(G_p, M) = 0\text{,}\) so the map \(\Norm_{G_p}: M \to M^{G_p}\) is surjective. In particular, for any \(x \in M^G\text{,}\) we can find \(y \in M\) such that \(x = \sum_{g \in G_p} y^g\text{.}\) Then \(\Norm_G(y) = [G:G_p]x\text{,}\) so the group \(H^0_T(G, M)\) is torsion and killed by \([G:G_p]\text{;}\) again varying over \(p\) shows that \(H^0_T(G, M) = 0\text{.}\)

Finally, we treat the case \(i < 0\) by dimension shifting. Make the exact sequence

in which \(m \otimes [g]\) maps to \(m^g\text{.}\) The term in the middle is acyclic, so \(H^{i+1}_T(H', N) \cong H^{i}_T(H', M)\) for any subgroup \(H'\) of \(G\text{.}\) In particular, \(H^1(H', N) = H^2(H', N) = 0\text{,}\) so the above argument gives \(H^i_T(G, N) = 0\) for \(i\geq 0\text{.}\) Now from \(H^0_T(G, N) = 0\) we get \(H^{-1}_T(G, M) = 0\text{;}\) since the same argument applies to \(N\text{,}\) we also get \(H^{-2}_T(G, M) = 0\) and so on.

Let \(M\) be the compositum of all cyclic \(\ell\)-extensions of \(K\text{.}\) The group \(K^*/(K^*)^{\ell}\) is finite (Exercise 1), and hence is isomorphic to \((\ZZ/\ell \ZZ)^n\) for some positive integer \(n\text{.}\) By Kummer theory (Theorem 1.2.9), we also have \(\Gal(M/K) \cong (\ZZ/\ell \ZZ)^n\text{.}\) By the local reciprocity law (Theorem 4.1.2), \(K^*/\Norm_{M/K}M^* \cong (\ZZ/\ell \ZZ)^n\text{;}\) consequently, on one hand \((K^*)^{\ell} \subseteq \Norm_{M/K}M^*\text{,}\) and on other hand these subgroups of \(K^*\) have the same index \(\ell^n\text{.}\) They are thus equal, proving the claim.

Let \(D_K\) be the intersection in question; note that \(D_K \subseteq \gotho_K^*\) by considering unramified extensions of \(K\text{,}\) so \(D_K\) is in particular a compact topological group. By Lemma 4.3.8, every element of \(D_K\) is an \(\ell\)-th power in \(K\) for every prime \(\ell\text{.}\) 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 \(n\text{.}\) This will then imply using Exercise 2 that \(D_K\) is the trivial group. (Alternatively, one can follow the suggestion of Remark 4.3.9 and prove that the conclusion of Lemma 4.3.8 retains true when \(\ell\) is replaced by an arbitrary positive integer \(n\text{,}\) and then apply Exercise 2 directly.)

We first need to verify something which might seem obvious but isn't quite: for \(L/K\) a finite extension,

This isn't obvious because for \(x \in D_K\text{,}\) for each individual finite extension \(M\) of \(K\) we can write \(x = \Norm_{M/K}(z)\) for some \(z \in M^*\text{,}\) but it is not apparent that we can force the elements \(y = \Norm_{M/L}(z)\) to all be equal. It is nonetheless true because, for any given \(M\) the set of such \(y\) is a nonempty compact subset of \(U_L\text{,}\) and any finite intersection of these subsets is nonempty (because we can pass to a large enough field to contain all of the \(M\) in question and bring an element from there); so the whole intersection is nonempty.

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 \(\Norm_{L/K} L^*\text{.}\) This set is finite (of cardinality at most \(\ell\)) and nonempty: we have \(x = \Norm_{L/K}(y)\) for some \(y \in D_L\) by the previous paragraph, so \(y\) has an \(\ell\)-th root \(z\) in \(L\) and \(\Norm_{L/K}(z) \in E(L)\text{.}\) 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 \(E(L) \neq \emptyset\text{.}\)

We note first that by the local reciprocity law (Theorem 4.1.2), it is enough to construct \(L\) so that \(U\) contains \(\Norm_{L/K}L^*\text{:}\) in this case, we will have \(\Gal(L/K) \cong K^*/\Norm_{L/K}L^*\text{,}\) and then \(U/\Norm_{L/K}L^*\) will corresponding to \(\Gal(L/M)\) for some intermediate extension \(M/K\) having the desired effect. We note next that by the norm limitation theorem (Theorem 4.1.7), it suffices to produce any finite extension \(L/K\text{,}\) not necessarily abelian, such that \(U\) contains \(\Norm_{L/K}L^*\text{.}\)

Let \(m\ZZ \subseteq \ZZ\) be the image of \(U\) in \(K^*/\gotho_K^* \cong \ZZ\text{;}\) by choosing \(L\) to contain the unramified extension of \(K\) of degree \(m\text{,}\) we may ensure that the image of \(\Norm_{L/K} L^*\) in \(K^*/\gotho_K^*\) is also contained in \(m\ZZ\text{.}\) It thus remains to further ensure that

Since \(\gotho_K^*\) is compact, each open subgroup \((\Norm_{L/K}L^*) \cap \gotho_K^*\) is also closed and hence compact. By Corollary 4.3.10, as \(L/K\) runs over all finite extensions of \(K\text{,}\) the intersection of the groups \((\Norm_{L/K} L^*) \cap \gotho_K^*\) is trivial; in particular, the intersection of the compact subsets

of \(\gotho_K^*\) is empty. By the finite intersection property (and taking a compositum), there exists a single \(L/K\) for which \((\Norm_{L/K} L^*) \cap \gotho_K^* \subseteq U \cap \gotho_K^*\text{,}\) as desired.