Notes on class field theory

Let \(f\) be a function of the form described above. By the linear independence of automorphisms (see Exercise 1), there exists \(x \in L\) such that \(t = \sum_{g \in G} x^g f(g)\) is nonzero. But now

Thus \(f\) is zero in \(H^1(G,L^*)\text{.}\)

On one hand, suppose that \(\alpha \in K^*\) is such that \(\alpha^{1/d} \notin K\) for any proper divisor \(d\) of \(n\text{.}\) Then every automorphism of \(K(\alpha^{1/n})\) over \(K\) must have the form \(\alpha \mapsto \alpha \zeta_n^r\) for some \(r \in \ZZ/n\ZZ\text{;}\) this defines a homomorphism \(\Gal(K(\alpha^{1/n})/K) \cong \ZZ/n\ZZ\) which is injective because any automorphism of \(K(\alpha^{1/n})\) over \(K\) is uniquely determined by its effect on \(\alpha^{1/n}\text{.}\) We claim that this map is also surjective. If \(n\) is prime, we can see this from the fact that by hypothesis \(K(\alpha^{1/p}) \neq K\text{,}\) so the map \(\Gal(K(\alpha^{1/n})/K) \cong \ZZ/n\ZZ\) is an injective map from a nonzero group into a prime cyclic group and hence must be surjective. In the general case, note that the definition of the map \(\Gal(K(\alpha^{1/n})/K) \cong \ZZ/n\ZZ\) is compatible with replacing \(n\) with one of its prime factors \(p\text{,}\) and this logic tells us that the image of \(\Gal(K(\alpha^{1/n})/K)\) in \(\ZZ/n\ZZ\) cannot be contained in \(p\ZZ/n\ZZ\) for any prime divisor \(p\) of \(n\text{.}\) So again we conclude that \(\Gal(K(\alpha^{1/n})/K) \cong \ZZ/n\ZZ\text{.}\) (As a corollary, we deduce that the polynomial \(x^n - \alpha\) is irreducible over \(K\text{;}\) see Exercise 4 and Exercise 5 for discussion of what happens when \(\zeta_n \notin K\text{.}\))

On the other hand, let \(L\) be an arbitrary \(\ZZ/n\ZZ\)-extension of \(K\text{.}\) Choose a generator \(g \in \Gal(L/K)\text{,}\) and let \(f: \Gal(L/K) \to L^*\) be the map that sends \(rg\) to \(\zeta_n^r\) for \(r \in \ZZ\text{.}\) Then \(f \in H^1(\Gal(L/K), L^*)\text{,}\) so there exists \(t \in L\) such that \(t^{rg}/t = f(rg) = \zeta_n^r\) for \(r \in \ZZ\text{.}\) In particular, \(t^n\) is invariant under \(\Gal(L/K)\text{,}\) so \(t^n = \alpha\) for some \(\alpha \in K\) and \(L = K(t) = K(\alpha^{1/n})\text{,}\) as desired.

The map comes from the pairing; we have to check that it is injective and surjective. If \(y \in K^* \setminus (K^*)^n\text{,}\) then \(K(y^{1/n})\) is a nontrivial Galois extension of \(K\text{,}\) so there exists some element of \(\Gal(K(y^{1/n})/K)\) that doesn't preserve \(y^{1/n}\text{.}\) Any lift of that element to \(\Gal(\overline{K}/K)\) pairs with \(y\) to give something other than 1; that is, \(y\) induces a nonzero homomorphism of \(\Gal(\overline{K}/K)\) to \(\ZZ/n\ZZ\text{.}\) Thus injectivity follows.

On the other hand, suppose \(f: \Gal(\overline{K}/K) \to \ZZ/n\ZZ\) is a homomorphism whose image is the cyclic subgroup of \(\ZZ/n\ZZ\) of order \(d\text{.}\) Let \(H\) be the kernel of \(f\text{;}\) then the fixed field \(L\) of \(H\) is a \(\ZZ/d\ZZ\)-extension of \(K\) with Galois group \(\Gal(\overline{K}/K)/H\text{.}\) By Kummer theory, \(L = K(y^{1/d})\) for some \(y\text{.}\) But now the homomorphisms induced by \(y^{mn/d}\text{,}\) as \(m\) runs over all integers coprime to \(d\text{,}\) give all possible surjective homomorphisms of \(\Gal(\overline{K}/K)/H\) to \(\ZZ/d\ZZ\text{,}\) so one of them must equal \(f\text{.}\) Thus surjectivity follows.

If \(a^g/a^{\omega(g)} \in (L^*)^n\) for all \(g \in \Gal(L/K)\text{,}\) then \(a\text{,}\) \(a^{\omega(g)}\) and \(a^g\) all generate the same subgroup of \((L^*)/(L^*)^n\text{.}\) Thus \(L(a^{1/n}) = L((a^g)^{1/n})\) for all \(g \in \Gal(L/K)\text{,}\) so \(M/K\) is Galois. Thus it suffices to assume \(M/K\) is Galois, then prove that \(M/K\) is abelian if and only if (1.2.1) holds. In this case, we must have \(a^g/a^{\rho(g)} \in (M^*)^n\) for some map \(\rho: \Gal(L/K) \to (\ZZ/n\ZZ)^*\text{,}\) whose codomain is cyclic by our assumption on \(n\text{.}\)

Note that \(\Gal(M/K)\) admits a homomorphism \(\omega\) to a cyclic group whose kernel \(\Gal(M/L) \subseteq \ZZ/n\ZZ\) is also abelian. Thus \(\Gal(M/K)\) is abelian if and only if \(g\) and \(h\) commute for any \(g \in \Gal(M/K)\) and \(h \in \Gal(M/L)\text{,}\) i.e., if \(h = g^{-1}hg\text{.}\) (Since \(g\) commutes with powers of itself, \(g\) then commutes with everything.)

Let \(A \subseteq L^*/(L^*)^n\) be the subgroup generated by \(a\text{.}\) Then the Kummer pairing gives rise to a pairing

which is bilinear and nondegenerate, so \(h = g^{-1}hg\) if and only if \(\langle h, s^g \rangle = \langle ghg^{-1}, s^g \rangle\) for all \(s \in A\text{.}\) But the Kummer pairing is equivariant with respect to \(\Gal(L/K)\) as follows:

because

(Here by \(s^{1/n}\) I mean an arbitrary \(n\)-th root of \(s\) in \(M\text{,}\) and by \((s^g)^{1/n}\) I mean \((s^{1/n})^g\text{.}\) Remember that the value of the Kummer pairing doesn't depend on which \(n\)-th root you choose.) Thus \(h = ghg^{-1}\) if and only if \(\langle h,s^g \rangle = \langle h,s \rangle^g\) for all \(s \in A\text{,}\) or equivalently, just for \(s=a\text{.}\) But

Thus \(g\) and \(h\) commute if and only if \(\langle h, a^g \rangle = \langle h, a^{\omega(g)}\rangle\text{,}\) if and only if (by nondegeneracy) \(a^g/a^{\omega(g)} \in (L^*)^n\text{,}\) as desired.