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Notes on class field theory

Kiran S. Kedlaya

Section 1.2 Kummer theory

Reference.

[46] Chapter X; [37] section IV.3; or just about any advanced algebra text (e.g., [32]). The last lemma is from [56], Chapter 14.
Before attempting to classify all abelian extensions of \(\QQ_p\text{,}\) we recall an older classification result. This result will continue to be useful as we proceed to class field theory in general, and the technique in its proof prefigures the role to be played by group cohomology down the line. So watch carefully!

Remark 1.2.1.

A historical note (due to Franz Lemmermeyer): while the idea of studying field extensions generated by radicals was used extensively by Kummer in his work on Fermat’s Last Theorem, the name Kummer theory for the body of results described here was first applied somewhat later by Hilbert in his Zahlbericht [21], a summary of algebraic number theory as of the end of the 19th century.

Subsection Theorem 90

We start with a fundamental result from field theory, which will crop up time and again in our work. To state it, let me introduce some terminology which marks the tip of the iceberg of group cohomology, which we will treat in a more comprehensive way Chapter 3.

Definition 1.2.2.

If \(G\) is a group and \(M\) is an abelian group on which \(G\) acts (written multiplicatively), one defines the group \(H^1(G,M)\) as the set of functions \(f: G \to M\) such that \(f(gh) = f(g)^h f(h)\text{,}\) modulo the set of such functions of the form \(f(g) = x (x^g)^{-1}\) for some \(x \in M\text{.}\)

Proof.

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
\begin{equation*} t^h = \sum_{g \in G} x^{gh} f(g)^h = \sum_{g \in G} x^{gh} f(gh) f(h)^{-1} = f(h)^{-1} t. \end{equation*}
Thus \(f\) is zero in \(H^1(G,L^*)\text{.}\)

Remark 1.2.4.

Lemma 1.2.3 derives its unusual name from the fact that in the special case where \(G\) is cyclic, this statement occurs as Theorem (Satz) 90 in [21]. The general case first appears in Emmy Noether’s 1933 paper [40] on the principal ideal theorem (Theorem 2.3.1), where Noether attributes it to Andreas Speiser.

Subsection Kummer extensions

Definition 1.2.5. Jargon watch.

If \(G\) is a group, a \(G\)-extension of a field \(K\) is a Galois extension of \(K\) with Galois group \(G\text{.}\)

Proof.

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.

Remark 1.2.7.

Another way to state Theorem 1.2.6 is as a bijection
\begin{equation*} \mbox{\((\ZZ/n\ZZ)^r\)-extensions of \(K\)} \longleftrightarrow \mbox{\((\ZZ/n\ZZ)^r\)-subgroups of \(K^*/(K^*)^n\)}, \end{equation*}
where \((K^*)^n\) is the group of \(n\)-th powers in \(K^*\text{.}\) (What we proved above was the case \(r=1\text{,}\) but the general case follows at once.) Another way is in terms of the absolute Galois group of \(K\text{,}\) as in Theorem 1.2.9 below.

Subsection The Kummer pairing

Definition 1.2.8.

Define the Kummer pairing
\begin{equation*} \langle \cdot, \cdot \rangle: \Gal(\overline{K}/K) \times K^* \to \{1, \zeta_n, \dots, \zeta_n^{n-1} \} \end{equation*}
as follows: given \(\sigma \in \Gal(\overline{K}/K)\) and \(z \in K^*\text{,}\) choose \(y \in \overline{K}^*\) such that \(y^n = z\text{,}\) and put \(\langle \sigma, z \rangle = y^\sigma/y\text{.}\) Note that this does not depend on the choice of \(y\text{:}\) the other possibilities are \(y \zeta_n^k\) for \(k=0, \dots, n-1\text{,}\) and \(\zeta_n^\sigma = \zeta_n\) by the assumption on \(K\text{,}\) so it drops out.

Proof.

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.

Subsection Cyclic extensions without roots of unity

Remark 1.2.10.

If \(K\) is of characteristic coprime to \(n\) but \(\zeta_n \notin K\text{,}\) then an extension of the form \(K(a^{1/n})\) is in general not Galois. For some analysis of such extensions, see Exercise 4 and Exercise 5.
By the same token, \(\ZZ/n\ZZ\)-extensions of a field that does not contains \(\zeta_{n}\) are harder to describe than Kummer extensions, and indeed describing such extensions of \(\QQ\) is the heart of this course. One statement that ties this together with the previous point is that if \(L/K\) is a \(\ZZ/n\ZZ\)-extension, then \(L(\zeta_n)/K(\zeta_n)\) is a \(\ZZ/d\ZZ\)-extension for some divisor \(d\) of \(n\text{,}\) and the latter is a Kummer extension.
We will use the following elaboration of Remark 1.2.10 in the proof of the Kronecker-Weber theorem (Theorem 1.3.4).

Proof.

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
\begin{equation*} \Gal(M/L) \times A \to \{1, \zeta_{n}, \dots, \zeta_n^{n-1}\} \end{equation*}
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:
\begin{equation*} \langle h,s \rangle^g = \langle g^{-1}hg, s^g \rangle, \end{equation*}
because
\begin{equation*} \left( \frac{(s^{1/n})^h}{s^{1/n}} \right)^g = \frac{((s^g)^{1/n})^{g^{-1}hg}}{(s^g)^{1/n}}. \end{equation*}
(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
\begin{equation*} \langle h,a \rangle^g = \langle h,a \rangle^{\omega(g)} = \langle h, a^{\omega(g)} \rangle. \end{equation*}
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.

Remark 1.2.12.

In what follows, we will only need one implication of Lemma 1.2.11: if \(M/K\) is Galois and abelian, then (1.2.1) holds. However, we chose to include both implications for completeness.

Exercises Exercises

1.

Prove Dedekind’s lemma on the linear independence of automorphisms: if \(g_1, \dots, g_n\) are distinct automorphisms of \(L\) over \(K\text{,}\) then there do not exist \(x_1, \dots, x_n \in L\) such that \(x_1 y^{g_1} + \cdots + x_n y^{g_n} = 0\) for all \(y \in L\text{.}\) (This is a key step in the proof of Artin’s lemma in Galois theory.)
Hint.
Suppose the contrary, choose a counterexample with \(n\) as small as possible, then make an even smaller counterexample.

2.

Prove the additive analogue of Lemma 1.2.3: if \(L/K\) is a finite Galois extension with Galois group \(G\text{,}\) then \(H^1(G, L) = 0\text{,}\) where the abelian group is now the additive group of \(L\text{.}\)
Hint.
By the normal basis theorem (see for example [32]), there exists \(\alpha \in L\) whose conjugates form a basis of \(L\) as a \(K\)-vector space.

3.

Prove the following extension of Lemma 1.2.3 (also due to Speiser). Let \(L/K\) be a finite Galois extension with Galois group \(G\text{.}\) Despite the fact that \(H^1(G, \GL(n,L))\) does not make sense as a group (because \(\GL(n,L)\) is not abelian), show nonetheless that “\(H^1(G, \GL(n,L))\) is trivial” in the sense that every function \(f: G \to \GL(n,L)\) for which \(f(gh) = f(g)^h f(h)\) for all \(g,h \in G\) can be written as \(x (x^g)^{-1}\) for some \(x \in \GL(n,L)\text{.}\)
Hint.
To imitate the proof in the case \(n=1\text{,}\) one must find an \(n \times n\) matrix \(x\) over \(L\) such that \(t = \sum_{g \in G} x^g f(g)\) is not only nonzero but invertible. To establish this, note that the set of possible values of \(t\) on one hand is an \(L\)-vector space, and on the other hand satisfies no nontrivial \(L\)-linear relation. (See Lemma 7.1.15 for a similar idea.)

4.

Let \(n\) be a positive integer, let \(K\) be a field of characteristic coprime to \(n\text{,}\) choose \(a \in K^*\text{,}\) and put \(L = K(a^{1/n})\text{.}\) Prove that \([L:K]\) divides \(n\text{.}\)
Hint.
Reduce to the case where \(n\) is prime. Since \([K(\zeta_n):K]\) is coprime to \(n\text{,}\) by taking norms we see that \(a \in (K^*)^n\) if and only if \(a \in (K(\zeta_n)^*)^n\text{;}\) we may thus reduce to the case \(\zeta_n \in K\text{,}\) to which Kummer theory applies.

5.

With notation as Exercise 4, prove that \([L:K] = n\) if and only if \(a \notin (K^*)^p\) for any prime divisor \(p\) of \(n\text{.}\)
Hint.
We may proceed by induction on \(n\) once we check that for any prime divisor \(q\) of \(n\text{,}\) if \(a \notin (K^*)^p\) for any prime divisor \(p\) of \(n\text{,}\) then \(a^{1/q} \notin ((K(a^{1/q})^*)^p\) for any prime divisor \(p\) of \(n\text{.}\) To prove this, assume the contrary; since \([K(a^{1/q}):K] = q\text{,}\) taking norms yields \(a \in (K^*)^p\text{,}\) a contradiction.
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