Math 254B, UC Berkeley, Spring 2002 (Kedlaya) The Hilbert Class Field

Reference. Milne, Introduction; Neukirch, VI.6.

Recall that the field ℚ has no extensions which are everywhere unramified (Minkowski’s theorem from last semester, which we used in proving the Kronecker-Weber theorem). This is quite definitely not true of other number fields; we begin with an example illustrating this.

In the number field K = ℚ(\sqrt{−5}), the ring of integers is ℤ[\sqrt{−5}] and the ideal (2) factors as 𝔭^{2}, where the ideal 𝔭 = (2, 1 + \sqrt{−5}) is not principal.

Now let’s see what happens when we adjoin a square root of −1, obtaining L = ℚ(\sqrt{−5}, \sqrt{−1}). The extension ℚ(\sqrt{−1})/ℚ only ramifies over 2, so ℚ(\sqrt{−5}, \sqrt{−1})/K can only be ramified over 𝔭. Moreover, in K_{𝔭}, −1 ≡ (2 + \sqrt{−5})^{2} (mod 8), so −1 is a square in K_{𝔭}. Thus L/K is also unramified over 𝔭!

We’ve now seen that ℚ(\sqrt{−5}) admits both a nonprincipal ideal and an unramified abelian extension. It turns out these are not unrelated events. Caution: until further notice, the phrase “L/K is unramified” will mean that L/K is unramified over all finite places in the usual sense, and that every real embedding of K extends to a real embedding of L. (Get used to this. The real and complex embeddings of a number field will be treated like primes throughout the course.)

Theorem 1. Let L be the maximal unramified abelian extension of a number field K. Then L/K is finite, and its Galois group is isomorphic to the ideal class group of K.

In fact, there is a canonical isomorphism, given by the Artin reciprocity law. We’ll see this a bit later. The field L is called the Hilbert class field of K.

Warning: there can be infinite unramified nonabelian extensions. In fact, Golod and Shafarevich used unramified abelian extensions to construct these! Namely, starting from a number field K = K_{0}, let K_{1} be the Hilbert class field of K_{0}, let K_{2} be the Hilbert class field of K_{1}, and so on. Then K_{i} is a unramified but not necessarily abelian extension of K_{0}, and for a suitable choice of K_{0}, [K_{i} : K_{0}] can be unbounded.

Exercises.

1. Let K be an imaginary quadratic extension of ℚ in which t finite primes ramify. Prove that #(Cl(K)/2 Cl(K)) = 2^{t−1}, assuming Theorem 1. (Hint: if an odd prime p ramifies in K, show that K(\sqrt{p})/K is unramified; if 2 ramifies in K, show that K(\sqrt{−1})/K is unramified.) This was first proved by Gauss using binary quadratic forms, as alluded to last semester.

2. Give an example, using a real quadratic field, to illustrate that

(a) Theorem 1 fails if we don’t require the extensions to be unramified above the real place;

(b) The previous exercise fails for real quadratic fields.

(Hint: see the hint for the previous exercise.) Optional: prove that the previous exercise works if you replace the class group by the narrow class group, in which you only mod out by principal ideals having a totally positive generator. This gives an example of a ray class group; more on those next time.

3. The field ℚ(\sqrt{−23}) admits an ideal of order 3 in the class group and an unramified abelian extension of degree 3. Find both; feel free to use a computer if that helps. (Hint: the extension contains a cubic field of discriminant -23.)