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Notes on class field theory

Kiran S. Kedlaya

Section 2.1 The Hilbert class field

Reference.

[36], Introduction; [37], VI.6.

Subsection An example of an unramified extension

Recall that the field \(\QQ\) has no extensions which are everywhere unramified (Theorem 1.1.9). This is quite definitely not true of other number fields; we begin with an example illustrating this.

Example 2.1.1. An unramified extension of a number field.

In the number field \(K = \QQ(\sqrt{-5})\text{,}\) the ring of integers is \(\ZZ[\sqrt{-5}]\) and the ideal \((2)\) factors as \(\gothp^2\text{,}\) where the ideal \(\gothp = (2, 1 + \sqrt{-5})\) is not principal.
Now let’s see what happens when we adjoin a square root of \(-1\text{,}\) obtaining \(L = \QQ(\sqrt{-5}, \sqrt{-1})\text{.}\) The extension \(\QQ(\sqrt{-1})/\QQ\) only ramifies over 2, so \(L/K\) can only be ramified over \(\gothp\text{.}\) On the other hand, if we write \(L = K(\alpha)\) where \(\alpha = (1 + \sqrt{5})/2\text{,}\) then modulo \(\gothp\) the minimal polynomial \(x^2-x-1\) of \(\alpha\) remains irreducible, so \(\gothp\) is unramified (and not split) in \(L\text{.}\)

Subsection Hilbert class fields

We’ve now seen that \(\QQ(\sqrt{-5})\) admits both a nonprincipal ideal and an unramified abelian extension. It turns out these are not unrelated events.

Definition 2.1.2. Jargon watch.

In class field theory, the phrase “\(L/K\) is unramified” is conventionally interpreted to 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 \(L\text{.}\)

Proof.

A canonical isomorphism will be given by the Artin reciprocity law (Theorem 2.2.6).

Remark 2.1.4.

While Theorem 2.1.3 implies that an abelian unramified extension must be finite, there can be infinite unramified nonabelian extensions. See Remark 2.3.12.

Remark 2.1.5.

At this point, it should now be apparent that class field theory is “class field” theory, i.e., the theory of class fields such as the Hilbert class fields (and other examples described in Definition 2.2.7) rather than a special type of “field theory”. Whether this affects your pronunciation of the entire phrase is up to you!

Exercises Exercises

1.

Let \(K\) be an imaginary quadratic extension of \(\QQ\) in which \(t\) finite primes ramify. Assuming Theorem 2.1.3, prove that \(\#(\Cl(K)/2\Cl(K)) = 2^{t-1}\text{;}\) this recovers a theorem of Gauss originally proved using binary quadratic forms.
Hint.
If an odd prime \(p\) ramifies in \(K\text{,}\) show that \(K(\sqrt{p^*})/K\) is unramified for \(p^* = (-1)^{(p-1)/2} p\text{;}\) if 2 ramifies in \(K\text{,}\) show that \(K(p^*)/K\) is unramified for one of \(p^* = -1, 2, -2\text{.}\)

2.

Give an example, using a real quadratic field, to illustrate that:
  1. Theorem 2.1.3 fails if we don’t require the extensions to be unramified above the real place;
  2. the previous exercise fails for real quadratic fields.

3.

Prove that Exercise 1 extends to real quadratic fields if one replaces 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 in Section 2.2.

4.

The field \(\QQ(\sqrt{-23})\) admits an ideal of order 3 in the class group and an unramified abelian extension of degree 3. Find both.
Hint.
The extension contains a cubic field of discriminant -23.

5.

Let \(L/K\) be an extension of number fields admitting no nontrivial abelian subextension \(M/K\) which is everywhere unramified (including at archimedean places). Assuming Theorem 2.1.3, prove that the class number of \(K\) divides the class number of \(L\text{.}\)

6.

A number field \(K\) is called a CM field if it is a totally complex quadratic extension of a totally real number field \(K_+\text{.}\) Using Exercise 5, show that the class number of \(K_+\) divides the class number of \(K\text{.}\) The ratio is called the relative class number.

7.

Let \(K\) be a number field of degree \(n\) with Galois group \(S_n\) whose discriminant \(D\) is squarefree. Prove that the Galois closure of \(K\) is unramified over all finite places of \(\QQ(\sqrt{D})\text{.}\) This gives an ample supply of everywhere unramified extensions (of various fields) which are nonabelian for \(n > 3\text{.}\)
Hint.
Let \(M\) be the Galois closure of \(K\text{.}\) For any odd prime \(p\) dividing the discriminant, use the restriction on \(D\) to show that there is exactly one prime of \(K\) above \(p\) which is ramified and that its ramification index is 2. Then deduce that the inertia group of a prime of \(M\) above \(p\) has order 2, and finally argue that said prime is unramified over its restriction to \(\QQ(\sqrt{D})\text{.}\)
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