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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 unramified abelian extension contains a cubic number field of discriminant -23. Compare Exercise 7.

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 \(P(x) \in \ZZ[x]\) be a polynomial whose discriminant \(D\) is squarefree (and hence odd).
  1. Prove that the field \(K = \QQ[x]/(P(x))\) has Galois group \(S_n\) over \(\QQ\text{.}\)
  2. Prove that the Galois closure of \(K/\QQ\) 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 \(n \gt 3\text{,}\) and nonsolvable for \(n \gt 4\text{.}\)
Hint.
Let \(M\) be the Galois closure of \(K/\QQ\text{.}\) Recall that \(M/K\) is unramified over a prime \(\gothp\) of \(K\) if and only if \(L/K\) is: if the decomposition group of some (hence any) prime of \(M\) above \(\gothp\) is nontrivial, then I can choose that prime so that the decomposition group is not contained in \(\Gal(M/K)\text{.}\) Next, note that for each odd prime \(p\) dividing \(K\text{,}\) there is a unique prime \(\gothp\) above \(p\) which is ramified with ramification index 2. Finally, using the different to compute the discriminant, show that \(L\) also has a unique prime ramified over \(p\) and that the ramification index is again 2, then deduce that this prime cannot ramify over \(\gothp\text{.}\) (This also shows that the decomposition group has order 2, so the Galois group is a transitive subgroup of \(S_n\) containing a transposition.)

8.

Show that the conclusion of Exercise 7 remains true if we only require \(K\) to be a number field of degree \(n\) with Galois group \(S_n\) whose discriminant \(D\) is squarefree (but not necessarily of the form \(\QQ[x]/(P(x))\) where \(P(x) \in \ZZ[x]\) has discriminant \(D\)).
Hint.
First show that if there are two distinct primes of \(L\) that ramify over \(p\text{,}\) then \(p^2\) must divide \(D\text{.}\) Then use the fact that \(L\) is locally monogenic at the unique ramified prime over \(p\) to emulate the solution of Exercise 7.
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