## Section 1.1 The Kronecker-Weber theorem

### Reference.

Our approach follows [56], Chapter 14. A variety of other methods can be found in other texts.

### Subsection Abelian extensions of \(\QQ\)

#### Definition 1.1.1.

An abelian extension of a field is a Galois extension with abelian Galois group. An example of an abelian extension of \(\QQ\) is the cyclotomic field \(\QQ(\zeta_n)\) (where \(n\) is a positive integer and \(\zeta_n\) is a primitive \(n\)-th root of unity), whose Galois group is \((\ZZ/n\ZZ)^*\text{,}\) or any subfield thereof. Amazingly, Theorem 1.1.2 implies that there are no other examples!

#### Theorem 1.1.2. Kronecker-Weber.

If \(K/\QQ\) is a finite abelian extension, then \(K \subseteq \QQ(\zeta_n)\) for some positive integer \(n\text{.}\)

#### Proof.

See Lemma 1.1.10.

#### Example 1.1.3.

A fundamental example of Theorem 1.1.2 is that every quadratic extension of \(\QQ\) is contained in a cyclotomic field. This was known to Gauss via what we now call Gauss sums, and forms the basis of one of his proofs of quadratic reciprocity. It is this proof in particular that generalizes to Artin reciprocity (Theorem 2.2.6). See also Exercise 2 and Exercise 3.

#### Definition 1.1.4.

The smallest \(n\) such that \(K \subseteq \QQ(\zeta_n)\) is called the conductor of \(K/\QQ\text{.}\) It plays an important role in the splitting behavior of primes of \(\QQ\) in \(K\text{,}\) as we will see a bit later.

We will prove Theorem 1.1.2 in the next few lectures. Our approach will be to deduce it from a local analogue (see Theorem 1.3.4).

#### Theorem 1.1.5. Local Kronecker-Weber.

If \(K/\QQ_p\) is a finite abelian extension, then \(K \subseteq \QQ_p(\zeta_n)\) for some \(n\text{,}\) where \(\zeta_n\) is a primitive \(n\)-th root of unity.

#### Proof.

See Theorem 1.3.4.

Before proceeding, it is worth noting explicitly a nice property of abelian extensions that we will exploit below.

#### Remark 1.1.6.

Let \(L/K\) be a Galois extension with Galois group \(G\text{,}\) let \(\gothp\) be a prime of \(K\text{,}\) let \(\gothq\) be a prime of \(L\) over \(\gothp\text{,}\) and let \(G_{\gothq}\) and \(I_{\gothq}\) be the decomposition and inertia groups of \(\gothq\text{,}\) respectively. Then any other prime \(\gothq'\) over \(\gothp\) can be written as \(\gothq^g\) for some \(g \in G\text{,}\) and the decomposition and inertia groups of \(\gothq'\) are the conjugates \(g^{-1} G_{\gothq} g\) and \(g^{-1} I_{\gothq} g\text{,}\) respectively. (Note: my Galois actions will always be *right* actions, denoted by superscripts.)

If \(L/K\) is *abelian*, though, these conjugations have no effect. So it makes sense to talk about *the* decomposition and inertia groups of \(\gothp\) itself!

### Subsection A reciprocity law

Assuming the Kronecker-Weber theorem, we can deduce strong results about the way primes of \(\QQ\) split in an abelian extension.

#### Definition 1.1.7.

Suppose \(K/\QQ\) is abelian, with conductor \(m\text{.}\) Then we get a surjective homomorphism

On the other hand, suppose \(p\) is a prime not dividing \(m\text{,}\) so that \(K/\QQ\) is unramified above \(p\text{.}\) As noted above, there is a well-defined decomposition group \(G_p \subseteq \Gal(K/\QQ)\text{.}\) Since there is no ramification above \(p\text{,}\) the corresponding inertia group is trivial, so \(G_p\) is generated by a Frobenius element \(F_p\text{,}\) which modulo any prime above \(p\text{,}\) acts as \(x \mapsto x^p\text{.}\) We can formally extend the map \(p \mapsto F_p\) to a homomorphism from \(S_m\text{,}\) the subgroup of \(\QQ\) generated by all primes not dividing \(m\text{,}\) to \(\Gal(K/\QQ)\text{.}\) This is called the Artin map of \(K/\QQ\text{.}\)

The punchline is that the Artin map factors through the map \((\ZZ/m\ZZ)^* \to \Gal(K/\QQ)\) we wrote down above! Namely, note that the image of \(r\) under the latter map takes \(\zeta_m\) to \(\zeta_m^r\text{.}\) For this image to be equal to \(F_p\text{,}\) we must have \(\zeta_m^r \equiv \zeta_m^p \pmod{\gothp}\) for some prime \(\gothp\) of \(K\) above \(p\text{.}\) But \(\zeta_m^r (1 - \zeta_m^{r-p})\) is only divisible by primes above \(m\) (see Exercise 4) unless \(r-p \equiv 0 \pmod{m}\text{.}\) Thus \(F_p\) must be equal to the image of \(p\) under the map \((\ZZ/m\ZZ)^* \to \Gal(K/\QQ)\text{.}\)

#### Remark 1.1.8.

The Artin reciprocity law states that a similar phenomenon arises for any abelian extension of any number field; that is, the Frobenius elements corresponding to various primes are governed by the way the primes “reduce” modulo some other quantity. There are several complicating factors in the general case, though.

Prime ideals in a general number field are not always principal, so we can't always take a generator and reduce it modulo something.

There can be lots of units in a general number field, so even when a prime ideal is principal, it is unclear which generator to choose.

It is not known in general how to explicitly construct generators for all of the abelian extensions of a general number field.

Thus our approach will have to be a bit more indirect. See Chapter 2 for the beginning of the story.

### Subsection Reduction to the local case

Our reduction of Kronecker-Weber to local Kronecker-Weber relies on a key result typically seen in a first course on algebraic number theory.

#### Theorem 1.1.9. Minkowski.

There are no nontrivial extensions of \(\QQ\) which are unramified everywhere.

#### Proof.

See for instance [37] III.2.

Using Minkowski's theorem, let us deduce the Kronecker-Weber theorem from the local Kronecker-Weber theorem.

#### Lemma 1.1.10.

The local Kronecker-Weber theorem (Theorem 1.1.5) implies the Kronecker-Weber theorem (Theorem 1.1.2).

#### Proof.

For each prime \(p\) over which \(K\) ramifies, pick a prime \(\gothp\) of \(K\) over \(p\text{;}\) by local Kronecker-Weber (Theorem 1.1.5), \(K_{\gothp} \subseteq \QQ_p(\zeta_{n_p})\) for some positive integer \(n_p\text{.}\) Let \(p^{e_p}\) be the largest power of \(p\) dividing \(n_p\text{,}\) and put \(n = \prod_p p^{e_p}\text{.}\) (This is a finite product since only finitely many primes ramify in \(K\text{.}\))

Write \(L = K(\zeta_n)\text{;}\) we will prove that \(K \subseteq \QQ(\zeta_n)\) by proving that \(L = \QQ(\zeta_n)\text{.}\) Form the completion \(L_\gothq\) for some prime \(\gothq\) over \(p\text{;}\) it is contained in \(\QQ_p(\zeta_{\lcm(n,n_p)})\text{.}\) Let \(I_p\) be the inertia group of \(p\) in \(L\text{;}\) the fixed fixed \(U\) of \(I_p\) on \(L_{\gothq}\) is the maximal unramified subextension of \(L_\gothq\text{.}\) Since \(\QQ_p(\zeta_e)\) is unramified over \(\QQ_p\) for any positive integer \(e\) coprime to \(p\text{,}\) we have \(L_{\gothq} = U(\zeta_{p^{e_p}})\) and so \(I_p \cong \Gal(L_\gothq/U) \subseteq (\ZZ/p^{e_p}\ZZ)^*\text{.}\) Let \(I\) be the group generated by all of the \(I_p\text{;}\) then

On the other hand, the fixed field of \(I\) is an everywhere unramified extension of \(\QQ\text{,}\) which can only be \(\QQ\) itself by Minkowski's theorem. That is, \(I = \Gal(L/\QQ)\text{.}\) But then

and \(\QQ(\zeta_n) \subseteq L\text{,}\) so we must have \(\QQ(\zeta_n) = L\) and \(K \subseteq \QQ(\zeta_n)\text{,}\) as desired.

### Exercises Exercises

#### 1.

Prove that the ring of integers in \(\QQ(\zeta_n)\) equals \(\ZZ[\zeta_n]\text{.}\)

For \(n\) a power of a prime \(p\text{,}\) the minimal polynomial of \(\zeta_n\) is the cyclotomic polynomial \(\Phi_n(x) = x^{(p-1)n/p} + \cdots + x^{n/p} + 1\text{;}\) use the polynomial \(\Phi_n(x-1)\) to show that \(1-\zeta_n\) generates a prime ideal. For the general case, show that if \(K\) and \(L\) are linearly disjoint extensions of \(\QQ\) with coprime discriminants, then \(\gotho_{KL} = \gotho_K \gotho_L\text{.}\)

#### 2.

For \(m \in \ZZ\) not a perfect square, determine the conductor of \(\QQ(\sqrt{m})\text{.}\)

First show that for \(p\) prime, \(\QQ(\sqrt{(-1)^{(p-1)/2} p})\) has conductor \(p\text{.}\)

#### 3.

Using the previous exercise, recover the law of quadratic reciprocity from the Artin reciprocity law.

#### 4.

Prove that if \(m,n\) are coprime integers in \(\ZZ\text{,}\) then \(1 - \zeta_m\) and \(n\) are coprime in \(\ZZ[\zeta_m]\text{.}\)

Look at the polynomial \((1-x)^m-1\) modulo a prime divisor of \(n\text{.}\)

#### 5.

Prove that if \(m\) is not a prime power, then \(1-\zeta_m\) is a unit in \(\ZZ[\zeta_m]\text{.}\)

#### 6.

Let \(p\) be an odd prime. Prove that \(\ZZ[\zeta_p]^*\) is generated by \(\zeta_p\) and \(\ZZ[\zeta_p + \zeta_p^{-1}]^*\text{.}\)

By Dirichlet's units theorem, the index \([\ZZ[\zeta_p]^*:\ZZ[\zeta_p+\zeta_p^{-1}]^*]\) is finite. For \(\alpha \in \ZZ[\zeta_p]^*\text{,}\) the ratio \(\alpha/\overline{\alpha}\) is an algebraic integer having absolute value \(1\) under each complex embedding, and hence is a root of unity by Kronecker's theorem.