Artin’s lemma | Climbing Mount Bourbaki

This is the lemma we shall use in the proof of the reciprocity law, to reduce the cyclic case to the cyclotomic case:

Lemma 4 (Artin) Let ${L/k}$ be a cyclic extension of degree ${n}$ and ${\mathfrak{p}}$ a prime of ${k}$ unramified in ${L}$ . Then we can find a field ${E}$ , a subextension of ${L(\zeta_m)}$ , with ${E \cap L = k}$ such that in the lattice of fields

we have:

1. ${\mathfrak{p}}$ splits completely in ${E/k}$

2. ${E(\zeta_m) = L(\zeta_m)}$ , so that ${LE/E}$ is cyclotomic

3. ${\mathfrak{p}}$ is unramified in ${LE/k}$

Moreover, we can choose ${m}$ such that it is divisible only by arbitrarily large primes.

The proof of this will use the previous number-theory lemmas and the basic tools of Galois theory.

So, first of all, we know that ${E}$ is a subextension of some ${L(\zeta_m)}$ . We don’t know what ${m}$ is, but pretend we do, and will start carrying out the proof. As we do so, we will learn more and more about what ${m}$ has to be like, and eventually choose it.
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So, the key to proving the Artin reciprocity law in general is to reduce it to the cyclotomic case (which has already been handled, cf. this). To carry out this reduction, start with a number field ${k}$ and a cyclic extension ${L}$ . We will prove that there is an extension ${k'}$ of ${k}$ and ${L'}$ of ${L}$ such that we have a lattice of fields

(where lattice means ${L \cap k' = k}$ , ${Lk' = L'}$ ) such that ${L'/k'}$ is cyclotomic and a given prime ${\mathfrak{p}}$ of ${k}$ splits completely in ${k'}$ . This means that, in a sense, the Artin law for ${L/k}$ becomes reduced to that of ${L'/k'}$ , at least for the prime ${\mathfrak{p}}$ , in that ${( \mathfrak{p}, L'/k') = (\mathfrak{p}, L/k)}$ . From this, we shall be able to deduce the Artin law for cyclic extensions, whence the general case will be a “mere” corollary.

1. A funny lemma

The thing is, though, finding the appropriate root of unity to use is not at all trivial. In fact, it requires some tricky number-theoretic reasoning, which we shall carry out in this post. The first step is a lemma in elementary number theory. Basically, we will need two large cyclic subgroups of multiplicative groups of residue classes modulo an integer, one of which is generated by a fixed integer known in advance. We describe now how to do this.

Proposition 1 Let ${a, n \in \mathbb{N} - \{1\}}$ . Then there exists ${m \in \mathbb{N}}$ , prime to ${a}$ , such that ${a \in (\mathbb{Z}/m\mathbb{Z})^*}$ has order divisible by ${n}$ . In addition, there exists ${b \in (\mathbb{Z}/m\mathbb{Z})^*}$ of order divisible by ${n}$ such that the cyclic subgroups generated by ${a,b}$ have trivial intersection.

Finally if ${N \in \mathbb{N}}$ , we can assume that ${m}$ is divisible only by primes ${>N}$ .
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Climbing Mount Bourbaki

The proof of Artin’s lemma

Towards a lemma of Artin: some elementary number theory

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