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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