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}.
(more…)

So, now to the next topic in introductory algebraic number theory: ramification. This is a measure of how primes “split.”  (No, definitely wrong word there…)

e and f 

Fix a Dedekind domain {A} with quotient field {K}; let {L} be a finite separable extension of {K}, and {B} the integral closure of {A} in {L}. We know that {B} is a Dedekind domain.

(By the way, I’m now assuming that readers have been following the past few posts or so on these topics.)

Given a prime {\mathfrak{p} \subset A}, there is a prime {\mathfrak{P} \subset B} lying above {\mathfrak{p}}. I hinted at the proof in the previous post, but to save time and avoid too much redundancy I’ll refer interested readers to this post.

Now, we can do a prime factorization of {\mathfrak{p}B \subset B,} say {\mathfrak{p}B = \mathfrak{P}_1^{e_1} \dots \mathfrak{P}_g^{e_g}}. The primes {\mathfrak{P}_i} contain {\mathfrak{p}B} and consequently lie above {\mathfrak{p}}. Conversely, any prime of {B} containing {\mathfrak{p}B} must lie above {\mathfrak{p}}, since if {I} is an ideal in a Dedekind domain contained in a prime ideal {P}, then {P} occurs in the prime factorization of {I} (to see this, localize and work in a DVR). (more…)

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