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}.

The proof is not terribly difficult, but we first need a lemma that will allow us to construct {m} so that {a} has a large order.

Lemma 2 Let {a, d \in \mathbb{N} - \{1\}} and let {q} be a prime. Then there exists a prime {p} such that the order of {a} modulo {p} is {q^{d'}} for some {d' \geq d}.

The order of {a} is {q^{d'}} modulo {p} if

\displaystyle p \mid a^{q^{d'}} - 1 , \quad p \not\mid a^{q^{d'-1}}-1.

We will find such a {p} with a little trick. Consider the ratio

\displaystyle T(d') = \frac{ a^{q^{d'}} - 1}{ a^{q^{d'-1}} - 1} ,

which we write as {(x^q -1) /( x -1)} for {x = a^{q^{d'-1}}}. I claim that there is a prime dividing {T(d')} that does not divide {x-1}.

Indeed, we have

\displaystyle T(d') = \sum_{i=0}^{q-1} x^i = \sum c_i (x-1)^i + q

for suitable integral coefficients {c_i}. This means that the greatest common divisor of {x-1} and {x^q-1} must be a power of {q}.

So if we show that {T(d')} is not a power of {q}, then we get a prime {p \neq q} with {p \mid T(d')}, implying (by the above reasoning) that {p \not\mid x-1} and {p} is our prime. But if {T(d')} were a power of {q} for all {d' \geq d}, we’d then have {x-1} divisible by {q^2} when {d' \geq d+1}. So a look at the expansion of {T(d')} in powers of {x-1} implies that for {d'} very large, we have {T(d')} divisible only to order 1 by {q}. In particular, {x^q - 1 = q(x-1)}. Since {x \rightarrow \infty} as {d' \rightarrow \infty}, this is evidently impossible.

This reasoning can be improved to yield a better result, but we don’t really need it. Cf. Lang’s book. I’m being rather lazy here and taking the path of least proof length.

We can now prove the first proposition.

We first factor {n = r_1^{s_1} \dots r_k^{s_k}} where the {r_i} are prime. By the previous lemma, we can choose for each {i}, a prime {m_i} such that the order of {a} in {(\mathbb{Z}/m_i \mathbb{Z})^*} is {r_i^{s_i'}}, where {s_i' > s_i}. If we take {s_i'} large (which can be done by the previous lemma). then {m_i} will be a large prime. We may assume that no two of the {m_i} are the same.

Now the order of {a} in {(\mathbb{Z}/\prod m_i\mathbb{Z})^*} is divisible by {n}, by the Chinese remainder theorem. However, we don’t have {b} yet. For this we will add in more primes.

So choose even bigger primes {l_i} such that the order of {a} in {(\mathbb{Z}/l_i \mathbb{Z})^*} is {r_i^{s_i''}} for {s_i''>s_i'}.

We let {m = \prod m_i l_i}. Then by the Chinese remainder theorem, we can find {b} such that {b \equiv a \mod \prod m_i, b \equiv 1 \mod \prod l_i}. I claim that {m,b} satisfy the hypotheses of the lemma.

First, we have {(\mathbb{Z}/m\mathbb{Z})^* = (\mathbb{Z}/\prod m_i\mathbb{Z})^* \times (\mathbb{Z}/\prod l_i\mathbb{Z})^*}, so it follows that {a,b} both have order divisible by {n}.

Suppose now that we had {a^c \equiv b^d \mod m}. Then {a^c \equiv 1 \mod \prod l_i} since {b \equiv 1 \mod \prod l_i}. Thus {\prod r_i^{s_i''} \mid c}, and in particular {a^c \equiv 1 \mod m} because the order of {a} is { \prod r_i^{s_i''}}. Thus {b^d \equiv 1 \mod m} too, and the cyclic groups generated by {a,b} have trivial intersection.

Advertisements