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 and a cyclic extension . We will prove that there is an extension of and of such that we have a lattice of fields

(where lattice means , ) such that is cyclotomic and a given prime of splits completely in . This means that, in a sense, the Artin law for becomes reduced to that of , at least for the prime , in that . 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 . Then there exists , prime to , such that has order divisible by . In addition, there exists of order divisible by such that the cyclic subgroups generated by have trivial intersection.

Finally if , we can assume that is divisible only by primes .

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

** Lemma 2** Let and let be a prime. Then there exists a prime such that the order of modulo is for some .

The order of is modulo if

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

which we write as for . I claim that there is a prime dividing that does not divide .

Indeed, we have

for suitable integral coefficients . This means that the greatest common divisor of and must be a power of .

So if we show that is not a power of , then we get a prime with , implying (by the above reasoning) that and is our prime. But if were a power of for all , we’d then have divisible by when . So a look at the expansion of in powers of implies that for very large, we have divisible only to order 1 by . In particular, . Since as , 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 where the are prime. By the previous lemma, we can choose for each , a prime such that the order of in is , where . If we take large (which can be done by the previous lemma). then will be a large prime. We may assume that no two of the are the same.

Now the order of in is divisible by , by the Chinese remainder theorem. However, we don’t have yet. For this we will add in more primes.

So choose even bigger primes such that the order of in is for .

We let . Then by the Chinese remainder theorem, we can find such that . I claim that satisfy the hypotheses of the lemma.

First, we have , so it follows that both have order divisible by .

Suppose now that we had . Then since . Thus , and in particular because the order of is . Thus too, and the cyclic groups generated by have trivial intersection.

June 7, 2010 at 7:05 pm

[…] So, let . Choose divisible only by super large primes and such that has order in dividing , and such that there also exists of order dividing such that the cyclic groups generated by respectively are disjoint. We can do this by yesterday’s lemma. […]