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. […]