Today, we will prove the second inequality: the norm index of the ideles is at most the degree of the field extension. We will prove this using ideles (cf. the discussion of how ideles and ideals connect to each other), and some analysis.
1. A Big Theorem
We shall use one key fact from the theory of L-series. Namely, it is that:
Theorem 1 If is a number field, we have
as . Here ranges over the primes of . The notation means that the two differ by a bounded quantity as .
This gives a qualititative expression for what the distribution of primes must kinda look like—with the aid of some Tauberian theorems, one can deduce that the number of primes of norm at most is asymptotically for , i.e. an analog of the standard prime number theorem. In number fields. We actually need a slight refinement thereof.
Theorem 2 More generally, if is a character of the group , we have
if , and otherwise it tends either to a finite limit or .
Instead of just stating this as a random, isolated fact, I’d like to give some sort of context. Recall that the Riemann-zeta function was defined as . There is a generalization of this to number fields, called the Dedekind zeta function. The Dedekind-zeta function is not defined by summing over for in the ring of integers (minus 0). Why not? Because the ring of integers is not a unique factorization domain in general, and therefore we don’t get a nice product formula.
2. Dedekind zeta and L-functions
We do, however, have unique factorization of ideals. And we have a nice way to measure the size of an ideal: the norm (to ). So, bearing this in mind, define the Dedekind zeta function
I claim that this extends analytically to with a simple pole at . The reasons are too detailed for me to sketch here without really leaving the topic that I’m trying to give an exposition of, namely class field theory, but nonetheless, here goes. So the key fact is that the number of ideals (integral ideals, that is) with is asymptotically proportional to when is large. This fact, utterly trivial when , takes much more work for general . The idea is to look at each ideal class separately (and there are finitely many of them!) and show that the number of with and in said class is asymptotically proportional to .
But such are basically in one-to-one correspondence with principal ideals with by multiplying each by a representative of the class . Then the question becomes of counting integers in the domain of having norms . That domain is a homogeneously expanding smooth domain and the integers are a lattice, so the number of lattice points contained in it is approximately proportional to the volume of the domain. (I am ignoring a subtletly because one has to look not just at integers, but integers up to the action of the units.) So in this way we get an estimate for the number of such lattice points, which when plugged back in yields the approximate linear growth in the number of ideals .
Whence, if we express as a Dirichlet series , we find that is basically proportional to . And from this it is possible to show using some analysis that minus a suitable multiple of extends to the full half-plane . And thus has a pole of order 1 at . By using the product formula (a corollary of unique ideal factorization)
we find that
and this implies the claim made in the first theorem. Similarly, for such characters of , we can define ; there is a similar product expansion, and we find
It now turns out that is actually analytic in a neighborhood of if is not the unit character. This follows by further lemmas on Dirichlet series, and summation by parts; basically the idea is that the partial sums are uniformly bounded because falls into each ideal class approximately the same number of times when is large. As a result, we get the second theorem (whose variability depends on whether or not; we will show below that this is not the case when is a character of :
3. The proof of the second inequality
I’m going to stop vaguely babbling now and get down to a detailed proof. To prove the second inequality, it suffices to prove the idealic version, namely:
Theorem 3 Let be an admissible cycle for a finite extension of degree . Then has order at most .
To prove this theorem, we consider characters of and use a few facts about them. If is the order of this group, then there are precisely characters. Summing all the associated L-functions, we have
But by basic character theory for finite abelian groups (note that is finite), this becomes
I claim now that this last sum is as ; the notation here is analogous to , and means “greater than up to a constant added when .”
Indeed, to see this we will first describe what primes in can look like; we will show that there are a lot of them. Suppose we know that is a prime of dividing of and the residue class field degree over of is 1, then the same is true over , and that implies . In particular, we have
This last sum on the right, however, is a subsum of the owing to the properties of the Dedekind zeta function for . However, the terms for are bounded because the norm is large; in particular, we have
as . This is an easily proved but often-used fact: the terms with don’t contribute significantly. It follows that
Since is finite for and as , it follows that , and also that the for .
Corollary 4 whenever is a nontrivial character of .
It turns out that we can harvest from this a generalization of the Dirichlet theorem on arithmetic progressions, but we will wait until we do class field theory for that.