We continue (and finish) the proof started in the previous post of the second inequality.

3. Construction of ${E}$

So, here’s the situation. We have a cyclic extension ${L/k}$ of degree ${n}$, a prime number, and ${k}$ contains the ${n}$-th roots of unity. In particular, we can write (by Kummer theory) ${L = k(D^{1/n})}$ for ${D}$ a subgroup of ${k^*}$ such that ${(k^{*n} D: k^{*n}) = n}$, in particular ${D}$ can be taken to be generated by one element ${a}$.

We are going to prove that the norm index of the ideles is at most ${n}$. Then, by the reductions made earlier, we will have proved the second inequality.

3.1. Setting the stage

Now take a huge but finite set ${S}$ of primes such that:

1. ${J_k = k^* J_S}$

2. ${a}$ is an ${S}$-unit

3. ${S}$ contains all the primes dividing ${n}$

4. ${S}$ contains the ramified primes We will now find a bigger extension of ${L}$ whose degree is a prime power. We consider the tower ${k \subset L \subset M = k(U_S^{1/n})}$ for ${U_S}$ the ${S}$-units. We have the extension ${[M:k]}$ whose degree we can easily compute; it is

$\displaystyle (k^{*n} U_S: k^{*n}) = (U_S: U_S^n) = n^{|S|}$

So, it turns out there’s another way to prove the second inequality, due to Chevalley in 1940. It’s purely arithmetical, where “arithmetic” is allowed to include cohomology and ideles. But the point is that no analysis is used, which was apparently seen as good for presumably the same reasons that the standard proof of the prime number theorem is occasionally shunned. I’m not going into the proof so much for the sake of number-theory triumphalism but rather because I can do it more completely, and because the ideas will resurface when we prove the existence theorem. Anyhow, the proof is somewhat involved, and I am going to split it into steps. The goal, remember, is to prove that if ${L/k}$ is a finite abelian extension of degree ${n}$, then

$\displaystyle (J_k: k^* NJ_L) \leq n.$

Here is an outline of the proof:

1. Technical abstract nonsense: Reduce to the case of ${L/k}$ cyclic of a prime degree ${p}$ and ${k}$ containing the ${p}$-th roots of unity

2. Explicitly construct a group ${E \subset J_k}$ and prove that ${NJ_L \supset E}$

3. Compute the index ${(J_k: k^* E)}$. The whole proof is too long for one blog post, so I will do step 1 (as well as some preliminary index computations—yes, these are quite fun—today). (more…)

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 ${k}$ is a number field, we have

$\displaystyle \sum_{\mathfrak{p}} \mathbf{N} \mathfrak{p}^{-s} \sim \log \frac{1}{s-1} (*)$

as ${s \rightarrow 1^+}$. Here ${\mathfrak{p}}$ ranges over the primes of ${k}$. The notation ${\sim}$ means that the two differ by a bounded quantity as ${s \rightarrow 1^+}$.

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 ${N}$ is asymptotically ${N/\log N}$ for ${N \rightarrow \infty}$, 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 ${\chi}$ is a character of the group ${I(\mathfrak{c})/P_{\mathfrak{c}}}$, we have$\displaystyle \sum_{ \mathfrak{p} \not\mid \mathfrak{c}} \chi(\mathfrak{p}) \mathbf{N} \mathfrak{p}^{-s} \sim \log \frac{1}{s-1}$

if ${\chi \equiv 1}$, and otherwise it tends either to a finite limit or ${-\infty}$.

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 ${\zeta(s)=\sum_n n^{-s}}$. There is a generalization of this to number fields, called the Dedekind zeta function. The Dedekind-zeta function is not defined by summing over ${\sum |N(\alpha)|}$ for ${\alpha}$ 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. (more…)