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).

1. Some index computations

We are going to need to know what the index ${(k: k^{*n})}$ looks like for a local field ${k}$; this is finite, because (by Hensel’s lemma) anything close enough to 1 is an ${n}$-th power. Suppose the ${n}$-th roots of unity are in ${k^*}$. To do this, we shall use the magical device called the Herbrand quotient. We have an action of the cyclic group ${G=\mathbb{Z}/n\mathbb{Z}}$ on ${k^*}$, where each element acts as the identity. Then ${k^* = \mathbb{Z} \times U_k}$ in a noncanonical manner (depending on a uniformizing parameter), so we have (the denominator in the first term is ${n}$ because ${k}$ contains the ${n}$-th roots of unity)

$\displaystyle \frac{( k^*: k^{*n} ) }{n} = Q(k^*) = n Q(U_k).$

Now ${U_k}$ has a subgroup of finite index which is isomorphic to the ring of integers ${O_k}$ by the exponential map. (Cf. the computations of the local norm index.) So we have that the index in question, ${(k^*: k^{*n})}$ is computed as ${n^2 Q(O_k)}$. But this last Herbrand quotient is easy to compute directly. The denominator in the Herbrand quotient is 1 because we are in characteristic zero. The numerator is ${(O_k: nO_k) = \frac{1}{|n|}}$ with respect to the absolute value on ${k}$ (assuming it is suitably normalized, of course). So we find

$\displaystyle \boxed{( k^*: k^{*n} ) = \frac{n^2}{|n|}.}$

Next, we need to do the same thing for the units ${U_k}$. In this case we find

$\displaystyle \boxed{( U_k^: U_k^{n} ) = \frac{n}{|n|}.}$

The reason there is no ${n^2}$ is that when computing ${Q(k^*)}$, we had a decomposition ${k^* = \mathbb{Z} \times U_k}$, while here there is no additional ${\mathbb{Z}}$.

2. Reduction to the cyclic case

The case of a cyclic extension ${L/k}$ (of number fields) of prime degree ${n}$ where the ${n}$-th roots of unity are in the ground field is much nicer, because it means that the extension is obtained by adjoining the ${n}$-th root of something in ${k}$. In general, it isn’t known how to generate the abelian extensions of a number field ${k}$ (this is a Hilbert problem, number 12), though it can be done in certain cases (e.g. ${k=\mathbb{Q}}$, where every abelian extension is contained in a cyclotomic extension—this is the famous Kronecker-Weber theorem that will be a corollary of class field theory).

So, let’s first formally state the theorem:

Theorem 1 (The second inequality) Let ${L/k}$ be a finite abelian extension of degree ${n}$ with Galois group ${G}$. Then ${(J_k: k^* NJ_L) = |H_T^0(G, J_L)| \leq n}$.

The second inequality is also true for non-Galois extensions. We will later see what the norm index actually means for non-abelian extensions. This more general result can also be reduced to the cyclic prime order case, but we shall not do so (since it requires a few facts about group cohomology that I haven’t covered). For a cyclic extension, this together with the first inequality implies that the global norm index is precisely the degree of the field extension—which hints closely at the reciprocity law (for which there is still much more work, however).

2.1. Dévissage

We will prove this weaker result by dévissage: i.e., an unscrewing technique.

So, here is the key lemma.

Lemma 2 (Dévissage) Suppose the result is true for ${M/L}$ and ${L/k}$, where ${M/k}$ is abelian. Then it is true for ${M/k}$.

This is a straightforward multiplicativity computation:

$\displaystyle (J_k: k^* N^M_k J_L) = (J_k: k^* N^L_k J_L )(k^* N^L_k J_L: k^* N^M_k J_M).$

In the product, the first term is bounded by ${[L:k]}$ by assumption. I claim that the second is bounded by ${[M:L]}$. Indeed, it is

$\displaystyle (k^* N^L_k J_L : k^* N^M_k (L^* J_M)) \leq (N^L_k J_L : N^L_k N^M_L (L^* J_M))$

by a general (and easy) inequality ${(AC:BC) \leq (A:B)}$ for ${B,C}$ subgroups of an abelian group ${A}$. This last term is bounded by

$\displaystyle (J_L: L^* N^M_L J_M) \leq [M:L]$

since ${(f(A):f(B)) \leq (A:B)}$ and by assumption the inequality is true for ${M/k}$.

Well, now any finite abelian group ${G}$ has a filtration whose quotients are cyclic of degree ${p}$. By the Galois correspondence between subgroups and subfields, it now follows that if we prove the second inequality for extensions ${L/k}$ of prime degree ${n}$, then it is true in general.

2.2. Roots of unity

We need to make one more extension: that the ${n}$-th roots of unity are in ${k}$. This we do next.

So consider a cyclic extension ${L/k}$ of degree ${n}$. Let ${\zeta_n}$ be an ${n}$-th root of unity, and consider ${L'=L(\zeta_n), k'=k(\zeta_n)}$, and the extensions ${L/k}$, ${L'/k'}$. Because ${k'/k}$ must have order prime to ${n}$ (indeed, at most ${n-1}$), it follows that ${[L':k']=[L:k]}$. In particular ${L'/k'}$ is cyclic.

Now the extension ${L'/k'}$ is precisely in the form we want: cyclic of prime order ${n}$, with the ${n}$-th roots of unity in the ground field.

So:

Lemma 3 If the second inequality is true for ${L'/k'}$ it is true for ${L/k}$.

First off, it is easy to see that ${J_k/k^*NJ_L}$ is an ${n}$-torsion group, because any ${n}$-th power is a norm. It is also finite; this is true because the Herbrand quotient ${Q(J_L)}$ is defined (and has been computed!). So the order is a power of ${n}$.

We will now prove that the order of the norm idele class group of ${L/k}$ is at most that of ${L'/k'}$ by exhibiting it as a factor group.

To abbreviate, I will change notation slightly, and write ${C_k}$ for ${J_k/k^*}$. There are maps:

$\displaystyle C_k/NC_L \rightarrow C_{k'}/NC_{L'} \rightarrow C_k/NC_L.$
Here the first one is the inclusion, and it sends ${NC_L}$ to ${NC_{L'}}$ because ${G(L'/k') \simeq G(L/k)}$ in the natural way (restriction to ${L}$), and the second is the norm map on the ideles. The composition is raising to the power ${[k':k]}$, so it is surjective. As a result, the norm map ${C_{k'}/NC_{L'} \rightarrow C_k/NC_L}$ is surjective; since the former group has order at most ${n}$ (since the second inequality is assumed true for it), then so does the latter.

The lemma is now proved. And we are now in the following situation. We need to prove the second inequality of a cyclic extensio of prime degree, where the roots of unity are contained in the ground field. Then, we will have proved (by the above reasoning) the second inequality in general. We will finish this in the next post.