The algebraic proof of the second inequality II

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

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

because ${U_S}$ is up to roots of unity a free abelian group of rank ${|S|-1}$ , and the units are a cyclic group of order divisible by ${n}$ (since ${k}$ contains the ${n}$ -th roots of unity). In particular, ${[M:L] = n^{|S|-1}}$ . The extension ${M}$ is only of an auxiliary nature in constructing the group ${E}$ .

3.2. First attempt

We will now describe a lower bound on the norm subgroup of ${L}$ . Then, we will modify this to get a lower bound on the norm subgroup of ${L}$ . It is easy to see that

$\displaystyle NJ_L \supset \prod_S {k_v^{*n}} \prod_{v \notin S} U_v.$

Indeed, ${L/k}$ is unramified outside ${S}$ because we are adjoining ${n}$ -th roots of a ${S}$ -unit ${a}$ : the equation ${x^n - a=0}$ has no multiple root when reduced modulo a place not in ${S}$ . Also, ${[L:k]=n}$ so ${n}$ -th powers are norms.

3.3. ${E}$ , at last

This is too small a group for our ${E}$ , though. For it, we will need more than the ${U_v}$ ‘s outside ${S}$ . This we tackle next; we shall use the extension ${M}$ to obtain places where ${L/k}$ splits completely. So, choose places ${w_1, \dots, w_{s-1} \notin S}$ of ${k}$ whose Frobenius elements ${\sigma_1, \dots, \sigma_{s-1}}$ form a basis for the Galois group ${G(M/L) \subset G(M/k)}$ . We can do this by surjectivity of the Artin map into ${G(M/k)}$ . Note that ${G(M/k)}$ is a ${\mathbb{Z}_n}$ -vector space, and this is what I mean by a basis. These places form a set ${T}$ . This will be important in constructing the set ${E}$ . Now, I claim that the group

$\displaystyle E = \prod_S k_v^{*n} \times \prod_T k_v^* \times \prod_{(S \cup T)^c} U_v$

contains the norms ${NJ_L}$ . Indeed, this is clear for the factors ${k_v^{*n}}$ (indeed, we are working with an extension of degree ${n}$ ). I claim now that the places ${T}$ split completely in ${L}$ , which would mean that the ${k_v^*, v \in T}$ are local norms from ${L}$ too. Let ${w, w'}$ be places of ${M,L}$ extending ${v \in T}$ . Then ${M_{w} \supset L_{w'} \supset k_v}$ , and this is a cyclic extension, since these are local fields and the extension ${M/k}$ is unramified. The extension ${[M_w: L_{w'}]}$ has order ${n}$ because the Frobenius element is nonzero, whence ${L_{w'} = k_v}$ or otherwise we could not have a cyclic group ${G(M_w/k_v)}$ since ${n}$ is prime. Finally, it is clear that ${U_v, v \notin S \cup T}$ is contained in the norm group.

4. The index of ${E}$

We will show that

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

which will prove the result, because we know that ${E}$ contains the norm subgroup. Now we have ${J_k = k^* J_{S \cup T}}$ , so by the general formula (ADD THIS)

$\displaystyle (J_k : k^* E) = (J_{S \cup T}: E) / (U_{S \cup T}: k^* \cap E).$

Indeed, there is a general formula, proven by a bit of diagram chasing:

$\displaystyle (AC:BC)(A \cap C: B \cap C) = (A:B)$

for ${B,C}$ subgroups of the abelian group ${A}$ .

4.1. Computation of the first index

This is the easiest. We just have

$\displaystyle (J_{S \cup T}: E) = \prod_{v \in S} (k_v^*: k_v^{*n}) = n^{2|S|}$

because ${\prod_{v \in S} |n|_v =1}$ by the product formula. The second index, by contrast, will be considerably more subtle; we don’t have an obvious expression for ${k^* \cap E}$ .

4.2. The second index

The point is that we will develop a nice expression for ${k^* \cap E}$ ; namely we will show that it is equal to ${U(S \cup T)^{n}}$ . By the unit theorem, this clearly implies that

$\displaystyle (U_{S \cup T}: k^* \cap E) = n^{2|S|-1}.$

This will, in particular, yield the global bound for the norm index and prove the second inequality.

This is a bit tricky, and will use the first inequality. First, let’s suppose ${b \in k^* \cap E}$ ; the key is to prove that ${k(\sqrt[n]{b}) = k}$ or that ${b}$ is. actually an ${n}$ -th power. We see this supposed equality of fields is true locally at the primes in ${S}$ because ${b}$ is an ${n}$ -th power there. Also, ${k(\sqrt[n]{b})}$ is unramified outside ${S \cup T}$ , so it follows that

$\displaystyle N J_{k(\sqrt[n]{b})} \supset \prod_S k_v^* \times \prod_T U_v^{*n} \times \prod_{(S \cup T)^c} U_v .$

Call the group on the right ${W}$ . I claim that ${k^* W = J_k}$ . This will follow from the next

Lemma 4 The map

$\displaystyle U(S ) \rightarrow \prod_{v \in T} U_v / U_v^n$

is surjective.

Indeed, once we have proved this lemma, the claim ${k^* W = J_k}$ follows easily, because any idele class, we can find a representative idele in ${J_{S \cup T}}$ . This would be in ${W}$ except for the fact that it may not be an ${n}$ -th power at ${T}$ . But by multiplying by some ${S}$ -unit, we can arrange this.

Thus ${k^* W = J_k}$ , implying ${b}$ is an ${n}$ -th power, and completing the claim that ${k^* \cap E = U(S \cup T)^{n}}$ —but for the lemma. So it all boils down to proving this lemma.

4.3. The last technical step

We must now prove this lemma. But it is—how fitting—another index computation! Consider the kernel ${H}$ of the map out of ${U(S )}$ . We will obtain a clean description of it. Then we will show that the image of ${U(S)}$ has the same cardinality as the product ${\prod_{v \in T} U_v / U_v^n}$ .

Namely, suppose ${b}$ is a ${S}$ -unit and a power at the primes of ${T}$ ; I claim then that ${b}$ is an ${n}$ -th power in ${L}$ . For ${b^{1/n} \in M}$ is then fixed by ${\sigma_1, \dots, \sigma_{s-1}}$ (because ${b^{1/n}}$ is in the appropriate local fields corresponding to ${T}$ ), and these ${\sigma}$ ‘s generate ${G(M/L)}$ . Conversely, suppose ${b \in U(S) \cap L^{*n}}$ ; then ${b}$ is an ${n}$ -th power at the primes of ${T}$ (because they split completely in ${L/k}$ ), and an ${S}$ -unit, so it belongs to the kernel. We have obtained our description of ${H}$ .

In particular, the kernel ${H}$ consists of ${U(S) \cap L^{*n}}$ , and we have

$\displaystyle (U(S): H) = (L^{*n} U(S)/L^{*n}) = [M:L] = n^{|S|-1}$

by Kummer theory. We have computed the index of ${H}$ .

Meanwhile the cardinality of ${\prod_{v \in T} U_v / U_v^n}$ is ${n^{|T|} = n^{|S|-1}}$ in view of the power index computations for the units. So we must have surjectivity, which proves the lemma.

The proof of the second inequality is thus complete.

Climbing Mount Bourbaki