Z_p-extensions of Number Fields, Part II

This is continuation of the last post. We claimed there that we were going to deform a totally real number field of degree n into a field with signature (r,s) with r+2s = n, and pass information about Leopoldt’s conjecture from one field to the other.

How does one “deform” a number field? One natural way is to think of a finite etale map of varieties X->Y defined over Q, and then consider the fibres. More prosaically, write down some family of polynomials and then vary the coefficients. Most of the time, the unit group doesn’t behave so well in such families. For example, consider the equation:

x^2 - D = 0.

If one varies D, even with some local control at primes dividing infinity (that is, keeping D positive), then it is not at all clear how the fundamental unit varies. In fact, one knows that the height of the fundamental unit is very sensitive to the size of the class number, which changes somewhat irregularly with D. On the other hand, consider the equation:

x^2 - Dx = 1.

Here one is in much better shape: as D varies, the element x will always be a unit, and moreover always generates a finite index subgroup of the full unit group. How might one use this for arguments concerning Leopoldt’s conjecture? The idea is to consider (as above) a family of number fields in which some finite index subgroup of the full unit group is clearly visible, and is deforming “continuously” in terms of the parameters. Then, by Krasner’s Lemma, we see that Leopoldt’s conjecture for one number field (and a fixed prime p) will imply the same for all sufficiently close number fields. To start, however, one needs to have such nice families.

Ankely-Brauer-Chowla Fields: One nice family of number fields that deforms nicely is the class of so-called Ankeny-Brauer-Chowla fields (from their 1956 paper):

\prod (x - a_i) - 1 = 0

It is manifestly clear that in the field Q(x), the elements x – a_i are all units, and that (generically) there is only one multiplicative relation, namely that the product over all such units is trivial. In this way, we get a family of number fields (with generic Galois group) S_n and with a family of units generating a free abelian group of rank n-1. With a little tweak, we can also ensure that the prime p splits completely. Concretely, consider the equations

\prod (X - a_i) - \prod (X - b_i) = 1,

where X is a formal variable. The corresponding variety Y is connected of dimension n, and the projection to A^n given by mapping to {b_i} is a finite map, and so, generically, the values of b_i on Y are all distinct. In particular, for sufficiently large primes p, Y has points over F_p where all the b_i are distinct modulo p. Fix such a point {a_i,b_i} over F_p. Lift the a_i in F_p to arbitrary integers in Z. Then, by Hensel’s lemma, there exist p-adic integers v_i congruent to b_i mod p such that

\prod (x - a_i) - 1 = \prod (x - v_i),

and so p splits completely in our field as long as the a_i satisfy some suitable non-empty congruences mod p.

Deforming the signature: Suppose we assume that, for a fixed choice A = (a_1,..,a_n), the corresponding field F satisfies Leopoldt’s conjecture. Then we see that, in a sufficiently small neighbourhood of A, we obtain many other fields which are totally real with Galois group S_n that also satisfy Leopoldt’s conjecture. On the other hand, our goal is to study fields of signature (r,s) with r+2s=n. So we want to deform our fields to have non-trivial signature. I learnt this trick by reading a paper of Bilu: we deform the fields in a slightly different way, by making the replacement

(x - a_i)(x - a_j) \Rightarrow (x - a_i)(x - a_j) + u,

where u has very small p-adic valuation, and yet is a large positive integer. The corresponding field no longer has n obvious units (whose product is one), but now only n-1 obvious units (whose product is one), where one of the units is now the quadratic polynomial above. On the other hand, one can also see that the signature of the number field is now (n-2,1). So we still have a nice finite index subgroup of the unit group. Moreover, p-adically, if our original units are written as {u_i}, then we get (p-adically) something very close (by Krasner), except now u_i and u_j have been replaced by u_i + u_j. By combining other pairs of units in the same way, we can reduce the signature to (r,s) with any r+2s = n and still have a nice p-adically continuous finite index family of global units.

Proposion Suppose that Leopoldt’s conjecture holds for the original field K at p. Then, by deforming suitably chosen pairs of roots, we obtain a (infinitely many) fields L with Galois group S_n and signature (r,s) with r+2s=n such that, for a choice of r+s-1 primes above p in L, the p-adic regulator of the units at those r+s-1 primes is non-zero.

As a consequence, for that choice of r+s-1 primes, the corresponding maximal Z_p-extension has rank zero. This proves that (L,p) is rigid for this choice of S. However, since S_n is n-transitive, the same result applies for any such choice of r+s-1 primes. It’s an elementary lemma to see that this also implies the result for sets S which are either larger or smaller than r+s-1.

The argument is exactly as you expect: Given the original field K, the assumption of Leopoldt’s conjecture for K implies that at least one of the corresponding (r+s-1) x (r+s-1) minors must be non-zero. We then deform the field globally so that the corresponding units in L of signature (r,s) are related to this minor, which (by Krasner) will still be non-zero. QED

Questions: The starting point of this construction was the assumption that K satisfied Leopoldt’s conjecture. Can one prove this directly? That is, can one find a choice of a_i such that the field

\prod (x - a_i) - 1 = 0.

satisfies Leopoldt at p? This seems quite plausible, after all, we have seen above that there are n nice units of finite index in the unit group whose regulator varies p-adically. So, it suffice to show that the regulator is not zero in the entire family. This certainly seems like an easier problem, because it’s easier to prove a function is non-zero rather than the special value of a function (for example, by looking at the derivative). Still, I confess that I don’t know how to prove this.

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