Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In their paper Computing Systems of Hecke Eigenvalues Associated to Hilbert Modular Forms, Greenberg and Voight remark that

...it is a folklore conjecture that if one orders totally real fields by their discriminant, then a (substantial) positive proportion of fields will have strict class number 1.

I've tried searching for more details about this, but haven't found anything.

Is this conjecture based solely on calculations, or are there heuristics which explain why this should be true?

share|improve this question
    
Are you bounding the degree of the field? If you don't then it's hard to make sense of this question, because I don't see why there will only be finitely many fields of a given discriminant. –  Kevin Buzzard Jan 11 '10 at 20:06
    
@Buzzard: Yes, the degree of the field is bounded. The paper works with a totally real field F of degree n and it is assumed throughout that F has strict class number 1. I believe that this remark is meant to justify this restriction. – Ben Linowitz 0 secs ago –  Ben Linowitz Jan 11 '10 at 20:17
    
Isn't it a classical result of Hermite that there are finitely many fields of bounded discriminant? –  Dror Speiser Jan 11 '10 at 21:53
    
No :-) In fact it's a result of Golod and Schaferevich that there aren't :-) You need to bound the degree too. Unless I've slipped up. –  Kevin Buzzard Jan 11 '10 at 23:14
3  
@buzzard: in a Golod-Shafarevic tower, the discriminant is exponential in the degree: it's the so-called root discriminant that's constant. (Describing the set of number fields with bounded root discriminant is an extremely interesting and mysterious problem!) The set of number fields with discriminant < X, on the other hand, is indeed finite. In an appendix to a paper of Belolipetsky, Venkatesh and I show that the log-size of this set is at most (log X)^{1+eps} (the finiteness is much older, as Ben Linowitz points out.) –  JSE Jan 12 '10 at 1:15
show 4 more comments

4 Answers 4

up vote 6 down vote accepted

One heuristic is the following: if one imagines that the residue at $s = 1$ of the $\zeta$-function doesn't grow too rapidly, then the value is a combination of the regulator and the class number. I don't know any reason for the regulator not to also grow (there are a lot of units, after all!), and hence one can imagine that the class number then stays small.

This is part of a general heuristic that in random number fields there tends to be a trade-off between units and class number, so especially in the totally real case, when there are so many units, the class number should often be 1.

I learnt some of these heuristics from a colleague of mine who regards it more-or-less as an axiom that a random number field has very small class number. I think this view was formed through a mixture of back-of-the-envelope ideas of the type described above, together with a lot of experience computing with random number fields. So the answer to your question might be that it is a mixture of heuristics and computations.

Incidentally, in the real quadratic case, it is compatible with Cohen--Lenstra, but I think it goes back to Gauss. Also, there are generalizations of Cohen--Lenstra to the higher degree context, and I'm pretty sure that they are compatible with the class group/unit group trade-off heuristic described above.

share|improve this answer
    
A totally real field should have strict class number 1 if and only if it has class number 1 and units of every possible sign combination, so assuming that Cohen-Lenstra generalizes to higher degrees (as you say),I suppose I have my answer. Thanks! –  Ben Linowitz Jan 11 '10 at 21:31
add comment

Maybe it's worth a word about why Cohen-Lenstra predicts this behavior. Suppose K is a field with r archimedean places. Then Spec O_K can be thought of as analogous to a curve over a finite field k with r punctures, which is an affine scheme Spec R. Write C for the (unpunctured) curve. Then the class group of R is the quotient of Pic(C)(k) by the subgroup generated by the classes of the punctures -- or, what is the same, the quotient of Jac(C)(k) by the subgroup generated by degree-0 divisors supported on the punctures. (This last subgroup is just the image of a natural homomorphim from Z^{r-1} to Jac(C)(k).)

The Cohen-Lenstra philosophy is that these groups and the puncture data are "random" -- that is, you should expect that the p-part of the class group of R looks just like what you would get if you chose a random finite abelian p-group (where a group A is weighted by 1/|Aut(A)|) and mod out by the image of a random homomorphism from Z^{r-1}. (There are various ways in which this description is slightly off the mark but this gives the general point.)

It turns out that when r > 1 the chance is quite good that a random homomorphism from Z^{r-1} to A is surjective. In fact, the probability is close enough to 1 that when you take a product over all p you still get a positive number. In other words, when r > 1 Cohen-Lenstra predicts a positive probability that the class group will have trivial p-part for all p; in other words, it is trivial. (In fact, it predicts a precise probability, which fits experimental data quite well.)

When r = 1, on the other hand, the class group is just A itself, and the probability its p-part is trivial is on order 1-1/p. Now the product over all p is 0, so one does NOT expect to see a positive proportion of trivial class groups. And in fact, when there is just one archimedean place -- i.e. when K is imaginary quadratic -- this is just what happens!

share|improve this answer
add comment

I think the best way for one to become convinced that class numbers of real quadratic fields tend to be small, is to look at the continued fraction expansion of $\sqrt{D}$.

The period length of the continued fraction is about the regulator (up to a factor of $\log{D}$). One can easily compute some random continued fractions, and see that for most numbers, the length really is a small factor away from $\sqrt{D}$.

Numbers that have small continued fraction period length are very scarce. I believe it is not hard to prove that the amount of numbers up to $X$ that have period length of $\sqrt{D}$ less than a fixed integer $n$ is $O(X^{1-\epsilon})$ for some $\epsilon > 0$ (I think 1/2 should always work).

In a sense (very strict actually) the regulator counts how many numbers $n$ with $|n| < 2\sqrt{D}$ can be represented as $n = x^2-Dy^2$ for some integers $x, y$. Well, if $D$ is large and random, then it seems reasonable that many should. So the regulator should be around $\sqrt{D}$, and hence, by Dirichlet's class number formula, the class number should be very small.

Once you become convinced of the real quadratic case, the rest immediately follow, because you already believe things people said while waving their hands a lot. (This is a general philosophy in mathematics)

share|improve this answer
add comment

For ordinary class number 1 in the real quadratic case, see Cohen and Lenstra's Heuristics on Class Groups of Number Fields https://openaccess.leidenuniv.nl/retrieve/2845/346_069.pdf

Maybe it's not so much of a jump from their to see a heuristic arguing that a substantial positive portion have strict class number 1.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.