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

Dirichlet showed that if q is a prime number and q == 3 mod 4 then the sum of the nonresidues of q minus the sum of the quadratic residues of q in the range 1, 2, …, q − 1 is an odd multiple of q. What coverage of the odd numbers is given by these odd mulitiples?

share|improve this question
add comment

2 Answers

It is very likely that every (positive) odd number is covered by a sum of this type.

As Robin Chapman points out, this is equivalent to asking, for a given odd number $h$, whether there exists an odd prime $q\equiv3\pmod{4}$ such that $\mathbb{Q}(\sqrt{-q})$ has class number $h$. Let $N(h)$ be the number of quadratic imaginary number fields with class number $h$ (for odd $h$) -- note that such a field is already necessarily of the form $\mathbb{Q}(\sqrt{-q})$ for $q\equiv3\pmod{4}$ by genus theory. This rephrases your question as "Is $N(h)>0$ for all odd $h$"?

By the calculation of mark Watkins mentioned in Stoppie's answer, it is known that $N(h)>0$ for all $h\leq 100$. More importantly, $N(h)$ gets rather large -- $N(1)=9$ is Heegner-Stark, and this is the smallest value of $N(h)$ in this range. For instance, $N(h)>100$ for all $h>37$, and $N(99)=289$. So we'd like an assurance of some sort of a rough upward trend to guarantee that $N(h)$ does not hit zero at some point down the road. Enter Soundararajan's article "The number of imaginary quadratic number fields with a given class number", which studies asymptotics of $N(h)$. Soundararajan remarks that $N(h)$ should be on the order of $h$, and conjectures more precisely that $$ \frac{h}{\log h}\ll N(h)\ll h\log h. $$ This is accompanied by a probabilistic argument as to why this is a reasonable conjecture. I have not worked through the analysis, but it seems likely to me that if one's only goal was to prove that $N(h)>0$ for all $h$ (a lower bound was not a goal of the paper), this heuristic argument could be made sufficiently rigorous, especially when combined with the data for $h<100$ above, to rule out pathologies for small inputs.

share|improve this answer
add comment

These odd multiples are (save for $q=3$) the class numbers $h_{-q}$ of the fields $\mathbb{Q}(\sqrt{-q})$. (This follows from the analytic class number formula,)

By the Brauer-Siegel theorem, $\log h_{-q}\sim\log\sqrt{q}$. From this it looks plausible that all possible odd $h$ could occur. I'm not aware of any proof that this is the case (nor of any potential counterexample).

share|improve this answer
Is every natural number $h$ the class number of a complex quadratic field? This does seem very plausible, but I'm not aware of any existing conjecture. It's not mentioned in Narkiewicz' "Elementary and Analytic Theory of Algebraic Numbers", which is quite detailed on open problems. What's the largest $h$ for which this is still open? (I know only that $h\ge 100$.) –  Stopple Oct 5 '10 at 21:36
Do the Cohen-Lenstra heuristics imply anything about the natural numbers that show up as class numbers of complex quadratic fields? –  Gerry Myerson Oct 5 '10 at 23:43
@Gerry: I don't think so. The C-L heuristics predict statements about p-ranks of class groups, so I think you'd have a hard time getting C-L to predict that a class number was exactly a given natural number. –  Cam McLeman Oct 6 '10 at 1:00
add comment

Your Answer


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.