I have the following conjecture involving a possible new formula for the class number of the quadratic field $\mathbb Q(\sqrt{(-1)^{(p-1)/2}p})$ with $p$ an odd prime.

Conjecture. Let $p$ be an odd prime and let $p^*=(-1)^{(p-1)/2}p$. Then the class number $h(p^*)$ of the quadratic field $\mathbb Q(\sqrt{p^*})$ coincides with the number
$$D(p):=\frac{(\frac{-2}p)}{2^{(p-3)/2}p^{(p-5)/4}}\det\left[\cot\pi\frac{jk}p\right]_{1\le j,k\le (p-1)/2},$$ where $(\frac{\cdot}p)$ is the Legendre symbol.

This is Conjecture 5.1 in my preprint arXiv:1901.04837. I have checked it for all odd primes $p<29$. Note that $h(p^*)=1$ for each odd prime $p<23$, and $h(-23)=3$.

Here I invite some of you to check this conjecture further. My computer cannot check it even for $p=29$.

Edit: F. Brunault computered $D(p)$ for $p=29,31,37,41,43,47$ and noted that the conjecture is false. However, I believe that $h(p^*)\mid D(p)$ for any odd prime $p$.

Now I explain why $D(p)\in\mathbb Q$ by Galois theory. The Galois group $\text{Gal}(\mathbb Q(e^{2\pi i/p})/\mathbb Q)$ consisits of those authormorphisms $\sigma_a$ $(1\le a\le p-1)$ with $\sigma_a(e^{2\pi i/p})=e^{2\pi ia/p}$. By Gauss' Lemma, $$\left(\frac ap\right)=(-1)^{|\{1\le j\le(p-1)/2:\ \{aj/p\}>1/2\}|}.$$ For $j=1,\ldots,(p-1)/2$ let $\pi_a(j)$ be the unique $r\in\{1,\ldots,(p-1)/2\}$ with $aj\equiv \pm r\pmod p$. Then, for $D=\det[\cot\pi\frac{jk}p]_{1\le j,k\le(p-1)/2}$, using $p$th roots of unity we get \begin{align}\sigma_a\left(\frac D{i^{(p-1)/2}}\right)=&\frac1{i^{(p-1)/2}}\det\left[\cot\pi\frac{ajk}p\right]_{1\le j,k\le(p-1)/2} \\=&\frac{(\frac ap)}{i^{(p-1)/2}}\det\left[\cot\pi\frac{\pi_a(j)k}p\right]_{1\le j,k\le(p-1)/2} \\=&\left(\frac ap\right)\left(\frac ap\right)^{(p+1)/2}\frac{D}{i^{(p-1)/2}}=\left(\frac ap\right)^{(p-1)/2}\frac{D}{i^{(p-1)/2}}\end{align} since $\text{sign}(\pi_a)=(\frac ap)^{(p+1)/2}$ as pointed by Pan in arXiv:0601026. Thus, if $p\equiv1\pmod4$ then $\sigma_a(D)=D$ for all $a=1,\ldots,p-1$ and hence $D\in\mathbb Q$. When $p\equiv3\pmod4$, we have $$\sigma_a\left(\frac D{\sqrt{p}}\right)=\left(\frac ap\right)\frac D{i^{(p-1)/2}}\cdot\frac1{\sigma_a(\sqrt{p^*})}.$$ Using Gauss' sums we see that $$\sigma_a(\sqrt{p^*})=\sum_{x=0}^{p-1}e^{2\pi iax^2/p}=\left(\frac ap\right)\sqrt{p^*}.$$ Therefore $\sigma_a(D/\sqrt p)=D/\sqrt p$ for all $a=1,\ldots,p-1$, and hence $D/\sqrt p\in\mathbb Q$.

  • 2
    $\begingroup$ Is it clear that this expression with det is integer? $\endgroup$ – Fedor Petrov Jan 17 at 11:00
  • 3
    $\begingroup$ Note that in my formula the class number $h(p)$ for $p\equiv1\pmod4$ is not related to the fundamental unit of $\mathbb Q(\sqrt p)$. For $p\equiv3\pmod4$, the determinant is not an integer. $\endgroup$ – Zhi-Wei Sun Jan 17 at 13:34

If by

"the class number $h(p^*)$ of the quadratic field $\mathbb{Q}(\sqrt{p^*})$"

you mean

"the minus class number $h^{-}$ of $\mathbf{Q}(\zeta_p)$"

and if by

" a possible new formula for the class number"

you mean

"an elementary formula for $h^{-}$ known to Kummer (obtained by considering the ratio of the zeta functions of $\mathbf{Q}(\zeta_p)$ and the totally real subfield $\mathbf{Q}(\zeta_p)^{+}$ at s = 1)"

Then you are correct.

  • 2
    $\begingroup$ Could you add a reference for this (especially for the connection to Kumner)? $\endgroup$ – Peter LeFanu Lumsdaine Jan 17 at 18:35
  • 22
    $\begingroup$ Why so snarky? Formulas like this lose none of their magic just by being incorrectly phrased and over a century old. $\endgroup$ – Dustin Clausen Jan 17 at 19:24
  • 10
    $\begingroup$ @DustinClausen Your second comment is certainly true. As to the first, the phrasing should be viewed in the context of the quality and frequency of the OPs recent questions, together with the self-promotional reference to the OP's own paper, and finally due to being written by a curmudgeon. $\endgroup$ – user134696 Jan 18 at 0:32
  • 2
    $\begingroup$ What you say is interesting (more interesting than most of the OP's questions). Can you please provide more detail? $\endgroup$ – GH from MO Jan 18 at 5:18
  • 7
    $\begingroup$ @user134696 you of course do not, but maybe somebody else do? After all, this site is exactly for concerning other's failures. $\endgroup$ – Fedor Petrov Jan 18 at 8:26

The conjecture is not true, as some examples show.

Let $D(p)$ denote your number. For primes $p \equiv 1 \textrm{ mod } 4$, we have $D(29)=8$, $D(37)=37$, $D(41)=121$ while $h(29)=h(37)=h(41)=1$.

For primes $p \equiv 3 \textrm{ mod } 4$, we have $D(23)=3$, $D(31)=9$, $D(43)=211$, $D(47)=695$ while $h(23)=h(31)=3$, $h(43)=1$, $h(47)=5$.

The following Pari/GP script calculates $D(p)$:


(This is a floating-point computation, but it is easy to be rigorous by replacing the entries of the matrix by algebraic numbers). It seems that $D(p)$ grows fast.

EDIT. For primes $p \equiv 1 \textrm{ mod } 4$, divisibility does not hold, for example $h(229)=h(257)=3$ but $D(229) \equiv D(257) \equiv 1 \textrm{ mod } 3$.

For primes $p \equiv 3 \textrm{ mod } 4$, one should be able to relate $D(p)$ with the minus class number $h(\mathbb{Q}(\zeta_p))^-$ as alluded to by user134696 using the analytic class number formula and the fact that $D(p)$ looks like a group determinant on $(\mathbb{Z}/p\mathbb{Z})^\times$, so should be a product over Dirichlet characters mod $p$.

The point is that Dirichlet's class number formula relates $h(p^*)$ with $L(\chi_p,1)$ where $\chi_p$ is the Legendre symbol, but $\chi_p$ is even in the case $p \equiv 1 \textrm{ mod } 4$ so it should have nothing to do with $h(\mathbb{Q}(\zeta_p))^-$ in this case.

  • $\begingroup$ Thank you for your computation. Though $D(p)$ is not exactly equal to $h(p^*)$, it seems that $h(p^*)$ divides $D(p)$. Can you find an odd prime $p$ with $h(p^*)\nmid D(p)$? $\endgroup$ – Zhi-Wei Sun Jan 17 at 21:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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