0
$\begingroup$

Gauss's class number one problem for imaginary quadratic fields, now a theorem due to Heegner, Stark, and Baker (independently), asserts that the set of imaginary quadratic fields having class number equal to one correspond exactly to the discriminants $\{-3,-4,-7,-8,-11,-19,-43,-67,-163\}$. Of course, except for $-4,-8$ each of the discriminants which appear are prime: this is of course because genus theory tells us that otherwise the 2-part of the class group is non-trivial (and in particular the class number is even), and hence cannot equal one.

Since we have such concrete information coming from genus theory in for the 2-part of the class group, a natural question is the following: for which fundamental (negative) discriminants $-d$ is it the case that $h_{-d} = h_{-d}[2]$? Here $h_{-d}$ is the class number of the imaginary quadratic field $\mathbb{Q}(\sqrt{-d})$ and $h_{-d}[2]$ is the size of the 2-part of the class group.

If we let $S^{-}$ denote the set of fundamental discriminants $-d$ with $h_{-d} = h_{-d}[2]$, then clearly

$$\displaystyle \{-3,-4,-7,-8,-11,-19,-43,-67,-163\} \subseteq S^{-}.$$

Moreover, by the Brauer-Siegel theorem and the fact that imaginary quadratic fields have no units of infinite order (i.e., the regulator is bounded) it follows that $h_{-d} \gg_\varepsilon d^{1/2 - \varepsilon}$. Since genus theory tells us that the 2-part of the class group is essentially the number of divisors of $d$, which gives $h_{-d}[2] = O(d^{1/\log \log d})$, it follows that $S^{-}$ is surely finite. However the Brauer-Siegel theorem is ineffective, so it still requires some work to compute the set $S^{-}$.

Is it known exactly what $S^{-}$ is? If not, what progress has been made towards computing it?

$\endgroup$
3
  • $\begingroup$ In terms of progress: presumably you know about the work on classifying $-d$ with $h(-d) = n$ for small $n$. This allows you to determine much more of $S^-$. $\endgroup$
    – Kimball
    Aug 4, 2022 at 0:35
  • $\begingroup$ PS You seem to be missing the class number criterion in your definition of $S^-$. $\endgroup$
    – Kimball
    Aug 4, 2022 at 0:36
  • $\begingroup$ @Kimball yes I noticed that; thank you for pointing it out! $\endgroup$ Aug 4, 2022 at 1:17

1 Answer 1

1
$\begingroup$

This question is actually older than Gauss's class number question. These numbers a called Euler's idoneal numbers and they are of interest because this is exactly when the prime numbers of the form $x^2 + d y^2$ are determined by congruence conditions (equivalently when the Hilbert class field is abelian over $\mathbb{Q}$. They are determined under the assumption of GRH but not otherwise. See https://www.mast.queensu.ca/~kani/papers/idoneal-f.pdf or https://en.wikipedia.org/wiki/Idoneal_number for more references.

$\endgroup$
1
  • 1
    $\begingroup$ Even without GRH, it is know there exists at most one such 'exceptional' fundamental discriminant in addition to the known examples. (This is due to Peter Weinberger.) $\endgroup$
    – Stopple
    Aug 4, 2022 at 2:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .