2
$\begingroup$

In the paper Zum Beweise des Starkschen Satzes Siegel considers the function

$$L_q(s)=\sum_{n=1}^{\infty}\left(\frac{q}{n}\right)n^{-s},$$

where $q$ is a discriminant of a quadratic number field and the character is the Kronecker symbol. Then he writes that "according to Dirichlet" we have, in case $G>0$,

$$L_G(1)=2G^{-1/2}h_G\log \varepsilon_G,$$

where $h_G$ is the corresponding class number and $\varepsilon_G$ the fundamental unit.

However, according to the book Zetafunktionen und quadratische Körper the formula reads

$$h_G=\frac{G^{1/2}}{\log \varepsilon_G}L_G(1).$$

Which one is correct? Am I making some mistake?

$\endgroup$
6
  • $\begingroup$ you mean where does the factor of two in the first formula come from? $\endgroup$ – Carlo Beenakker Mar 21 '20 at 16:08
  • $\begingroup$ @CarloBeenakker Yes, that is the problem. $\endgroup$ – Shimrod Mar 21 '20 at 16:08
  • 1
    $\begingroup$ Dirichlet's class number formula is quoted with this factor of two in several other sources, for example here. $\endgroup$ – Carlo Beenakker Mar 21 '20 at 16:27
  • $\begingroup$ Maybe it would be wiser to read Dirichlet's proof itself. $\endgroup$ – Sylvain JULIEN Mar 21 '20 at 16:52
  • 1
    $\begingroup$ Siegel is always right. $\endgroup$ – GH from MO Mar 23 '20 at 0:16
7
$\begingroup$

Dirichlet proved his class number formula for quadratic forms; in particular he was working with class numbers $h^+$ in the strict sense, and his unit $\varepsilon$ was the fundamental solution of the Pell equation $t^2 - Du^2 = 1$, not the fundamental unit $\eta$ of the corresponding number field. The relation $$ \eta^{2h} = \varepsilon^{h^+} $$ encodes the two cases

  • $\varepsilon = \eta$, $h^+ = 2h$
  • $\varepsilon = \eta^2$, $h^+ = h$.
$\endgroup$
1
  • $\begingroup$ Thank you for your answer! $\endgroup$ – Shimrod Mar 24 '20 at 16:30

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.