# Is this Siegel's formula correct?

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?

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

## 1 Answer

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

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