For a number field $K/\mathbb{Q}$, put $h_K, R_K$ respectively for the class number (of the ring of integers $\mathcal{O}_K$ of $K$) and the regulator of $K$ respectively. Moreover, put $\zeta_K(s)$ for the Dedekind zeta function of $K$. We then have the class number formula:

$$\displaystyle \lim_{s \rightarrow 1} (s-1) \zeta_K(s) = \frac{2^{r_1(K)} (2\pi)^{r_2(K)} h_K R_K}{w_K \sqrt{|\Delta_K|}},$$

where $w_K$ is the number of roots of unity inside $K$, $r_1(K)$ the number of real embeddings of $K$, $r_2(k)$ the number of pairs of complex embeddings, and $\Delta_K$ the discriminant of $K$.

By the Brauer-Siegel theorem, we know that if $\mathcal{K}_n$ is the sequence of cyclic degree $n$ fields over $\mathbb{Q}$, ordered by discriminant, then

$$\displaystyle \lim_{K \in \mathcal{K}_n} \frac{\log(h_K R_K)}{(\log |\Delta_K|)/2} = 1.$$

However, the logarithm hides a lot of the information about the precise asymptotic behaviour of $\Delta_K$ and $h_K R_K$.

For $n = 2$ Mertens and Siegel obtained a more precise asymptotic relation, namely the asymptotic formulae:

$$\displaystyle \sum_{-X < D < 0} h_{\mathbb{Q}(\sqrt{D})} R_{\mathbb{Q}(\sqrt{D})} = c_1 X^{3/2} + O \left(X \log X\right), $$

$$\displaystyle \sum_{0 < D < X} h_{\mathbb{Q}(\sqrt{D})} R_{\mathbb{Q}(\sqrt{D})} = c_2 X^{3/2} + O \left(X \log X\right)$$

for absolute constants $c_1, c_2 > 0$.

Is there an analogous formula for $n = 3$? That is, let $\mathcal{C}_3$ be the set of cyclic cubic fields. Is there an asymptotic formula for

$$\displaystyle \sum_{m \leq X} \sum_{\substack{K \in \mathcal{C}_3 \\ D_K = m^2 }} h_K R_K?$$

  • $\begingroup$ How did they simplify $\sum_{0< D < X} h_{\mathbb{Q}(\sqrt{D})} R_{\mathbb{Q}(\sqrt{D})}$ ? $\endgroup$
    – reuns
    May 7, 2019 at 18:33
  • $\begingroup$ Note that these formulas were also known to Gauss: mathoverflow.net/questions/109330/… though he did not provide proofs (and IIRC he had an incorrect error term for the real quadratic case). $\endgroup$ May 7, 2019 at 20:10


Your Answer

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

Browse other questions tagged or ask your own question.