Simplest cubic fields with small class number

Abstract

Let $m\in \mathbb{Z}$ be an integer and $L_m=\mathbb{Q}(\hat{I}\pm )$ be the simplest cubic field with class number $h_m$ and conductor ${\mathfrak{f}}_m$ where $\hat{I}\pm$ is a root of $f_m(X)=X^3-m{X}^2-(m+3)X-1$. Let ${\mathcal{O}}_{L_m}$ be the ring of integers of $L_m$. By using PARI/GP, we confirm that if $[{\mathcal{O}}_{L_m}:\mathbb{Z}[\hat{I}\pm ]]=1$ $($resp. $3$, $27$$)$, i.e. $m^2+3m+9={\mathfrak{f}}_m$ $($resp. $3{\mathfrak{f}}_m$, $27{\mathfrak{f}}_m$$)$, then there exist exactly $581$ (resp. $80$, $142$) integers $m\ge -1$ such that $h_m\le 1000$. We also show that if $-1\le m\le 10^7$, then $h_m<16$ holds for $138=26+31+11+10+36+21+3$ integers $m$. More precisely, there exist $26$ $($resp. $31$, $11$, $10$, $36$, $21$, $3$$)$ integers $m$ with $-1\le m\le 10^7$ such that $h_m=1$ $($resp. $3$, $4$, $7$, $9$, $12$, $13$$)$ which are given explicitly.

23 pages. We would like to thank Stephane Louboutin for helpful explanations concerning Theorem 1.8 which improved the proof of Theorem 1.13. For Theorem 1.14, the result is also improved as integers m up to 10^7 instead of 4x10^6. Only one additional case was found as in Table 3