# Class number parity in pure cubic number fields

Consider the family of pure cubic number fields $K = {\mathbb Q}(\sqrt{m})$ for $m = a^3 \pm 3$.

Proposition. If $4 \mid a$ and $m$ is cubefree, then the class number of $K$ is even.

Proof. Let $\omega = \sqrt{m}$; the element $\alpha = a - \omega$ has norm $\pm 3$. Since $3$ is completely ramified, the element $\varepsilon = \alpha^3/3$ is a unit.

If $m = a^3 + 3$, then $\varepsilon = 1 - 3a^2\omega + 3a\omega^2$. If $4 \mid m$, then $\varepsilon \equiv 1 \bmod 4$, hence $K(\sqrt{\varepsilon}\,)/K$ is an unramified quadratic extension.

Experiments seem to suggest that if $m = a^3+3$ and $a \equiv 2 \bmod 4$, then $h$ is also even, but there is no explanation, class field theoretic or otherwise. In fact, the class number is even for all cubefree values of $m$ for $a = 2, 4, \ldots, 2 \cdot 88$, but is odd for $a = 2 \cdot 89$.

This cannot be an accident; the parity of the class number in the case $m = a^3 - 3$ for $a \equiv 2 \bmod 4$ shows a more typical (i.e. more random) behaviour in that the class number is odd quite often.

Question: How can this behaviour in the case $m = 8a^3+3$ be explained?

My first guess would be that, for fields in this family, there is a family of ideals ${\mathfrak a}$ such that ${\mathfrak a}^2$ is principal, but I can't seem to find anything in this direction.

Edit. Dror's comment made me look at the family of elliptic curves $y^2 = x^3 - m$. These have rank $\ge 1$, and by the parity conjecture rank $\ge 2$. An inequality due to Billing now shows that $K$ has even class number. For details, see this pdf file. Actually, Paul Monsky stumbled across something similar for pure quartic fields; see here.

• Not helping really, but confirming the peculiarity and unexplainedness of the phenomenon: since the early-mid 1970s, I've seen (and created, myself) a few examples wherein unramified abelian extensions (especially quadratic, or 2,2,... type) are explicitly/ad-hoc-ly constructed, proving (by classfield theory) something corresponding about the class group of the base. The goofiness of the methods does not portray any deeper principle... as in your argument... But/and empirical evidence (as yours) often seems to show that various ad-hoc arguments are... definitive? I don't get it... – paul garrett Jul 12 '11 at 0:51
• Let $E_a$ be the elliptic curve $Y^2=X^3+(8a^3+3)$. A rational point $(x/z^2,y/z^3)$ gives rise to an ideal $I=(y,x+z^2\sqrt{8a^3+3})$ of order dividing 2. Liverance's 95' paper shows that the root number of $E_a$ is $-c_3(-1)^{\omega(8a^3+3,2,3)}$, where $c_3$ is 1 or $-1$ according to whether $a=1\ (\text{mod}\ 3)$ or not, and $\omega(n,2,3)$ is the number of distinct primes dividing $n$ that are $2\ (\text{mod}\ 3)$. Up to 90 there are 6 values of $a$ with positive root number: 44, 56, 68, 69, 86, 89. 69 and 86 have positive rank, the others don't. In fact, under 200000, only 9742 have.. – Dror Speiser Jul 13 '11 at 20:38
• .. positive root number. The number of $a$ with positive root number seems to converge to around 1/20, and this should be provable. But, the above is a bit disingenuous: some of the $I$'s constructed can be principal, and hence don't directly account for the even class number. My Magma broke when trying to compute the Mordell-Weil group of $E_{13}$, so I really don't know what to expect. – Dror Speiser Jul 13 '11 at 20:49
• @Dror: the elliptic curve connection does it (I should have thought of that myself, but ...). I'll attach a pdf file to my question when I've written down what I have so far. Thanks! – Franz Lemmermeyer Jul 20 '11 at 8:48
• @Franz: cool, looking forward to reading it. The reason I thought of elliptic curves was the sheer analogie between the situation above and some results, pending parity conjecture and otherwise unexplained, described by Tim Dokchitser in his Sage Days 22 lecture "Parity Predictions" (available at MSRI). These results shocked me; still do. – Dror Speiser Jul 29 '11 at 23:19

Let $f(x) = x^3 + ax^2 + bx + c \in {\mathbb Z}[x]$ be irreducible, and consider the elliptic curve $E: y^2 = f(x)$. Let $K$ be the cubic number field generated by a root $\alpha$ of $f$, and let $E_K$ be its unit group. Write $({\mathcal O}_K: {\mathbb Z}[\alpha]) =: m_f^2$. Then $$r \ \le \ r_2(K) + r_E(K) + 2n_+ + n_-,$$ where $r$ is the Mordell-Weil-rank of $E({\mathbb Q})$, $r_2(K)$ is the $2$-rank of the ideal class group of $K$, $r_E(K)$ is the ${\mathbb Z}$-rank of the unit group $E_K$ of $K$, $n_+$ is the number of primes $p \mid m_f$ that split in $K$,
and $n_-$ is the number of primes $p \mid m_f$ that decompose as $p {\mathcal O}_K = {\mathfrak p}{\mathfrak p}'$ or as $p {\mathcal O}_K = {\mathfrak p}^2 {\mathfrak p}'$.
If ${\mathbb Q}(\sqrt{m})$ is a pure cubic field with $m \not\equiv \pm 1 \bmod 9$ cubefree, then the index is trivial, and $n_+ = n_- = 0$ provides us with the bound $$r \ \le \ r_2(K) + 1.$$
On the other hand, the parity conjecture (see the article by Liverance pointed out by Dror) implies that the Mordell-Weil rank of $E$ is even for squarefree values of $m = 8b^3 + 3$, and the family of nontorsion points $$P_b\Big( \frac{2b^3+1}{b^2}, \frac{3b^3+1}{b^3} \Big)$$ shows that $r \ge 1$. Thus the parity conjecture implies $r \ge 2$, and Billing's bound finally gives $r_2(K) \ge 1$.