Class number parity in pure cubic number fields

Question

Consider the family of pure cubic number fields $K = {\mathbb Q}(\sqrt[3]{m})$ 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 $\omega = \sqrt[3]{m}$; 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.

Answer 1

Billing (Beiträge zur arithmetischen Theorie der ebenen kubischen Kurven vom Geschlecht Eins, R. Soc. Scient. Uppsala (4) 11, Nr. 1. Diss. 165 S. Uppsala 1938; see Ian Connell's Handbook for elliptic curves for a modern presentation of the result) proved the following result:

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 ${\mathbb Q}(\sqrt[3]{m})$ 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 $r_2(K) \ge 1$.