Why is the class number of \(\mathbb {Q}(\root 3\of {11})\) even? (Q2856917)

This paper is written in a very nice style, almost being a detective story: We can see that the class numbers of the pure cubic fields \(K_b=\mathbb {Q}(\root 3\of m)\), where \(b\in \mathbb {Z}\) and \(m=8b^3+3\) is cube-free, have a tendency to be even. But what is responsible for that?NEWLINENEWLINEIf \(b\) is even then the author shows almost immediately that \(K_b(\sqrt {\varepsilon })/K_b\), where \(\varepsilon =-13/(2b-\root 3\of m)\), is an unramified quadratic extension. But the case of \(b\) being odd appears to be much more difficult. The main result of this paper is the following theorem: Let \(b\) be an odd integer such that \(m=8b^3+3\) is square-free. Then the \(2\)-rank \(s\) of the \(2\)-class group \(\text{Cl}_2(K_b)\) of \(K_b=\mathbb {Q}(\root 3\of m)\) and the rank \(r\) of the Mordell-Weil group \(E(\mathbb {Q})\) of \(E\: y^2=x^3-m\) satisfy the inequality \(r\leq s+1\).NEWLINENEWLINEAs the author explains, the tendency of the class numbers of \(K_b\) to be even could be explained by citing the parity conjecture and Billing's bound (which is even more general than the previous theorem). But the presented proof is different, it is given by an explicit construction of subfields of the Hilbert class field of \(K_b\) using rational points on the elliptic curve.