Possible extensions of a conjecture (perhaps now a theorem) proposed by Sylvester

Question

Let $\omega$ be a primitive cube root of unity, $\mathcal O$ be $\Bbb Z[\omega]$, and $K$ be $\Bbb Q[\omega]$. I am asking about the $K$-rank of the elliptic curve $x^3 + y^3 = M$ for certain cube-free $M$ in $\mathcal O$ that are not in $\Bbb Z$.

Corresponding questions for certain cube-free $M$ in $\Bbb Z$, about the $Q$-rank of $x^3 + y^3 = M$ were asked and partly answered by Sylvester implicitly using infinite descent in $\mathcal O$. He showed that if $M$ is a prime congruent to $2$ or $5$ mod $9$ (or its square) the $\Bbb Q$-rank of the curve is $0$, and conjectured that it is $1$ when $M$ is a prime (or the square of a prime) congruent to $4$, $7$, or $8$ mod $9$.

I've recently shown the following.

Let $p$ be a prime congruent to $4$ or $7$ mod $9$, and let $M$ be an irreducible factor of $p$ (or the square of such a factor) in $\mathcal O$. Then $$ x^3 + y^3 = M $$ has no $K$-rational solutions (apart from the three $K$-rational points at infinity).

THE QUESTIONS:

Suppose $M$ is either $\omega p$, or $\omega p^2$ with $p$ a prime congruent to $8$ mod 9, or is an irreducible factor in $\mathcal O$ of a prime congruent to $1$ mod $9$ (or the square of such a factor).

• What Selmer-like upper bounds are guaranteed for the $K$-rank of the curve?
• What can be said assuming the Swinnerton-Dyer conjecture?
• What does experimental evidence suggest about the K-rank?

(The LMFDB site gives next to no information.)

(The Sylvester conjectures are now perhaps theorems, due to unpublished work of Elkies and Kriz. Some self-promotion: My proofs of the rather simple rank $0$ results are very elementary and can be presented in an undergrad number theory course.)

I'm not feeling too optimistic about this now. Since there's CM by Z[omega] for these curves, the Mordell-Weil groups are Z[omega] modules and their ranks as Z-modules are even. And Heegner point constructions,I understand, only work when the rank is 1. Might the techniques Zagier and Villegas employ for primes that are 1 mod 9 lead anywhere? I've gotten as far as I can, I think, with non-existence results; higher descents would involve too heavy algebraic number theory for me. But perhaps the computer-savvy could look for K-rational points on these curves and see if a pattern emerges in the cases I asked about. (Apart from the lone non-existence result I gave in my (modified) answer, when p is a prime congruent to 1 mod 9, and M = m, m^2, or their negatives, and m is 1 mod 3, but not 1 mod 9, with norm p). I feel as if I'm back in the days when those with nothing better to do searched for solutions to (x^3) + (y^3) = p, but at least the computers and programmers are better now.

Answer 1

This is a very partial answer when p is 1 mod 9, together with some not well substantiated guesses. There are, up to GL(2)(Z) equivalence of forms, two primitive forms of discriminant -243, the principal form, which can be taken to be (1,9,81), and the form (7,3,9). My methods show that if p is 1 mod 9 and represented by the second form, and pi is an irreducible factor of p, then no associate of pi or pi^2 is a sum of two cubes in K. When p is represented by the first form the situation is less clear. The first few such p are 73, 271, 307, 523 and 577. For a number of these p I can write pi and/or pi^2 as a sum of two cubes; in each case pi (or pi^2) is congruent to 1 mod 3 in O. Perhaps this situation holds whenever p is represented by the first form? (By the way, a theorem of Gauss and/or Eisenstein says that a p that is 1 mod 9 is represented by the first form if and only if 3 is a cube in Z/p--does anyone know an elegant proof of this?) I'd also be grateful if someone would fix an omission of mine, adding Birch to accompany Swinnerton-Dyer. And I'd like to know what the parity conjecture predicts for the K-rank of our elliptic curve, in the above cases, and when M is the product of omega by a prime congruent to 8 mod 9 (or by the square of the prime.)

EDIT

My statement above is not correct. Suppose for example that p is 109, and so is not represented by the principal form (1,9,81). For convenience of notation write u and v for omega and omega^2. Then my argument shows that neither 5u - 7v nor 5v - 7u, (or their negatives) is a sum of two cubes in K. But for the other 8 elements of norm 109, a little experimentation allows one to write down sum of two cubes representations---the same applies when p is 379 with 5u-7v replaced by -22u-7v. And the guess I hazarded for p represented by (1,9,81) isn't quite right either. Though the divisors in O of 73,271,307,523 and 577 that are 1 or -1 mod 3 all are sums of two cubes, all 12 elements of O of norm 577 admit such representations.

What I do show is that if p is 1 mod 9 and m has norm p, then if m is 1 mod 3, but is not 1 mod 9, then neither m nor m^2 (or their negatives) is a sum of two cubes. But I now believe that this congruence restriction is the only restriction to the existence of a sum of two cubes representation for m whose norm is a prime that is 1 mod 9. At any rate there don't appear to be any local obstructions to the 3-descent when these restrictions hold. Of course it must be next to impossible to construct solutions in general--but might the existence follow from the parity conjecture?

Edit 2 (More mistakes)

Once again, I claimed to have proved more than I really did, though it's likely that my claims are true.

When p=19, and m is 5u + 2v, then although m is not a sum of two cubes, both um and vm are. And though m^2 isn't, u*(m^2) is, but I don't know about v*(m^2).

Similarly when m is 5v-7u of norm 109 or -22u-7v of norm 379, then though m isn't such a sum, and um is, I don't know about vm. When m is -19u+8v of norm 577, I know that m and um are such sums but can make no claims about vm. My trouble is that I assumed that when m is 1 or -1 mod 3, then if um is a sum of two cubes, the same must hold for vm---I forgot that m was not in Z. More empirical investigations to see what's going on seem to be in order, but they're beyond my capacity. Might Noam Elkies be interested?