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Probably this is well known to those who know it.

Got an argument and numerical support that over number fields elliptic curves in minimal models might have unbounded number of integral points, the number depending on the degree of the field.

Set $f(x)=x^3+ax+b$ and consider the curve $E: y^2=f(x)$.

Chose $x_1 \ldots x_n$ such that $f(x_n)$ is prime and work in $K=\mathbb{Q}[\sqrt{f(x_1)},\ldots\,\sqrt{f(x_n)}]$.

$E$ has the obvious $n$ points $(x_n,\sqrt{f(x_n)})$.

Experimentally for $f(x)=x^3-x+1$ over $\mathbb{Q}[\sqrt{7},\sqrt{61},\sqrt{211},\sqrt{337},\sqrt{991}]$ the five points are linearly independent according to sage so the rank is at least $5$.

Computing the absolute field is not efficient for me.

Over the rationals there is a conjecture relating the number of integral points to the rank, is there a similar conjecture for number fields?

Is there an example (with few primes) when in this construction the points are linearly dependent?

The same argument works for higher genus.

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  • $\begingroup$ If the field $\mathbb{Q}(\sqrt{f(x_1)},\ldots,\sqrt{f(x_r)})$ has degree $2^r$, then it's an elementary exercise to prove that the points $P_i=(x_i,\sqrt{f(x_i)})$ are independent in $E(\overline{\mathbb{Q}})$. Just apply various elements of Galois to a linear combination $\sum n_iP_i$ to change some, but not all, of the signs of the $n_i$, add, repeat. Eventually you'll end up showing that the $P_i$ are torsion. (It's much more interesting to get rank $r$ over fields of degree $O(r)$, rather than degree $O(2^r)$.) $\endgroup$
    – Joe Silverman
    Commented Apr 28, 2014 at 17:58

2 Answers 2

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joro asks: "Over the rationals there is a conjecture relating the number of integral points to the rank, is there a similar conjecture for number fields?"

Yes, the conjecture is that for a given field $K$, on a quasi-minimal Weierstrass equation for $E/K$, the number of $S$-integral points satisfies $$ \# E(R_S) \le C(K)^{1+\#S+\text{rank} E(K)}. $$ This is known to be a consequence of the $abc$-conjecture for $K$; see [1]. It is also known unconditionally in the weaker form $$ \# E(R_S) \le C(K,\nu(j_E))^{1+\#S+\text{rank} E(K)}, $$ where $\nu(j_E)$ is the number of primes $\mathfrak{p}$ of $R_K$ such that $\text{ord}_{\mathfrak{p}}(j_E)<0$; see [2].

[1] M. Hindry, J.H. Silverman, The canonical height and integral points on elliptic curves, Invent. Math. 93 (1988), 419-450.

[2] J.H. Silverman, A quantitative version of Siegel's theorem: Integral points on elliptic curves and Catalan curves, J. Reine Angew. Math. 378 (1987), 60-100.

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If the 5 points had a linear dependence, their coordinates could not generate a (Z/2Z)^5 - extension of Q. But they visibly do.

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    $\begingroup$ Thank you. Does your argument mean the rank is unbounded in certain number fields? $\endgroup$
    – joro
    Commented Apr 28, 2014 at 15:44
  • $\begingroup$ It should do. Just take your r points defined over disjoint quadratic fields, whose compositum is K; the MW rank over THAT field is finite, so you can certainly choose an integer x such that (sqrt(f(x))) generates a further quadratic extension of K, then you have rank r+1 and you just keep going. $\endgroup$
    – JSE
    Commented Apr 28, 2014 at 17:33
  • $\begingroup$ @JSE -- It sounds like joro was asking if it means there is a number field $K$ and we know that ranks are unbounded for elliptic curves over $K$... but in order to make a curve with larger rank in this way, you have to enlarge $K$. It does show that ranks are unbounded over some big field that contains all those guys, like the compositum of all quadratic extensions of the compositum of all quadratic extensions of $\mathbb{Q}$, which I would write as $\left(\mathbb{Q}^{(2)}\right)^{(2)}$. $\endgroup$
    – Bobby Grizzard
    Commented Apr 29, 2014 at 15:59
  • $\begingroup$ Oh I see. But if that's what joro was asking, I don't see how his construction gives any hint of how to get unbounded rank over a fixed K. $\endgroup$
    – JSE
    Commented Apr 29, 2014 at 19:39
  • $\begingroup$ I agree. Just wanted to make sure no one got the wrong idea. $\endgroup$
    – Bobby Grizzard
    Commented Apr 29, 2014 at 21:34

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