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Let $A(n), B(n) \in \mathbb{Z}[n]$ be polynomials, not both constant, such that $4A^3(n) + 27B^2(n)$ is not the zero polynomial and the polynomial (in variables $x, y$) $$y^2 - x^3 - A(n)x - B(n) \in \mathbb{C}(n)[x, y]$$

has no zeroes in $\mathbb{C}(n) \times \mathbb{C}(n)$. Let $K$ be a number field. Furhter, let $Z$ denote the common complex zeroes of the above polynomials when $n$ runs through $\mathbb{N} = 1, 2, 3, ...$

I wonder if it is known whether there always exists an $n_0 \in \mathbb{N}$ such that when $n = n_0$, all the zeroes of the respective polynomial that belong to $\mathcal{O}_K \times \mathcal{O}_K$ must also belong to $Z$.

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  • $\begingroup$ Edit note: I added the requirement that the family of curves had no rational parametrization which I initially forgot. $\endgroup$
    – Albertas
    Commented Oct 22, 2013 at 18:11
  • $\begingroup$ P.s. One may assume that $Z = \emptyset$, were that of any convenience (the question does not become less interesting). $\endgroup$
    – Albertas
    Commented Oct 22, 2013 at 18:33

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Let $E_n$ denote your elliptic curve. It's probably easier to ask for an integer $n_0$ such that the Mordell-Weil groups $E_{n_0}(\mathbb{Q})$ and $E_{n_0}(K)$ coincide. There has been a fair amount of attention given to the question of elliptic curves $E/K$ and extensions $L/K$ such that both $E(K)$ and $E(L)$ have rank 1, because this turns out to be useful in studying Hilbert's 10th problem. So the following article (and its reference list) might be helpful for your problem:

MR2041072: Bjorn Poonen, Using elliptic curves of rank one towards the undecidability of Hilbert's tenth problem over rings of algebraic integers. Algorithmic number theory (Sydney, 2002), 33–42, Lecture Notes in Comput. Sci., 2369, Springer, Berlin, 2002

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  • $\begingroup$ Thank you for the reference. As the problem that you refer to is not only interesting, but also very difficult and I am not able to see if the question that I asked indeed requires an insight about ranks of $E_{n_0}(\mathbb{Q})$ and $E_{n_0}(K)$, I would appreciate very much some more elaboration (N.b.: this question is also valid when $K = \mathbb{Q}$). $\endgroup$
    – Albertas
    Commented Oct 22, 2013 at 18:01
  • $\begingroup$ My first point was that I don't see any easy way to check that $E(\mathbb{Z})=E(\mathcal{O}_K)$ other than showing that $E(\mathbb{Q})=E(K)$. And since it shouldn't be that hard to figure out what's happening with torsion, the problem is to check that $\hbox{rank }E(\mathbb{Q})=\hbox{rank }E(K)$. And the question of whether rank grows or not in extensions is a much-studied problem. $\endgroup$
    – Joe Silverman
    Commented Oct 22, 2013 at 19:34
  • $\begingroup$ Thank you again. Do I infer correctly then, that the answer to the above question is not known already in the case $K = \mathbb{Q}$? $\endgroup$
    – Albertas
    Commented Oct 23, 2013 at 8:27
  • $\begingroup$ Problem 1: Given $K$, find an $E/\mathbb{Q}$ such that $E(\mathbb{Q})$ and $E(K)$ have the same rank. Problem 2: Given $K$ and given an algebraic family of curves $E_T/\mathbb{Q}$, find a $t\in\mathbb{Z}$ such that $E_t(\mathbb{Q})$ and $E_t(K)$ have the same rank. I expect that Problem 1 can be solved, even to force both ranks to be 0. But Problem 2, which is your setup, seems significantly more difficult. Offhand I don't know how to solve it, nor do I know a specific reference dealing with it. (But I haven't thought about such problems for a while, so I don't know the recent lit.) $\endgroup$
    – Joe Silverman
    Commented Oct 23, 2013 at 13:48
  • $\begingroup$ Nevermind my previous comment! (I delete it, as I incorrectly assumed that you were adressing the more difficult problem because my question was not clear enough, while you did point out above that you do not see how to solve the "easier" question without solving the more difficult one.) As, conjecturally, there exists a one-parameter family of elliptic curves that all have positive rank over $\mathbb{Q}$, then, in view of the above remarks, it seems that one has no good basis to expect a positive answer to the above question already when $K = \mathbb{Q}$. $\endgroup$
    – Albertas
    Commented Nov 8, 2013 at 15:06

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