The following is probably well-known (I'd appreciate a link):

for a field $K$ that is a finite extension of the field of rational numbers, give a polynomial $f(x,y) ∈ Q[x,y]$ of the form $y^2 − x^3 − Ax − B$, s.t. $4A^3 + 27B^2 ≠ 0$ and $f(x,y)$ has no zero in $K^2$.

Thank you - Albertas

  • 4
    $\begingroup$ An elliptic curve over $K$ is by definition a projective curve of genus 1 with a distinguished point over $K$. $\endgroup$
    – Alex B.
    Dec 23, 2011 at 10:53
  • 5
    $\begingroup$ I'll assume you mean "genus 1 curve" not "elliptic curve". MO's own Pete Clark has various papers with much stronger results than this, for example "There are genus one curves of every index over every number field" - Crelle 594 (2006), 201-206. $\endgroup$ Dec 23, 2011 at 11:42
  • 11
    $\begingroup$ Or, if you mean that there exists $E/K$ with $E(K)=\{0\}$, this is a (recent) theorem of Mazur and Rubin (arxiv.org/abs/0904.3709) $\endgroup$ Dec 23, 2011 at 13:30
  • 3
    $\begingroup$ @Albertas: perhaps I am missing something, but Theorem 1.1 in the paper quoted by Tim Dokchister will give an elliptic curve with just the 'distinguished point' and not other point. Now, for each elliptic curve one can get an isomorphic one being defined via an equation of your form (Weierstrass form). There will be the 'point at infinity' but no other one, thus no 'affine' one. That is to say the equation has no solution. $\endgroup$
    – user9072
    Dec 23, 2011 at 22:32
  • 7
    $\begingroup$ The curve in the Mazur-Rubin theorem is defined over K; the proposer wants a curve defined over Q. So as far as I can see the problem is still open. $\endgroup$ Dec 24, 2011 at 3:36


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Browse other questions tagged or ask your own question.