Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
4  
An elliptic curve over $K$ is by definition a projective curve of genus 1 with a distinguished point over $K$. –  Alex B. Dec 23 '11 at 10:53
5  
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. –  Martin Bright Dec 23 '11 at 11:42
11  
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) –  Tim Dokchitser Dec 23 '11 at 13:30
3  
@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. –  quid Dec 23 '11 at 22:32
7  
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. –  paul Monsky Dec 24 '11 at 3:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.