A famous counterexample to the Hasse principle is the Selmer curve, with equation $$C: 3X^3+4Y^3+5Z^3=0,$$ which has a local point everywhere but no $\mathbb{Q}$-points.
Since an elliptic curve over a base field $k$ is defined to be a genus one smooth projective curve with a distinguished $k$-rational point, I'm wondering is it true that if $C$ has a $k$-point for some number field $k$, then $C/k$ is an elliptic curve? Is it then possible to write $C$ in Weierstrass form, or am I misunderstanding something?
Best Answer
Yes, for example you can adjoin an algebraic element $\alpha$ which is a root of $3X^3 + 4$ to $\mathbb Q$ to get a number field $k$, upon which this curve obtains a rational point $P = [\alpha:1:0]$. You can then follow the procedure in example 2.1 of this document to make a change of variables to turn the above equation into one in Weierstrass form. In this case, we get the affine equation $y^2 + 720y = x^3 -172800$ which, over $k$, ends up acquiring a rational point and thus is an elliptic curve.
If you are lazy like me, you can also ask Sage to do it for you as in here.