Fixing a number field $k$ as the base field and given a smooth affine genus one curve (with no $k$-rational points), is it possible that its smooth compactification is a projective curve with no $k$-rational points, i.e., not an elliptic curve?
The answer should be affirmative, since if $k = \mathbb{Q}$ and $C:3x^3+4y^3+5=0$ is the affine curve, then I've already seen on this site that the smooth compactification is done by adding three points at infinity, none of which are $\mathbb{Q}$-rational, to obtain the smooth projective (well-known) Selmer curve $3X^3+4Y^3+5Z^3=0$.
The issue I have is more to do with the reverse process. I know that given any projective curve (smooth or not), the removal of a finite number of points results in an affine curve, and thus their compactification is done by adding back the said points. But what are the points we are removing here? Must they be $k$-rational, or just general closed points?
Best Answer
Here is a nice, basic fact.
This is a somewhat easy exercise using Riemann--Roch to find for a given divisor $D\subseteq X$ (namely the complement of $U$) a large enough $m$ so that there is a rational function $f$ with poles precisely at the support of $mD$. You can see Lemma 11 here for the full proof (DISCLAIMER: this is my blog).
In particular, if $p$ is any closed point of $C$ (not necessarily a $k$-point) then $C-\{p\}$ is affine.
Question for you to ponder: Are the hypotheses I wrote above of smooth and goemetrically integral necessary?