Genus one affine curves

algebraic-curvesalgebraic-geometryelliptic-curves

This question is a follow up to Genus zero affine curves with the same goal in mind.

A little bit of work can show us that:

Theorem. For a smooth projective geometrically integral curve $C$ over a number field $k$ (or a general field of characteristic not 2), if $C(k) \neq \emptyset$, then $C$ has genus one if and only if it is isomorphic to a plane curve of the form

$$y^2 + a_1xy+a_3y=x^3 + a_2x^2 +a_4x+a_6,$$
with $a_i \in k$.

A corollary of this result would be that every genus one curve $C/k$ with a rational point is isomorphic to a plane cubic curve.

Of course, the key word here is projective. The equation that we see above is the "affine part" of the curve, upon projectivization, we obtain an elliptic curve after fixing some rational point as the point of origin.

Question 1. Does the characterization of genus one projective curves in the above theorem extend to that of genus one affine curves with a rational point?

If the answer is affirmative, then the compactification of an affine genus one curve $C$ with a rational point would be an elliptic curve of the same equation by an abuse of notation. Then the next thing to ask would naturally be:

Question 2. What about the case of genus one affine curves with no rational points? That is, do we have any characterization as above, and, what can be said about its compactification?

EDIT. The comment section of the accepted answer to this MSE post has a small discussion about the situation for a smooth projective genus curve with no $k$-rational points as mentioned by Mummy the turkey.

Best Answer

For question 1 the answer is yes. Take the projective closure $\bar{C}$ of $C$, then this is a genus $1$ projective curve with a rational point (and $C$ is a Zariski open subset). Then $\bar{C}$ is isomorphic to a curve in Weierstrass from, so that $C$ is isomorphic to a Zariski open subset of this.

For question 2 the answer is quite complicated in fact. Suppose that we are given a genus 1 curve (assumed projective) without a $K$-rational point. Then (see Siverman's Arithmetic of Elliptic Curves, Chapter X) $C$ corresponds to a class in $WC(E/K) \cong H^1(K, E)$ where $E = Jac(C)$. Let $[C]$ denote this class. A significant literature exists on how to represent such homogeneous spaces as varieties, I will mention one (there is a nice paper of Cremona-Fisher-O'Niel-Simon-Stoll which discusses these).

Since $WC(E/K)$ is torsion, $[C]$ has finite order - say $n$. Thus by the long exact sequence on cohomology comes from a unique element of $H^1(K, E[n])$. It is a fact that elements of this cohomology group are represented by isomorphism classes of "$n$-coverings", i.e., an (isomorphism class of) a pair $(C, \pi)$ equipped with an isomorphism $\phi$ over $\bar{K}$ making $\require{AMScd}$ \begin{CD} C@>\pi>> E\\ @V \phi V V @VV id V\\ E @>>[n]> E \end{CD} commute.

A final remark is that if $K$ is a number field and $C$ represents an element of the Shafarevich-Tate group, then it is possible to realise $C$ as a curve of degree $n$ in $\mathbb{P}^{n-1}$.