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.


1 Answer 1


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 $\mathbb{P}^{n-1}$.

  • $\begingroup$ Thank you for your detailed response. Based on what you said, I'm guessing that the abelian group $H^1(K,E)$ can be described as follows: each non-trivial (class of) $K$-torsor $C'$ under $E$ is the compactification of some genus one affine curve $C$ with no rational points. Some in some sense, we can classify isomorphism classes of genus one affine curves based on the Jacobian of their compactification. But is it possible for a genus one affine curve $C$ without rational point to have compactification $E$, i.e., correspond to the class of trivial torsors? $\endgroup$
    – oleout
    Commented Aug 7, 2021 at 0:57
  • $\begingroup$ @KelvinLian yes this is very easy to construct. Consider the affine part of an elliptic curve over a number field written in Weierstrass form, and suppose it has trivial Mordell-Weil group. $\endgroup$ Commented Aug 7, 2021 at 9:54
  • $\begingroup$ Oh I think I get it, you are saying we start off by considering an elliptic curve $E$ with trivial Mordell-Weil group, then write it in Weierstrass form and we make it affine by deleting the origin (rational point). The resulting cubic curve would have no rational points but its compactification is precisely $E$. $\endgroup$
    – oleout
    Commented Aug 7, 2021 at 11:04
  • $\begingroup$ @KelvinLian Yes, exactly. $\endgroup$ Commented Aug 7, 2021 at 11:22
  • $\begingroup$ Thank you so much, actually there's something I'm really confused about. For the affine part $C$ of the elliptic curve $E$, does it make sense to talk about the Neron-Severi group of $C$? Because certainly we can take quotient of its Picard group by Pic$^0$, wouldn't it give us the usual NS group $\mathbb{Z}$? I'm confused because I've never seen the Neron-Severi group mentioned for affine curves, so what's the difference? Or where will the misconception be? $\endgroup$
    – oleout
    Commented Aug 7, 2021 at 11:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .