Elliptic curves are usually defined as "smooth genus 1 projective curves having a distinguished base point". Specifically, the base point is served as the identity element for the group structure on a elliptic curve.

As a result of the above definition, if two elliptic curves share a common defining polynomial but have different distinguished base points, they are different elliptic curves (with different group structure).

So, given a smooth projective cubic and two different rational points on it, One can define two different elliptic curves with each of the two rational point act as the distinguished base point and identity element.

My question is: what is the relationship between these two elliptic curves?

Taking the projective curve $x^3-y^2z+2z^3=0$ as an example. It already looks like a Weierstrass form and one can easily take "the obvious choice" $(0:1:0)$ as the base point, which results in a "usual" elliptic curve with Weierstrass form $y^2=x^3+2$.

On the other hand, there is another rational point $(-1:1:1)$ on the said curve. What happens if one chooses that point instead as the base point?

  1. If one wants to bring the said genus 1 projective curve $x^3-y^2z+2z^3=0$ with an "unusual" base point $(-1:1:1)$ into a Weierstrass form (i.e., maps the said curve birationally to $y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$ and sends the chosen base point to the point at infinity), what will the result (and the map) look like?
  2. It seems, from both complex analytics and divisor algebra argument, that there should be an isomorphism between the "usual" elliptic curve and the "base-changed" one. What will such an isomorphism look like? Is it possible to write it down explicitly?
  • 1
    $\begingroup$ Let $0$ and $p$ be two base points. I guess it's not so obvious from the plane cubic perspective, but (at least over $\mathbb C$) thinking of your curve as $\mathbb C/\Lambda$, it's sort of "visually obvious" that the translation map induced by $\Lambda \mapsto \Lambda + p$ is an isomorphism of varieties that is additionally a homomorphism from $(\mathbb C/\Lambda, 0)$ to $(\mathbb C/\Lambda, p)$. Since it's an isomorphism of varities, its kernel is trivial hence it's also a group isomorphism. $\endgroup$ Jan 20, 2022 at 7:36
  • $\begingroup$ @Tabes Bridges Thanks for your comment. I do believe that there should be some kinds of isomorphism between two elliptic curves differing only by its distinguished base point, and you pointing out the analytic view point of a "base change" is very helpful. However, it would still be nice if such an isomorphism could be explicitly written down (i.e., determine the Weierstrass form after such a translation and the rational map between the two elliptic curves). Do you have any idea about how to do it algebraically? $\endgroup$
    – Tippisum
    Jan 20, 2022 at 7:55
  • $\begingroup$ I don't think you can find a projective morp`hism because with the usual group law the basepoint is an inflection point and projective morphisms preserve inflection points. However, in both cases the groups are isomorphic, I think, since you can see that the group law with any base point is isomorphic to the group of degree $0$ Weil divisors modulo linear equivalence $\endgroup$ Jan 20, 2022 at 16:36
  • $\begingroup$ birationally equivalence is not an isomorphism. There are points that that not mapped. $\endgroup$
    – kelalaka
    Jan 21, 2022 at 18:55

1 Answer 1


Let $E$ be a complete smooth curve of genus $1$. There is exactly one group structure on $E$ for each given choice of base point as the identity (there are many proofs of this, one uses Riemann-Roch to identify the group law on $E$ with the group law on degree zero divisors).

You are asking if $(E, p_1)$ and $(E,p_2)$ equipped with their canonical group laws are isomorphic as elliptic curves. The answer is always yes. Let $\phi : E \to E$ be the automorphism given by $x \mapsto x +_1 p_2$ where $+_1$ represents the addition with respect to the $(E, p_1)$ structure. I claim this is a group homomorphism. Indeed, consider the group law $+'$ defined by, $$ x +' y = \phi(\phi^{-1}(x) +_1 \phi^{-1}(y)) $$ Then $x +' p_2 = \phi(\phi^{-1}(x) +_1 p_1) = \phi(\phi^{-1}(x)) = x$ so $p_2$ is the identity for $+'$ and therefore $+' = +_2$ because there is a unique such group law. This shows that $\phi$ is a group isomorphism.

Therefore, we see that all $(E, p)$ are isomorphic with a unique isomorphism defined by translation in the group law.

In your question, you are asking about the Weierstrass equation and therefore about the embedding $E \to \mathbb{P}^2$. When you choose a different base point, you still have an embedded elliptic curve but it is no longer in Weierstrass form for the new group law / base point (i.e. the identity need not be a flex point). However, if you put $(E, p')$ in Weierstrass form meaning choosing a new embedding $E \to \mathbb{P}^2$ defined by the complete linear system $\mathcal{O}(3p')$ (the divisor $3p'$ explicitly means that $p'$ will be a flex point under the embedding) then you will get an equivalent Weierstrass form for $E$ with the new base point (thought of as a point of the abstract curve $E$).

  • $\begingroup$ So, if a "base-change" is made, the Weierstrass form of the new elliptic curve still have the same expression as before, while all points on it are remapped under a translation defined by elliptic curve addition, which is an elliptic curve isomorphism and sends the chosen new base point to the point at infinity. Did I got it right? $\endgroup$
    – Tippisum
    Jan 22, 2022 at 14:51
  • $\begingroup$ @Tippisum I believe that you might not get literally the same Weierstrass equation but you will get an equivalent one and then you can do a change of basis to get a minimal Weierstrass equation (there is some discussion of this in Silverman "Arithmetic of Elliptic Curves" chapters III and VII). It would likely be instructive to choose a simple curve and some point and actually do the computation of the new embedding. $\endgroup$
    – Ben C
    Jan 26, 2022 at 9:43
  • $\begingroup$ I should perhaps say you don't have to get the same Weierstrass equation but if you cook up your new embedding correctly then you will get the same equation. Indeed, let $\phi : (E,p') \to (E,p)$ be an automorphism taking the new base-point $p'$ to the old base-point $p$. Then composing $E \xrightarrow{\phi} E \to \mathbb{P}^2$ using the old embedding gives you a new embedding sending $p'$ to infinity. However, this embedding has literally the same image as the old one and thus the same Weierstrass equation. $\endgroup$
    – Ben C
    Jan 26, 2022 at 9:53

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