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?
- 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?
- 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?
Best Answer
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 $E$).