The definition of the elliptic curve: An elliptic curve over K is a curve given in $y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$ over K with discriminant not 0 and the point at infinity $O$. But if I use a birational transformation between two elliptic curves, what will happen with the point at infinity 0? Is there a convention for $O \to O$? Is $O$ not singular? If $O$ is not singular, why? Thank you very much :)

  • $\begingroup$ You can choose $O$ to be another point. On a smooth projective curve $C(k)$ we have the (abelian) group law $P+Q= S$ iff $[P]+[Q] \sim [S]+[O]$ ie. there is a meromorphic function $f : C(\overline{k}) \to \overline{k}$ whose zeros are at $P,Q$ and its poles are at $S,O$. See Milne's book p.22-23 $\endgroup$ – reuns May 25 '17 at 19:11

A couple of things:

  1. Most defintions of Elliptic curve $E/K$ require that $E$ is a smooth projective curve with a $K$-rational point on it. So any point $O$ you care to choose on it will be non-singular. And the point is that any rational map between smooth projective curves is automatically regular.
  2. Next, if $E$ and $E'$ are just algebraic curves to you and $f:E\to E'$ is a morphism, then there is no requirement that $f(O_E)=O_{E'}$. But for this $f$ to be a morphism of Elliptic Curves (also called an Isogeny), you do require this additional condition. The point is that just this (seemingly simple) requirement will force $f$ to respect the group structures on $E$ and $E'$, i.e. $f$ will be an actual homomorphism of groups $E$ and $E'$.

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