I dont understand the definition of rational maps.

Here is the definition:

Let $E_1$ and $E_2$ be elliptic curves over a field $K$. (projectively written). A rational map $\Phi:E_1\rightarrow E_2$ is an element $(\Phi_x,\Phi_y,\Phi_z)\in \mathbb P^2(k(E_1))$ such that for every $P\in E_1(\overline{k})$ where $\Phi_x(P),\Phi_y(P)$ and $\Phi_z(P)$ are defined and not all zero we have that the Point $(\Phi_x(P),\Phi_y(P),\Phi_z(P))\in E_2(\overline{K})$

Note: $k(E)$ is the function field of $E$ over $k$. An element of that field is of the form $g/h$ where $g$ and $h$ are homogenous polynomials with the same degree.

My question is: How the map $\phi:E_1\rightarrow E_2$, $\phi(x,y)=(x,-y)$ can be a rational map according to this definition?

Thanks in advance :)


Let's lift the curves into projective space. We then have the map $\phi:E \rightarrow E$, $\phi([x:y:z])=[x:-y:z]$. Assume that $[x:y:z] \neq [0:1:0]$, the point at infinity. This means $z \neq 0$ so we may divide by it.

Then we can rewrite the map as $\phi([x:y:z])=[x/z:-y/z:z/z]=[x/z:y/z:1]$. Now we see that $\Phi_x=x/z$ which is of the form $g/h$ where $g,h$ are homogeneous polynomials of degree 1 and similarly for $\Phi_y$ and $\Phi_z$.

For $z=0$, we have $y \neq 0$ (in fact we must have the point at infinity), which means we can divide by $y$ and apply the same argument to get that the map is defined everywhere.

  • $\begingroup$ Thanks for the answer! Can you explain me why we can write $\phi([x:y:z])=[x/z:-y/z:z:z]?$ Since we are in $\mathbb P^2(k(E_1))$. So we are only allowed to multiply with some $\lambda\in k(E_1)^{\times}$ but such an element is also of the form $g/h$ with $g$ and $h$ homogenous with the same degree. $\endgroup$ – Marc Sep 14 '15 at 15:44
  • $\begingroup$ I'm not sure that you can necessarily go from one representation to the other like this since $[x:-y:z]$ is defined everywhere whereas $[x/z:-y/z:z/z]$ is undefined at $z=0$. However, they do agree when they are both defined and are also dominant maps so the closure of their images should coincide which I believe suffices. $\endgroup$ – Matt B Sep 14 '15 at 18:10

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