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.