Basic computation for the degree of an isogeny I am trying to compute the degree of the isogeny $\phi:E_{1} \to E_{2}$
where $\phi(x,y)=(\frac{y^2}{x^2},\frac{y(b-x^{2})}{x^2})$ with
$E_{1} : y^{2} = x^{3} + ax^{2} + bx$,
$E_{2} : Y^{2} = X^{3} - 2aX^{2} + rX$,
$char(K) \neq 2$ and $a, b \in K$ verifying $b \neq 0$ and $r = a^{2} - 4b \neq 0$. 
This is an example mentioned in Silverman's Arithmetic of Elliptic Curves, page 70.
The book provides the expression of the dual isogeny as well, so I can just compose them and get the $[2]$ isogeny. 
However, I am interested in a general way of doing things. My guess here is to either
compute the degree of the field extension $[\bar{K}(E1):\phi^{\star}\bar{K}(E2)]$ or to try to use the sum of the ramification indexes for all the preimages of a point. My algebra skills are letting me down though, since I am not even able to compute $\phi^{\star}$. 
Any help or suggestions are greatly appreciated.
 A: In fact I wouldn't suggest to compute the degree of that field extension as a way to compute the degree of the isogeny, because it might be difficult. I think it is more reasonable to compute the ramification indexes. So here one would like to compute for example $\sum_{P\in\phi^{-1}(O')} e_{\phi}(P)=\deg\phi$. (Just to be clear, I will call $O$ the point at infinity of $E_1$ and $O'$ the one of $E_2$). You can see easily that $\phi^{-1}(O')=\{O,(0,0)\}$. Now let's say you want to compute the ramification index at $(0,0)$. This is by definition the valuation at $(0,0)$ of $\phi^*(t)$ for $t$ a uniformizer at $\phi((0,0))=O'$. So we have to find out what is $t$ and what is its pullback. It is a standard trick to move $O'$ at $(0,0)$ in order to make computation easier. So we homogeneize the equation of $E_2$ and then we dehomogeneize it w.r.t. $y$, getting $Z=X^3-2aX^2Z+rXZ^2$. Now $X,Z$ generate the maximal ideal at $(0,0)$ but since $Z(1+2aX^2+rXZ)=X^3$, we deduce that $X$ is a uniformizer at $(0,0)$. So we just need to compute $\phi^*(t)$. To do that we need to rewrite $\phi$ in such a way that it is clear how it works at $(0,0)$. If my computations are not too wrong, it should be something like $(x,y)\mapsto \left(\frac{y}{b-x^2},\frac{y^3}{(x^2+ax+b)^2(b-x^2)}\right)$, which tells you that $\phi^*(X)=\frac{y}{b-x^2}$. Now you prove in the same way as before that $y$ is a uniformizer at $(0,0)$ and you see that $e_{\phi}((0,0))=1$. To compute the ramification index at $O$, you have to basically do the same: you move $O$ to $(0,0)$, you do the same with $O'$ and you rewrite $\phi$. That shouldn't be hard...
