I have this exercise:
Let be $\phi:(E_1,O_1)\rightarrow (E_2,O_2)$ a isogeny. Let be $\phi=(r_1(x),r_2(x)y)$ its standard affine representation and write $r_1(x)=\frac{p(x)}{q(x)}$ with polynomials $p(x)$ and $q(x)$ that do not have a common factor. Prove that deg $\phi=$ max{ deg $p(x)$, deg $q(x)$}.
My professor provided me with this exercise but I don't understand which is the right way to proceed because on the book I'm consulting (Elliptic Curves Number Theory And Cryptography) the fact that I should demonstrate is precisely the definition that gives for 'degree of an isogeny'.
Instead the definition of degree given by my professor (degree of a homomorphism between elliptic curves) depends on the kenel of the pullback map. So I initially thought about doing this using this definition and then working with the kernel. This approach brings me closer to thinking that there is a relationship between the degree of $\phi$ and the degree of the polynomial $p(x)$ in the numerator, but I just don't understand how the degree of the polynomial $q(x)$ in the denominator can also come into play. So I thought the path I was following was wrong. Then I also thought of working with dual isogenies to study the degree but it's all too unknown with this particular affine writing of $\phi=(r_1(x),r_2(x)y)$.
Would any of you have any ideas on how to proceed? Thanks.
An isogeny $\phi:E_1\to E_2$ corresponds to an inclusion of function fields $\phi^*:K(E_2)\to K(E_1)$ and its degree is $d=|K(E_1):\phi^*K(E_2)|$.
Write $K(E_i)=k(x_i,y_i)$ where $x_i$ and $y_i$ satisfy the equation of $E_i$. Suppose the isogeny has the form $\phi:(x,y)\mapsto (u(x),yv(x))$ where $u$ and $v$ are rational functions. Then $\phi^*(x_2)=u(x_1)$. For each $i$, $|K(E_i):k(x_i)|=2$. Also $\phi^*k(x_2)\subseteq k(x_1)$. Therefore $d=|k(x_1):\phi^*k(x_2)| =|k(x_1):k(u(x_1))|$. But this degree is $\max(\deg p,\deg q)$ where we write $u(x)=p(x)/q(x)$ in lowest terms.