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I'm looking for a reference of an isogeny fact that I've used many times but am having a hard time proving formally.

One can define the degree of an isogeny as the degree of extension fields of the function fields of two elliptic curves using the pullback.

One can also write an isogeny $\phi$ as a map $\phi((x,y)) = (f(x),yg(x))$, define $f(x) = p(x)/q(x)$ and then say that the degree of $\phi$ is equal to the maximum of the degree of $p$ and $q$.

My question is why/how are these two definitions equivalent? Is there a standard reference that includes this proof?


For an online reference of the question I'm posing, see Andrew Sutherland's course notes on page 7 at

http://ocw.mit.edu/courses/mathematics/18-783-elliptic-curves-spring-2013/lecture-notes/MIT18_783S13_lec5.pdf

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First, I think you have a typo. You say "One can define an isogeny as the degree ...", but you probably mean that "One can define the degree of an isogeny as the degree ...".

Second, an isogeny $\phi$ is a homomorphism, so it commutes with $[-1]$, so $\phi$ induces a map on the quotient $E/\pm1$, which is just the projective line $\mathbb P^1$. Your rational function $f(x)$ is the induced map $\mathbb P^1\to\mathbb P^1$. Now the commutative diagram $$\begin{array}{ccc} E & \xrightarrow{\phi} & E \\ \downarrow && \downarrow \\ \mathbb P^1 & \xrightarrow{f} & \mathbb P^1\\ \end{array} $$

and multiplicativity of degrees shows that $\deg\phi$ (considered as a map of algebraic curves, and so the degree is by definition the degree of the field extension) is the same as $\deg f$, which is the degree of the map $f:\mathbb P^1\to\mathbb P^1$.

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  • $\begingroup$ Learned quite a bit from this post. Thanks! $\endgroup$ – user51725 Dec 20 '14 at 0:30

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