In the proof of Theorem III.6.2 (c) in Silverman's The Arithmetic Of Elliptic Curves it says:

Let $x_1, y_1 \in K(E_1)$ and $x_2, y_2 \in K(E_2)$ be Weierstrass coordinates. We start by looking at $E_2$ considered as an elliptic curve defined over the field $K(E_1) = K(x_1, y_1)$. Then another way of saying that $\phi$ is an isogeny is to note that $\phi(x_1, y_1) \in E_2(K(x_1,y_1))$, [...]

Why is that? I'm blind.

(Here, $E_1, E_2$ denote elliptic curves, $\phi \colon E_1 \to E_2$ is an isogeny, $K(E_1), K(E_2)$ denote the function fields of $E_1$ resp. $E_2$ and $E_2(K(x_1,y_1))$ are the points in $K(x_1,y_1)$ on the curve $E_2$.)

  • 1
    $\begingroup$ I doubt that it is an isogeny, but it is at least a rational map from $E_1$ (which by smoothness extends to a real map). A point in $E_2(K(x_1,y_1))$ is really an expression in $K(x_1,y_1)$ that satisfies the defining equation of $E_2$, so such a point defines a map from $E_1 \to E_2$. I am not sure how to argue $O$ must be mapped to $O$ though. $\endgroup$ – user27126 Jan 26 '13 at 9:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.