Silverman's proof of III.6.2(c) (dual of sum of two isogenies is the sum of the duals)

I am teaching a course on Silverman's nice book "The Arithmetic of Elliptic Curves", and I am puzzled by the proof of III.6.2(c), which asserts $$\widehat{\phi+\psi}=\hat{\phi}+\hat{\psi}$$ as title. In my version of the 2nd edition book it is on page 83-84. In short, I am not able to convince myself if it is rigorous using the tools in the book, and will appreciate any comment from anybody who have thought about this!

There are a few obstacles for me, but I will consider the main problem to be the displayed equation on page 84: $$\operatorname{ord}_{P_1}(f)=e_{\phi}(P_1).$$ The situation is that we have elliptic curves $$E_1$$ and $$E_2$$ over $$K$$ for which we may assume that $$K$$ is algebraically closed and it is also assumed that $$\operatorname{char}(K)=0$$. Denote by $$K(E_1)$$ the function field of $$E_1$$. Let $$f\in K(E_1\times E_2)=K(E_1)(E_2)=K(E_2)(E_1)$$. Here, $$K(E_1)(E_2)$$ may be viewed (and Silverman does) as the function field of the base change $$(E_2)_{K(E_1)}:=E_2\times_{\operatorname{Spec}K}\operatorname{Spec}K(E_1)$$. (I should say $$(E_2)_{K(E_1)}$$ is just the curve defined by the Weierstrass equation of $$E_2$$ but considered over the field $$K(E_1)$$, thanks to the suggestion of Mariano Suárez-Álvarez.)

Adapting this viewpoint, we talk about the divisor $$\operatorname{div}_{(E_2)_{K(E_1)}}(f)\in\operatorname{Div}_{K(E_1)}(E_2):=\operatorname{Div}((E_2)_{K(E_1)})$$ of $$f$$ on $$(E_2)_{K(E_1)}$$. Likewise, we can talk about $$\operatorname{div}_{(E_1)_{K(E_2)}}(f)\in\operatorname{Div}_{K(E_2)}(E_1)$$. I think the proof essentially claims that $$\operatorname{div}_{(E_1)_{K(E_2)}}(f)$$ is determined by $$\operatorname{div}_{(E_2)_{K(E_1)}}(f)$$ up to some pullback from $$\operatorname{Div}_K(E_1)$$. But how does one show it without some serious language of schemes and divisors on the surface $$E_1\times E_2$$?

Given that $$f$$ defines a rational map from $$E_1\times E_2$$ to $$\mathbb{P}^1$$ that becomes a morphism after we remove a codimension-$$2$$ subset, I think we can justify, as I feel like the proof does, that $$\operatorname{div}_{(E_2)_{K(E_1)}}(f)$$ and $$\operatorname{div}_{(E_1)_{K(E_2)}}(f)$$ have zeroes and poles on the same $$1$$-dimensional closed subvarieties of $$E_1\times E_2$$ modulo some vertical and horizontal ones. But I can't see how one shows that the orders (of zeroes and poles) are the same without some discussion about the surface $$E_1\times E_2$$.

• Well, I guess I should say that my $(E_2)_{K(E_1)}$ is just the elliptic curve defined by the same Weierstrass equation but considered over the field $K(E_1)$, the function field of $E_1$ over $K$. That should put everything back to the language of varieties. Sep 22, 2023 at 3:11