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 $E_1\times E_2$.

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    $\begingroup$ 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. $\endgroup$ Sep 22, 2023 at 3:11


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