1
$\begingroup$

I am currently studying Katz and Mazur's book "Arithmetic Moduli of Elliptic Curves" and I face difficulties understand a part of the proof of theorem 2.5.1, page 77 in the book (available here, page 44 in this pdf). The statement I would like to show is the following.

Let $S$ be a scheme and $E/S$ an elliptic curve. The structure of $S$-group-scheme on $E/S$ given by Abel's theorem (cf. 2.1.2, p.63 in the book, p.37 in the pdf) is the unique structure of $S$-group-scheme for which the section $"0"\in E(S)$ is the identity.

To show this, we consider $K:E\times_SE \rightarrow E$ a structure of $S$-group-scheme on $E/S$ having $"0"$ as the identity. We want to show that for any $S$-scheme $T$, for any $P,Q\in E(T)$, we have $$K(P,Q)=P+Q$$ To do this, we reduce to the case $T=S$ by base change. Then, we fix $P\in E(S)$ and define the $S$-morphism $f_P:E\rightarrow E$ by $$f_P(Q)=K(P,Q)-P$$ and we want to show that this is the identity on $E$.

After some arguing (cf. the book), we prove that $f_P$ actually is an automorphism of $E$ for its "Abel's theorem" structure. Then, the last part of the proof goes like this:

Viewing $P$ as a variable, ie making the f.p.p.f. base change $E\rightarrow S$, we may apply the rigidity theorem (cf. 2.4.2, p.76 in the book, p.44 in the pdf) to the homomorphism $$f_P-\operatorname{id}:E_E\rightarrow E_E$$ of elliptic curves over the base $E$. This homomorphism is zero for $P=0$, ie over the zero-section of the base $E$. As the zero section meets every connected component of $E$, this implies $f_P=\operatorname{id}$ over all of $E$, in particular for any $P\in E(S)$. Therefore, $K(P+Q)=P+Q$ as required.

This conclusion sounds to me somewhat tricky and I can't get a clear mind about what is going on here. Actually, I find this base change quite confusing. I do not understand to what extent it makes $P$ the variable, as stated. I do not get what object is now refered to as the "zero-section of the base $E$" (is it $0\in E(S)$? Or in $E(E)=E_E(E)$ after base change?). And lastly, I absolutely do not see why the zero section (in $E(S)$ ?) meets every connected component.

Could someone please provide some details about what is happening here? I would be truly thankful.

$\endgroup$

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.