Completion along zero section of an elliptic curve.

Question

I am trying to understand the intuition that I should have about the formal group of an elliptic curve. Say that I have an elliptic curve $E\to \text{Spec} R$ for some ring $R$, with section $0\colon \text{Spec} R\to E$. My first question is: when I hear speaking about the "completion of $E$ along $0$", should I think that such a thing is the formal scheme whose underlying topological space is $0(\text{Spec} R)$ and whose sheaf of rings is the completion of $\mathcal O_E$ with respect to the ideal sheaf defining $0(\text{Spec} R)$ in E? And what is the relation of this object with the formal group of $E$? My second question is: say that I have a nowhere vanishing differential $\omega \in H^0(E,\Omega_{E/R}^1)$. I somehow have this idea (but I can't understand how true is it) that completion along the zero section tells us about "Taylor expansion" of $\omega$. How does one formalize that? Also, is the sheaf $\Omega_{E/R}^1$ always globally isomorphic to $\mathcal O_E$? or is it just invertible? Thank you in advance if you're willing to help me!

Answer 1

Regarding the first question: You're right. If we have an elliptic curve over a base scheme $\operatorname{Spec}(R)$, the zero section corresponds to a surjective map $\mathcal{O}_E \to \mathcal{O}_{\operatorname{Spec}(R)}$ whose kernel is the ideal sheaf $I$ defining $\operatorname{Spec}(R)$ in $E$. The formal completion of $\mathcal{O}_E$ with respect to $I$ gives the formal completion, which is by definition the formal group of the elliptic curve $E$. (Note that $E$ is an abelian group)

Regarding the Taylor expansion I'm not sure, but be aware that if you have a nowhere vanishing differential $\omega$ as you say, this is a nowhere vanishing global section of the line bundle $\Omega_{E/R}^1$. Thus the line bundle is trivial and hence globally isomorphic to $\mathcal{O}_E$.