# Completion along zero section of an elliptic curve.

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!

This is just following up on Sebastian's answer to explain why your intuition is right: the formal completion of E along the "0"-section is like the Taylor series expansion of E about the origin.

As you know, Taylor series are usually of the form $f(x+y) = f(x) + yP_1(x) + y^2P_2(X) + ...$, where $P_i(x) = \frac{f^{(i)}(x)}{i!}$. However, this uses the exponential, which fails to make sense, even as a formal power series, in characteristic $p$.

In some sense, 'completion at a point $t=0$' gives you the Taylor series of a map in the algebraic-geometric setting, but I'd like to stress that it's more general than this. Instead of a plain old power series $\text{A}[[t]]$, we'll get a topologized power series with the $t$-adic topology, $\text{Spf } A[[t]]$.

Spf is the "formal spectrum of a ring," its worth noting explicitly that its topology is different from Spec: $$\text{colim }(\text{Spec } A[t]/t^n) =: \text{Spf } A[[t]]$$ $$\text{Spec } (\text{lim } A[t]/t^n) =: \text{Spec } A[[t]]$$

Note that a formal group law is the formal spectrum of a power series ring.

Keep in mind that a formal scheme can be thought of as an algebraic replacement of tubular neighborhoods (when we consider the formal completion of a subvariety inside the ambient variety).

Let's look at an elliptic curve $C \to \text{Spec } A$.

$\text{Spec } A[t]/t^2$ is like truncating the Taylor series after the information given by the first derivative, it's the the first infinitesimal neighborhood of $\text{Spec }A$.

Similarly, $\text{Spec } A[t]/t^3$ is the second infinitesimal neighborhood, and we need a slightly larger infinitesimal neighborhood to take the second derivative.

I want you to imagine:

• $\text{Spec } A$ as a line, sitting on $t=0$,
• $\text{Spec }A[t]/t^2$ as an infinitesimal normal bundle sticking out of that line
• $\text{Spec }A[t]/t^3$ as a slightly fatter infinitesimal normal bundle sticking out of that line, etc.

In the case of a formal group law, we want to write down all of the derivatives, so we take the colimit of $\text{Spec } A[t]/t^n$, which is by definition the formal scheme $\text{Spf } A[[t]]$.

Zariski locally, the completion of $C$ is isomorphic to the formal scheme, $\text{Spf }A[[t]]$.

For example: Zariski locally, over $\text{Spec } A$, we can put our elliptic curve in Weierstrass normal form $y^2 + a_1 xy = x^3 + a_2 x^2 + a_4 x + a_6$ (outside of characteristic 2 and 3, we can take $a_1 = 0$).

In this form, the variable $z = x/y$ is a uniformizer at the identity ($y$ has a pole of order 3 at the identity, $x$ has one of order 2, so $x/y$ has zero of order 1). So it's reasonable to expect $\text{Spf }A[[z]]$ to be the formal completion along the identity.

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 $\mathcal{O}_E$.