# Formal completion of an elliptic curve along the $0$ sectioin and the formal expansion of functions

Let $$S = \operatorname{Spec}A$$ be an affine scheme, $$f : E \to S$$ an elliptic curve and $$\mathscr{I}$$ the ideal sheaf of the $$0$$-section. (This is invertible since the section defines the effective relative Cartier divisor.)
Assume that $$f_* \Omega_{E/S}, f_*\mathscr{I}^n$$ are free over $$\mathscr{O}_S$$. ($$n = 1, \cdots , 6$$)
I want to show $$\hat{E} \cong \operatorname{Spf} A[[T]]$$. And I don't understand the formal expansion, of a basis $$\omega$$ of $$f_* \Omega_{E/S}$$ and a basis of $$f_*\mathscr{I}^n$$, along the $$0$$-section.

Here is what I have tried: Since the $$0$$-section is a regular immersion, for any $$x \in S$$, there exists affine opens $$x \in V \subseteq S$$, $$0(x) \in U \subseteq E$$ s.t. $$0(V) \subseteq U$$ and the diagram

$$\require{AMScd} \begin{CD} S @>{0}>> E \\ @VV{1}V @VV{f}V \\ S @>{1}>> S \end{CD}$$

corresponds to

$$\require{AMScd} \begin{CD} C @<{0}<< B \\ @A{\text{localization by one element}}AA @AAA \\ A @<{1}<< A, \end{CD}$$

where the kernel $$I$$ of $$B \to C$$ is generated by $$t \in B$$, a regular element.

I showed that $$\hat{B}$$ (the completion of $$B$$ along the kernel $$I$$) $$\cong C[[t]]$$. And $$\Omega_{B/A} \otimes_B \hat{B} = dt \hat{B} = dt C[[t]].$$
That is, I can show $$\hat{E} \cong \operatorname{Spf}A[[t]]$$ locally, and I can expand $$\omega$$ locally.

How can I extend these operation globally?
I also showed that the isomorphism $$\hat{B} \cong C[[t]]$$ is compatible with localization. So I think intuitively that these isomorphisms (at any points) are glued together, and we have $$\hat{E} \cong \operatorname{Spf}A[[t]]$$.

Any help will be much appreciated!

Yes, what you are saying is true, at least over an algebraically closed base $$k$$ of characteristic $$0$$. In fact, all you need is that $$S$$ is affine and that the normal bundle $$I/I^2$$ is trivial (as a bundle over $$S$$), which in your case is equivalent to $$f_*\Omega(E/S)$$ being free. The rest follows essentially from deformation theory.
There is a standard way to reduce questions about formal thickenings to questions in deformation theory, and this is called deformation to the normal cone. Here's the idea. Let $$R = O(\widehat{E})$$ be the algebra of function on the formal thickening, viewed as a formal ring over $$A$$. Then this algebra is filtered by $$R_n : = I^n,$$ with associated graded ring $$R_{gr}:=\bigoplus_n \frac{I^n}{I^{n-1}}\cong A[I/I^2].$$ Since $$I/I^2$$ is trivial and one-dimensional over $$A$$, we have $$R_{gr}\cong A[[t]].$$ Now the deformation to the normal cone is an algebra over $$\text{Spf} (k[[s]])$$ interpolating between $$R$$ and $$R_{gr}$$, given by the Rees construction $$R_{\text{rees}} : = R[s^{-1}\cdot I]^{\wedge} = \bigoplus_n^\wedge s^{-n}I^n,$$ where the completion is taken along (positive) powers of $$s$$. The key thing to observe here is that the fiber over $$s=0$$ of $$R_{\text{rees}}$$ is $$\bigoplus s^{-n}(I^n/I^{n+1})$$ (not completed), which is $$R_{gr}$$ and the generic fiber over $$k((s))$$ is isomorphic to $$R\otimes k((s))$$.
Now a standard result in the deformation theory of schemes tells us that flat deformations of a smooth scheme $$X$$ up to isomorphism (as well as of a smooth formal scheme) are classified by (what is noncanonically isomorphic to) a subset of $$H^1(X,T)^\infty,$$ i.e. infinite sequences of vectors in the first homology of the tangent bundle (deformations of order $$n$$ would be certain sequences of size $$n$$). In particular, if $$X$$ is affine (in your case — $$X = \text{Spf}A[[t]],$$ the special fiber of the Rees construction), the relevant $$H^1$$ is trivial, so there is only one deformation — the trivial one. Equivalently, $$R_{\text{rees}}\cong R_{gr}[[s]]$$.
Now you're done! Basechanging to the generic point gives $$R\otimes k((s)) \cong A[[t]]\otimes k((s)).$$ But since you are over an algebraically closed base of characteristic zero, such an isomorphism implies an isomorphism $$R\cong A[[t]]$$ over $$k$$.
• Thank you for your comment! I have a question: Why is the freeness of $\mathscr{I/I^2}$ equivalent to one of $f_* \Omega$? My text (Katz-Mazur) says that the freeness of $f_* \Omega$ is also equivalent to one of $f_* \mathscr{I}^{-n}$. I think these two statements follow from the same argument. – k.j. Jul 23 at 9:58
• Since $E/S$ is an elliptic curve, its (co)tangent bundle is constant on each fiber. So its pushforward is the same as the (co)tangent at the zero section. – Dmitry Vaintrob Jul 23 at 17:54