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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!

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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 $k$.

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  • $\begingroup$ 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. $\endgroup$ – k.j. Jul 23 at 9:58
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    $\begingroup$ 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. $\endgroup$ – Dmitry Vaintrob Jul 23 at 17:54

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