Let $A$ be a ring, $S$ the spectrum of $A$, $f : E \to S$ an elliptic curve. Then assuming $f_*\Omega_{E/A}$ is free over $S$, $\hat{E}$ (the formal completion along the $0$-section.) $ \cong \operatorname{Spf}(A[[T]])$.

I intuitively think that this isomophism means "$E$ is, locally at $0$, just the line", and that the formal group associated to $E$ is the group law of $E$ around $0$. (At least if $A$ is a field.)
So I think that the group scheme-structure of $E$ induces the "group formal scheme"(I don't know whether such thing exists or not...) -structure on $\hat{E}$, and it is closely related to the formal group of $E$.
But I can't get this relation even over a filed.

Here's what I have tried: Let $S_n$ be the scheme of "$n$-th thickened", i.e., the ringed space $(S, 0^{-1}(\mathscr{O}_E/\mathscr{I}^n))$. ($0$ is the $0$-section, $\mathscr{I}$ is the sheaf of ideal of $0 : S \to E$.) Then I think the restriction of the multiplication $E \times_S E \to E$ to $S_n \times_S S_n$ factors through $S_n$. (And so does the inverse.) So the group structure of $E$ induces one of $S_n$.
This is compatible with the closed immersion $S_n \to S_{n+1}.$ So this induces the "group structure" on $\hat{E} = \varinjlim S_n$.
Now $\hat{E} \cong \operatorname{Spf}(A[[T]])$. So the multiplication $\hat{E} \hat{\times}_S \hat{E} \to \hat{E}$ corresponds to $A[[T]] \to A[[X, Y]]$, an $A$-homomorphism, and the inverse corresponds to $A[[T]] \to A[[T]]$. So we have $F(X,Y) \in A[[X,Y]]$ and $i(T) \in A[[T]]$. (the image of $T$, under these maps.) It follows that these two power serieses satisfies the formal group actioms from the group-scheme law of $\hat{E}$. (except that $F(X,Y) = X + Y +$ higher terms.)

I want to check this intuitively argument. That is, my question is:
1. Does the multiplication and the inverse on $E$ induce the group scheme structure on $S_n$?
2. Do these group schemes $S_n$ induce the "group formal scheme"-structure on $\hat{E}$?
3. Does there exist the category equivalence between the category of affine formal schemes and one of rings? (or of complete rings?) (like the equivalence between the affine schemes and rings)
4. Under this equivalence, fibre product of formal schemes correnponds the completed tensor product of rings? (I don't know what completed tensor product is, very well.)
5. Why $F(X,Y) = X + Y + $ higher terms?

(I'm not familiar with the difficult theory of formal group and formal schemes.)

Any help will be much appreciated!

  • 1
    $\begingroup$ Actually the expression is "formal group scheme", not "group formal scheme". $\endgroup$ – user43326 Jul 25 '19 at 19:22
  • 2
    $\begingroup$ @DmitryVaintrob I don't think so. At least I posted these two questions for completely different purpose. $\endgroup$ – k.j. Jul 25 '19 at 23:50
  • 1
    $\begingroup$ @k.j. sorry about that -- I saw that and had retracted my flag, but the auto-generated comment stayed on $\endgroup$ – Dmitry Vaintrob Jul 27 '19 at 21:59

On questions 3-5:

  1. Yes, EGA I, new edition, comment after 10.2.2.

  2. Yes, EGA I, new edition, Proposition 10.7.2.

  3. This follows fro the fact that F is the comultiplication of a coalgebra, in particular the image of T is not invertible.

I guess your 1 and 2 follow from the description of a formal scheme as a certain direct limit of schemes, combined with the functorial description of formal schemes. This description is as follows: the functor associated to $A[[T]]$ takes $A$-algebras to their set of nilpotent elements (compare with the case of schemes, the functor associated to $A[T]$ takes any $A$-algebra to the set of all its elements).


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