For a commutative ring $R$, consider the formal group $\widehat{\mathbb{G}}_m$ over $R$ that is the completion of $\mathbb{G}_{m, R}$ along its identity section (naively, $\widehat{\mathbb{G}}_m$ is the formal group law $F(X, Y) = XY + X + Y \in R[[X, Y]]$). Now let $R$ be Noetherian and such that there is a nilpotent ideal $I \subset R$ with $I^n = 0$ such that $R/I = \mathbb{Z}$. Suppose one has a smooth commutative $R$-group scheme $G$ (of finite type if that's relevant) equipped with an isomorphism $G_{R/I} \cong \mathbb{G}_m$ and that one considers the formal completion $\widehat{G}$ of $G$ along its identity section. Is $\widehat{G}$ necessarily isomorphic to $\widehat{\mathbb{G}}_m$ as formal groups over $R$?
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2 Answers
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Are you trying to justify some of the assertions about the Tate curve in Katz-Mazur? (Would help to know the motivation for the question.) Anyway, the affirmative answer is part of the "standard" infinitesimal deformation theory for tori as developed in SGA3. In fact, the intervention of formal groups is a red herring, as is the hypothesis $R/I = \mathbf{Z}$ (the latter being what suggests to me that you are really trying to understand something about the Tate curve): $G$ is uniquely $R$-isomorphic to $\mathbf{G}_m$ in a manner lifting the given such isomorphism over $R/I$.
Indeed, clearly $G$ is affine, as $G_0 := G \bmod I$ is affine and $I^n=0$, so we can apply the deformation theory of tori to the given isomorphism $f_0: \mathbf{G}_m \simeq G_0$. By deformation theory considerations with split relative tori and Hochschild cohomology (in the recently-published proceedings of the Luminy summer school on SGA3, see either item 3(c) in Chapter IV of Oesterle's article on group schemes of multiplicative type, or see Corollary B.2.6 and the comment immediately after its statement in the article on reductive group schemes), this $f_0$ lifts uniquely to an $R$-homomorphism $f:\mathbf{G}_m \rightarrow G$, and $f$ is an isomorphism by the fibral isomorphism criterion (with source and target fppf over the base) since $f_0$ is an isomorphism.
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$\begingroup$ "The intervention of formal groups is a red herring": why is that? The question is about formal groups. $\endgroup$ Commented May 7, 2015 at 6:03
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1$\begingroup$ @LaurentMoret-Bailly: All I meant is that since the given data is an ordinary smooth group $G$ and there is a stronger conclusion valid even for $G$ itself (as explained in my answer), the core result can be formulated and proved without reference to formal groups. $\endgroup$– grghxyCommented May 7, 2015 at 10:44
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$\begingroup$ Oh, right. I had read the question too fast. $\endgroup$ Commented May 7, 2015 at 10:53
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This isn't a complete answer, but I think the general case is subtle.
If $R\supset\mathbf{Q}$ is (pro-)artinian, then [SGA 3, VIIB 3.2] tells us that formal groups over $R$ are uniquely determined by their Lie algebras. The Lie algebra $\mathfrak{gl}_1$ has no non-trivial deformations, so neither does $\widehat{\mathbf{G}}_\mathrm{m}$.
If $R$ is an artinian $\mathrm{W}(k)$-algebra, where $k$ is a perfect field of characteristic $p>0$, then the deformation theory of $\widehat{\mathbf{G}}_\mathrm{m}$ will be controlled by a Lubin-Tate space. For $\mathcal G$ of height $n$, the Lubin-Tate space is $\mathrm{Spf}(\mathrm{W}(k)[[ t_1,\dots,t_{n-1}]])$. So for $\mathcal G=\widehat{\mathbf{G}}_\mathrm{m}$ (the formal group of height $1$) the deformation space has coordinate ring $\mathrm{W}(k)$, which means that all deformations are trivial.
For $R$ not of the form above, I'm not sure that you can say very much.
answered May 6, 2015 at 23:17