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Let $R$ be a ring and $f \in R[[x]]$ a commutative formal group law over $R$, meaning $f(f(x, y), z)=f(x, f(y, z))$, $\ f(x, y)=f(y, x)$ and $f(x, y)=x+y + \text{higher order terms}$. Let $G=\operatorname{Spec}R[[x]]$ (considered as a scheme over $R$). Does $f$ induce a group structure on $G$ (over $\operatorname{Spec} R$), where multiplication $\mu:G \times G \to G$ induced by $R[[z]] \to R[[x, y]]$ sending $z \mapsto f(x, y)$?

I ask because it seems to me to be the case, but everywhere I read it says a formal group law induces a group structure on a formal group scheme, basically the one denoted by $Spf(R[[x]])$, which has $\operatorname{Spec} R$ as its underlying topological space, and $R[[x]]$ as its structure sheaf.

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The fibre product $\operatorname{Spec} R[[x]] \times_R \operatorname{Spec} R[[y]]$ is not isomorphic to $\operatorname{Spec} R[[x,y]]$. Indeed, the map \begin{align*} \phi \colon R[[x]] \otimes_R R[[y]] &\to R[[x,y]]\\ f \otimes g &\mapsto fg \end{align*} is injective (I think), but not surjective, even when $R = k$ is a field (and probably one can reduce to this case by choosing a maximal ideal).

Indeed, for $f = \sum_{n=0}^\infty a_n x^n \in k[[x]]$ and $g = \sum_{m=0}^\infty b_m y^m \in k[[y]]$, we have $$fg = \sum_{m=0}^\infty b_m f y^m.$$ Every element in the image of $\phi$ is a finite linear combination of such. Thus, for an element $$h = \sum_{m=0}^\infty h_m y^m \in k[[x,y]],\ \ \ \ \ \ \ h_m \in k[[x]]$$ to be in the image of $\phi$, the span of the $h_m$ in $k[[x]]$ needs to be a finite-dimensional subspace (contained in the span of the $f$'s occurring). Most elements do not satisfy this property. For example, consider $$h = 1 + xy + (xy)^2 + \ldots,$$ so $h_m = x^m$.

Remark. On the other hand, the fibre product of formal schemes is given by the $\operatorname{Spf}$ of the completed tensor product. Since $\phi$ is surjective onto $R[[x,y]]/(x,y)^n$ for every $n$, we see that $$\phi \colon R[[x]] \hat\otimes_R R[[y]] \stackrel\sim\to R[[x,y]].$$

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  • $\begingroup$ Fantastically clear answer, Remy. Thank you! $\endgroup$ – usr0192 May 6 '16 at 4:14
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    $\begingroup$ @user90219 This is just to compliment Remy's nice answer. A ($1$-dimensional) formal group is functorially assigning to adic rings $R$ a group structure on $R^{\circ\circ}$ (the group of topologically nilpotent elements) whereas an actual group structure on $R$ would be defining the structure on $R$ for any adic ring $R$—very differen structures (of course stemming from the different topological structure sheaves on $\text{Spec}(R)$ vs. $\text{Spf}(R[[T]])$ despite their equality on underlying spaces). $\endgroup$ – Alex Youcis May 7 '16 at 12:38

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