do formal group laws induce group structures on schemes (as opposed to formal schemes) 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.
 A: 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
$$\phi \colon R[[x]] \hat\otimes_R R[[y]] \stackrel\sim\to R[[x,y]].$$
