Warning: this is a soft question about usual terminology, to make sure I understand things correctly.
Let $R$ be any commutative ring and $n\geq 1$. Consider the $R$-algebra $\mathcal A = R[[X_1,\ldots,X_n]]$ of formal power series over $R$ in $n$ variables. In the following, I denote by $X$ the $n$-uple of variables $(X_1,\ldots,X_n)$ and similarly for $Y$ and $Z$.
A (commutative) $n$-dimensional formal group law over $R$ is the data of a family of $n$ formal power series over $R$ in $2n$ variables $F(Y,Z)=(F_1(Y,Z),\ldots,F_n(Y,Z))$ satisfying the conditions
- $X=F(X,0)=F(0,X)$
- $F(X,F(Y,Z))=F(F(X,Y),Z)$
- $F(X,Y)=F(Y,X)$
Such a data allows us to define a new commutative group law $\star$ on $\mathcal A^n$ via the following formula $$g(X)\star h(X) = F(g(X),h(X))$$ My question is the following:
Am I correct to think of a "(commutative) $n$-dimensional formal Lie group over $R$" as the abelian group $(\mathcal A^n,\star)$ associated to some formal law $F(Y,Z)$, that I have just described above ?
The reason for my confusion lies in Tate's paper about $p$-divisible group, where in order to define a formal Lie group $\Gamma$, only a formal group law is actually defined. Late on I see expressions such as "points in $\Gamma$", or arrows $\Gamma \rightarrow \Gamma$, even though the underlying set of $\Gamma$ has not been exactly defined. Thus I'm a little unsure about how to think of them.
You may find the paragraph in question here.
This is too long for a comment, and you should regard it as not being complete or particularly authoritative.
It’s not true that your $\mathcal A^n$ is a group under the $n$-dimensional formal group law $F$, since you can’t plug constants from $R$ into the series, unless maybe $R$ has some complete topological structure.
Here’s the way you can make groups by using your $n$-dimensional f.g.l. $F$ over $R$: Let $S$ be any $R$-algebra, and $\mathcal N_S$ be the ideal of nilpotent elements of $S$. Then $(\mathcal N_S)^n$ becomes a group under $F$.
Alternatively, if $S$ is an $R$-algebra with the additional property that it is complete under the topology given by powers of an ideal $I\subset R$, then $I^n$ becomes a group under $F$. Better, the ideal of all $z\in R$ for which there is $m$ with $z^m\in I$, I guess that’s $\sqrt I$, has the property that $(\sqrt I)^n$ becomes a group under $F$.
For example, if $R=\Bbb Z_p$, the $p$-adic integers, and $S$ is the ring of integers of a finite extension of $\Bbb Q_p$, with maximal ideal $\mathfrak m$, then $\mathfrak m^n$ becomes a group under the $n$-dimensional f.g.l. $F$ defined over $\Bbb Z_p$.
Another example: let $R$ be a finite field $k$, and $F$ be an $n$-dimensional f.g.l. over $k$. Then you can take points of $F$ with values in $k[[t]]$, power series in a single variable over $k$. You combine any two $n$-tuples of elements of $tk[[t]]$, that is, series with no constant term, by plugging them in to $F$, to get another $n$-tuple of series in $t$.