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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.

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  • $\begingroup$ I have been looking up a little bit more about formal schemes, notably with this related question math.stackexchange.com/questions/1773536/… If my understanding is correct, by $n$-dimensional formal Lie group over $R$, I should rather think of the formal scheme $\operatorname{spf}(\mathcal A)$, on which the formal group law alledgely defines a group structure. Please, correct me if I'm wrong or misguided. Any additionnal detail/reference would be appreciated. $\endgroup$
    – Suzet
    Commented Jan 7, 2020 at 0:22
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    $\begingroup$ About this the identity of $G=Spec(A),A=\Bbb{Z}_2[x]/(x^6-1)$ corresponds to the maximal ideal $(2,x-1)$ then the connected component are the prime ideals contained in it ie. $(x-1),(x+1)$, their intersection is $(x^2-1)$ so $G^0=Spec(A/(x^2-1))$ and $G/G^0$ can be identified with $Spec(A/(2))=G(\overline{\Bbb{F}}_2)$ $\endgroup$
    – reuns
    Commented Jan 10, 2020 at 6:36
  • $\begingroup$ @reuns It was a bad timing, I deleted my question because I have found justification for it another reference on internet. It was just at the moment you were writing your comment it seems. Sorry for this... Your example is very nice, thank you very much. It makes things clearer for me $\endgroup$
    – Suzet
    Commented Jan 10, 2020 at 6:43

1 Answer 1

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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 $t$.

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  • $\begingroup$ Thank you very much, this is a very informative answer and it helps making things clearer for me. $\endgroup$
    – Suzet
    Commented Jan 7, 2020 at 7:25

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