0

The definition of a formal group is given here.

Apologies in advance if this question sounds trivial/obvious. I am still trying to wrap my head around the idea of formal group.

Suppose $F$ is a commutative formal group, there is a lemma in my course stating that

$\exists! I(T)\in R[[T]]$ such that $F(T, I(T))=0.$ (where $R$ is the ring of concern.)

Now I like to think that this is saying $T$ has an additive inverse (sort of). However, suppose we have a general $g(T)\in R[[T]]$, am I right in thinking that it too has an additive inverse and it is given by $I(g(T))?$ In other words, would I be correct in believing that $F(g(T), I(g(T)))=0?$

Thank you so much in advance!

Descartes Before the Horse
  • 7,038
UnsinkableSam
  • 3,876
  • 1
    You need to assume that $g(0) = 0$ (i.e. no constant term). Then it is simply composing the power series $F(T, I(T))$ with the power series $g(T)$. – WhatsUp Dec 09 '21 at 23:03
  • @WhatsUp Just out of curious, what happens if $g$ has a constant term? does that mean it wont have an inverse or would the inverse be given in a different form? – UnsinkableSam Dec 09 '21 at 23:05
  • For a general ring $R$, it doesn't make sense to compose with something that has a constant term. Look at $f(T) = \sum_{n\geq 0} \frac {T^n}{n!}\in \Bbb Q[[T]]$ and $g(T) = 1 + T$. What would you expect $f(g(T))$ to be? What is its constant term? – WhatsUp Dec 09 '21 at 23:07
  • @WhatsUp Oh I see! One would expect the constant term $e$ but it is not in $\mathbb{Q}$. One follow up question though if it is ok please? Say if $g(T)$ has no constant term, why is it guaranteed that one of the coefficients of $f(g(T)$ wont run into the trouble that we encountered above? Thanks! – UnsinkableSam Dec 09 '21 at 23:13
  • 1
    If $g(T)$ has no constant term and $f(T) = \sum a_nT^n$, then we have $f(g(T)) = \sum a_n g(T)^n$. As all terms of $g(T)^n$ has degree $\geq n$, for each degree there are only finitely many terms in the sum that contributes. For example the coefficient of $T$ in $f(g(T))$ is totally determined by $a_1$, the coefficient of $T^2$ in $f(g(T))$ is totally determined by $a_1, a_2$, etc. A finite sum always makes sense in any ring. – WhatsUp Dec 09 '21 at 23:17
  • @WhatsUp Ohhh I see! Sorry I didn't see it at the first go! – UnsinkableSam Dec 09 '21 at 23:20

1 Answers1

1

Indeed, you seem to have hit on an important property of formal groups.

If $F=F(x,y)\in R[[x,y]]$ is a (one-dimensional) formal group over the commutative ring $R$, then the set $xR[[x]]$ becomes an ordinary abelian group by means of the law of combination $F$.

That is, if $f,g\in xR[[x]]$. we may use $F$ to add the two series: $$(f+_Fg)(x)=F(f(x),g(x))\,.$$ Now it’s up to you to show that you do in fact have a group, where the inverse of $f(x)$ is $I(f(x))$.

Lubin
  • 62,818