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!
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 $I(f(x))$.