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I have started learning about formal groups defined over a R1NG, $R$. Then, I studied multiplication by $m$ map on a formal group. The definition is as follows:

Def: Let $F$ be a formal group over $R$. Define the multiplication by $m$ map $[m](T)$ inductively as follows $[0](T)=0$ and $[m+1](T)=F([m](T), T).$ (So that each $[m](T)$ is a power series in $T$.)

Then the notes I am reading claims that such map defines a homomorphism from $F$ to $F$ given that $F$ is a commutative formal group but I cannot seem to figure why?

I suppose that by homomorphism, I should show the following equation holds:

$$[m](F(X,Y))=F([m](X), [m](Y))$$ for each $m$.

I can see this equation holds for $m=0,1.$ But in general situation, I tried to do an induction on $m$ and really got nowhere. I also thought about maybe writting out the terms but it was very very messy.

Wondering if there is a slick way of showing this homomorphism? Thank you so much in advance!

Edit 1: There was a bad notation that I used, which I should have noticed earlier in stating the equation I want to prove.

Edit 2: $F$ is meant to be a commutative formal group.

UnsinkableSam
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    The axioms of $F$ imply by induction on $m$ that $[m] F(X,Y)=F([m]X,[m]Y)$. It is defined by $[m]T= F(T,[m-1]T), [0]T=0$, where you can think to $T$ as the one of $R[[T]]$ and $[m]T$ will be an element of $T \ R[[T]]$. Then you will be able to take $T$ in any ideal $I$ of a ring $S$ which is "I-adically complete" and is a $R$-algebra. In particular you can take $T$ in the prime ideal $(X,Y)$ of $R[[X,Y]]$ (where the formal group law is defined) which is why $[m]F(X,Y)$ makes sense. – reuns Dec 09 '21 at 17:44
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    For $[m]$ to be a homomorphism, we need precisely that $[m]$ have no constant term, and that $[m]\bigl(F(x,y)\bigr) = F\bigl(m,m\bigr)$. There should be no third indeterminate. To understand the second requirement, translate back to additive abelian groups: you want $m(a+b)=ma+mb$, or if you like, $m(a)+m(b)$. – Lubin Dec 09 '21 at 18:46
  • I overlooked something: obviously your $F$ must be a commutative formal group law. The $m$-power is often not an homomorphism on non-commutative groups. – reuns Dec 09 '21 at 19:10
  • @reuns Sorry I should have included that in the main question and thank you! I am not entirely sure how the intended induction works and would it be possible that you could show me how? Also I am intrigued by your second equality, $[m]T=F(F(T, [m-1] T),0)$ (I think there was a pair of brackets missing and so is this the equality that was claimed?). Could you please explain why this hold? I am trying to figure out but I couldnt seem to do so. Thank you so much in advance! – UnsinkableSam Dec 09 '21 at 19:13
  • $F$ is commutative and associative. The steps to obtain $[m]F(X,Y)=F([m]X,[m]Y)$ are the same as when you prove that in a commutative group $g^m h^m = (gh)^m$ only the notation is different. – reuns Dec 09 '21 at 20:19
  • @reuns Ahh I see! That was stupid from me to not see it from the first place! – UnsinkableSam Dec 09 '21 at 20:33

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