Non-commutative formal groups- proof of non-commutativity

Question

I'm studying Hazewinkel's "Formal Groups" book, and example of a non-commutative group law is given by:

on the ring $\mathbb{F}_p[c]/(c^2)$ we define $$F(X,Y)=X+Y+cXY^p.$$ I see how to prove everything except non-commutativity- my problem is I tried to find example where it fails for $p=3$ and I keep getting that this is commutative (I am making mistake somewhere):

1. I think that elements in $\mathbb{F}_3[c]/(c^2)$ can be seen as polynomials of degree $1$ or $0$, ie. elements are of form $a\cdot c+b$.

2. Also, since characteristic of ring is $3$ we have $(a\cdot c+b)^3=a^3c^3+b^3=b$ so

3. if $F(X,Y)=F(Y,X)$ than since $X+Y=Y+X$ we must have $cXY^p=cYX^p$

4. Let $X=ac+b$ and $Y=Ac+B$ then $cXY^3=c(ac+b)(Ac+B)^3=$(step 2)$=c(ac+b)B=bBc$

5. and $cYX^3=c(Ac+B)(ac+b)^3=$(step 2)$=c(Ac+B)b=bBc$

which leads to conclusion it is commutative. Which steps are wrong?

Answer 1

You have misunderstood what it means for a formal group law to be commutative.

A formal group law $F(X,Y)\in R[[X,Y]]$ is commutative precisely when $F(X,Y)=F(Y,X)$. Nothing less and nothing more. Your series (happens to be a polynomial) does not satisfy the condition, and is therefore noncommutative.

Has nothing to do with whether the base field is the prime field or not.

Answer 2

To add to my comment: the problem is that in $\mathbb{F}_p$, $p$ prime, we'll have $x^p=x$ for all $x\in\mathbb{F}_p$. Therefore your group law is almost trivially commutative. Instead you can consider $\mathbb{F}_{p^n}$ for $n\geq2$. Below an example for $p=n=2$.

Let's construct $\mathbb{F}_4=\mathbb{F}_2[x]/(x^2+x+1)$. Then construct $\mathbb{F}_4[c]/(c^2)$. Now

$$F(x,1)=x+1+c\cdot x\cdot 1^2=x+1+cx$$

while $$F(1,x)=1+x+c\cdot 1\cdot x^2=x+1+c(x+1).$$

Since $x\neq x+1$ in $\mathbb{F}_4$, the result follows.