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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?

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    I think you mean $\mathbb{F}_3[c]/(c^2)$ in point (1), with $c$ instead of $a$. – 2017-01-24
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    @anomaly: yes, thank you – 2017-01-24
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    Hmm. May be I don't know the first thing, but isn't commutativity of the formal group exactly the requirement that $F(X,Y)=F(Y,X)$? And this is clearly not true because $XY^p\neq X^pY$? Looks like you should concentrate on the associativity law: $F(X,F(Y,Z))=F(F(X,Y),Z)$. Have you verified that? – 2017-01-25
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    And, the idea of a formal group law is that you do not substitute any elements of the ring $\Bbb{F}_p[c]/(c^2)$ in place of $X$ or $Y$. They should be treated as indeterminates. In other words, your listed items 2,4,5 feel misplaced. – 2017-01-25
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    So, even though substituing $X$ and $Y$ always leads to conclusion that $XY^p=YX^p$ over that ring, I only write that it is not commutative in general for some indeterminates $X$ and $Y$ and commutativity doesn't hold? – 2017-01-25
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    In $\mathbb{F}_p$ you'll always have $Y^p=Y$, so your group law will definitely be commutative. You may want to consider an infinite field instead. Or maybe just $\mathbb{F}_4$? – 2017-01-25
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    @JyrkiLahtonen:I believe you are right. Problem is I missunderstood formal groups. $X$ and $Y$ are elements from $R[[X]]$ and not from $R$. But CurveEntusiast example from below is interestinng example where commutativity fails for elements which are of degree 1. – 2017-01-25
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    Yes. You can turn any $\Bbb{F}_p[c]/(c^2)$-algebra $R$ into a group by plugging in elements of $R$ in place of $X$ and $Y$. If $R$ is a large enough extension ring, it may easily have elements such that $cXY^p\neq cX^pY$. For example if you use $R=\Bbb{F}_{p^2}[c]/(c^2)$. So the formal group gives you a functor from the category of $\Bbb{F}_p[c]/(c^2)$-algebras to the category of groups. It is similar to the idea of a group scheme in this regard. – 2017-01-26
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    Another thought: Since your group (**if** it is indeed a group) - as you already stated in your comment (1) - consists only of elements of the form $(a\cdot c + b) + (c^2)$, its order must be $p^2$. And it is well known that there aren't any non-commutative groups of that order. – 2017-01-27
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    @CurveEnthusiast: Your comment ($Y^p=Y$) does hold in $\mathbb{F}_p$, however, it is generally not true over $\mathbb{F}_p[c]/(c^2)$, so it is not so easy to see how the commutativity should follow from it. – 2017-01-27
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    @jpvee You are right that it is not as obvious as I perhaps claimed. But note that $cXY^p=c(c\cdot a+b)(c\cdot A+ B)^p=c\cdot b\cdot B^p$, since $c^2$ vanishes and the field has characteristic $p$. As $b,B\in\mathbb{F}_p$, this equals $c\cdot B\cdot b^p$, so we have commutativity. – 2017-01-27

2 Answers 2

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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.

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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.

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    I’m sorry, but this is really not right at all, at least not as a statement about formal group laws. – 2017-02-07
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    @Lubin You are correct! Looking back, I must've been under the impression that this was simply some example of a non-commutative group law. Yours is the much better answer. – 2017-02-08