Let $R$ be a ring ; if $(a+b)^2=a^2+2ab+b^2 , \forall a,b \in R$ , then we know that $R$ is commutative ; also if $R$ is commutative then we know that $(a+b)^n=\sum _{r=0}^n {n \choose r} a^{n-r}b^r , \forall a,b \in R , \forall n \in \mathbb Z^+ $ ; so I would like to ask , under what additional conditions on $R$ , does for a fixed given $n \in \mathbb Z^+$ , $(a+b)^n=\sum _{r=0}^n {n \choose r} a^{n-r}b^r , \forall a,b \in R$ implies $R$ is commutative ? what if we also include the condition for some positive integer (or all ) $m>1$ , $(ab)^m=a^mb^m , \forall a,b \in R$ ?

Is it true that for every integer $n>2$ , there exist a non-commutative ring $R$ such that $(a+b)^n=\sum _{r=0}^n {n \choose r} a^{n-r}b^r , \forall a,b \in R$?

  • $\begingroup$ Actually, I think I have seen something along the lines of the $(ab)^m=a^mb^m$ conjecture... I'll have to look into finding it... $\endgroup$
    – rschwieb
    Jan 15, 2015 at 11:15

1 Answer 1


Let $R$ be the ring of $3\times3$ real matrices of the form $\begin{bmatrix} 0&0&0\\ x&0&0 \\ y&z&0\end{bmatrix}$.

This ring is not commutative; for example $\begin{bmatrix} 0&0&0\\ 1&0&0\\ 0&0&0\\ \end{bmatrix}$ and $\begin{bmatrix} 0&0&0\\ 0&0&0\\ 0&1&0\\ \end{bmatrix}$ do not commute.

Every triple product is $0$, so, for $n\ge3$, the binomial theorem reduces to $0=0$.

  • 2
    $\begingroup$ I like this example since it very clearly generalizes to all $n > 3$. This suggests that one of the conditions one might need to impose is that the elements in the ring are non-nilpotent in order to get commutativity, though I'm not sure how far that line of thought can get you. $\endgroup$ Jan 16, 2015 at 15:18
  • $\begingroup$ And, if you like your rings to have $1$, you can modify this example by imposing that the diagonal elements all be equal, rather than all be $0$. $\endgroup$ Jan 19, 2015 at 19:38
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    $\begingroup$ David, I have some doubts. If the ring has unity and the binomial theorem holds true for $n=3$, then the expansion of $\big(1+(a+b)\big)^3-(a+b)^3$ yields $3(ab-ba)=0$. $\endgroup$
    – G.Kós
    Jan 20, 2015 at 5:46
  • $\begingroup$ (Hope I did not miss it.. :-) ) $\endgroup$
    – G.Kós
    Jan 20, 2015 at 5:54
  • $\begingroup$ @user141614 Can you elaborate a bit on the algebra in your comment? I'm not sure which you are expanding out and which you are applying the binomial theorem to. I can't get 3(ab - ba) = 0 like you did. $\endgroup$ Feb 5, 2015 at 16:50

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