Justify: $R$ is commutative

Question

Let $R$ be a ring. If $(ab)^n=a^nb^n$ holds for all $n$ and all $a, b\in R$, then justify that $R$ is commutative ring.

No idea how to prove it. But also fail to get one counter example. Please help

Answer 1

By assumption $$ (a(b+1))^2=a^2(b+1)^2=a^2b^2+2a^2b+a^2 $$ and by distrubutive law $$ (a(b+1))^2=(ab+a)^2=(ab)^2+aba+a^2b+a^2 $$ From this it follows that $$ a^2b=aba $$ and this holds for all $a,b \in R$. Especially for $a+1$ and $b$ $$ (a+1)^2b=(a+1)b(a+1)=(ab+b)(a+1)=aba+ab+ba+b $$ but compute the left side to get $$ (a+1)^2b=(a^2+2a+1)b=a^2b+2ab+b=aba+ab+ab+b $$ now compare

Answer 2

You do not really need the equality to be true for all $n$..

All you need to see is :

• $(ab)^2=abab$

• $a^2b^2=aabb$

Answer 3

The claim is false if $R$ is not assumed to be unital. Here is an example:

Let $K$ be a field, and $A:=K\langle x,y\rangle$ the free (non-commutative non-unital) algebra on two indeterminates. Let $I$ be the two sided ideal generated by $\{(ab)^n-a^nb^n|a,b\in A,n\in\mathbb{N}\}$, and set $R=A/I$. By construction, $R$ clearly satisfies the given condition. However, since $xy-yx\not\in I$ (as we will prove in the next paragraph), $R$ is not commutative.

We explain shortly why $xy-yx\not\in I$. Observe a general generator $(ab)^n-a^nb^n$ of $I$. If $n=1$, this generator vanishes, and this is also the case if $a,b$ commute. It follows that all non-vanishing generators are of degree $\geq4$, and the claim follows since $xy-yx$ has degree $2$.

Answer 4

Let $R$ be the free ring on two generators $X$ and $Y$, modulo the relations $X^2 = Y^2 = XYX = YXY = 0$. The additive group of $R$ is a free abelian group with generators $X,Y,XY,YX$. Clearly $R$ is not commutative.

Note that the product of two elements of degree $\geq 2$ is identically zero.

Fix $n\geq 2$. For any $a,b \in R$, $ab$ has degree $2$, so $(ab)^n = 0$. But also $a^n$ and $b^n$ have degree $\geq 2$, so $a^n b^n = 0$, verifying the condition.

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Note that Blah's proof appears to show that such an example is impossible when $R$ has an identity. Also note that my example is a quotient of Amitai's example (and is, in some sense, the same proof).

Answer 5

We only need $\forall a, b\in R, (ab)^2=a^2b^2$ to conclude $R$ is commutative if $R$ is unital. The following might help better understand the accepted solution.

$(ab)^2=a^2b^2$ is equivalent to $a([a,b])b=0$ where $[a,b]=ab-ba$ is the commutator.

It's easy to check that $[a+1, b]=[a,b+1]=[a+1,b+1]=[a,b]$.

Therefore replace $a$ by $a+1$, or $b$ by $b+1$ or both, in all we get $$\begin{cases} a[a,b]b=0 & (0) \\ (a+1)[a,b]b=0 & (1) \\ a[a,b](b+1)=0 & (2) \\ (a+1)[a,b](b+1)=0 & (3) \end{cases}$$

Now $(3)-(1)-(2)+(0)$ shows $[a,b]=0$.

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Here is a more concrete counter-example in the non-unital case where $(ab)^n=a^nb^n$ holds for all $n$ while $R$ is non-commutative. Consider the ring of the matrices $$R:=\{ \begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix} \mid a, b\in \mathbb Q \}$$

For each $M=\begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix}$, define $\tau_M = a$, then it's easy to check that $$MN = \tau_M N$$

Now $MN=\tau_M N$ is not equal to $NM=\tau_N M$ when $M,N$ are linearly independent. So $R$ is not commutative. However,

$$(MN)^n=MNMN\cdots MN = \tau_M\tau_N\cdots \tau_M N = (\tau_M)^n(\tau_N)^{n-1} N = M^nN^n$$

From the above, it's clear that the following stronger identity holds: $$MNL=NML\Leftrightarrow [M,N]L=0, \forall M, N, L\in R$$

Also $R$ has many left identities, but not no right one.