Let R be a ring in which $x^3 = x$ for all $x \in R$. Prove that if $a, b \in R$ with $ab = 0$, then $ ba = 0$ also.

Question

Let $R$ be a ring in which $x^3 = x$ for all $x \in R$. Prove that if $a, b \in R$ with $ab = 0$, then $ba = 0$ also.

$$ab = (ab)^3 = ababab = 0$$

$$ba = (ba)^3 = bababa = b(abab)a = b(0)a = 0$$

Does this seem correct?

Already proved here in the dupe. – Bill Dubuque Apr 15 '23 at 20:24 — Bill Dubuque, Apr 15 '23 at 20:24

score 2 · Accepted Answer · answered Nov 20 '16 at 18:37

2

Suppose $ab = 0$. Since $x^3 = x$ for all $x$, $$ ba = bababa = b(ab)^2 a = b\cdot 0^2\cdot a = 0, $$ so $ba = 0$ as well.

answered Nov 20 '16 at 18:37

Stahl

I was editing mine right as you commented. Thank you. – Zed1 Nov 20 '16 at 18:39

Shiksharthi Sharma · Answer 2 · 2016-11-20T18:54:33.233

0

If x^3=x then ba=(ba)^3 for all a,b belongs to R. Now ba=bababa=b(abab)a=b(ab)^2a=b0a=0. Thus ba=0.

edited Nov 20 '16 at 18:54

answered Nov 20 '16 at 18:38

Not really. You could conclude that $x^2=1$ if you knew that $x$ is not a divisor of zero. – Mariano Suárez-Álvarez Nov 20 '16 at 18:46
While it is true that $x^3 = x$ for all $x\in R$ forces $R$ to be commutative, you cannot claim that $x^2 = 1$ for all $x$, as there may not be a multiplicative identity in $R$. Even if there is, $x$ might not have a multiplicative inverse in $R$. Consider the following: $0^3 = 0$, but $0^2\neq 1$. – Stahl Nov 20 '16 at 18:46
Even if there is a multiplicative identity, you cannot conclude that. For example, in the ring $\mathbb Z/\mathbb 2Z\times\mathbb Z/2\mathbb Z$ every element is equal to its cube. – Mariano Suárez-Álvarez Nov 20 '16 at 18:47

2 Answers2