If $S$ is a ring with the property that $s=s^2$ for each $s \in S$, then is $S$ commutative?

Question

If $S$ is a ring with the property that $s=s^2$ for each $s \in S$, then is $S$ commutative?

I know $st=(st)^2=stst$. On the other hand, $st=s^2t^2$. Hence, $$stst=sstt$$

If $S$ is a division ring, then by cancellation we can have $st=ts$.

But how to make sure that any non-zero element in $S$ is not a zero-divisor?

You can't, since in fact all elements except $1$ are zero-divisors. — Tobias Kildetoft, Mar 31 '17 at 14:23

score 9 · Accepted Answer · answered Mar 31 '17 at 14:43

9

Since $x+x = (x+x)^2 = x^2+x^2+x^2+x^2 = 4.x$ we have that $2.x = 0$ and $-x = x$ for all $x \in S$. Now $(x+y)^2 = x+y$ so $$x^2+xy+yx+y^2 = x+xy+yx+y = x+y \Rightarrow xy = -yx = yx$$ So it is indeed commutative, for another reason than you state.

answered Mar 31 '17 at 14:43

Jef

3,821

So $S$ has zero-divisors but is also commutative? – SHBaoS Mar 31 '17 at 14:48
yes, that's right. A ring can happily have these two properties at the same time! For example, $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ – Jef Mar 31 '17 at 15:56

score 3 · Answer 2 · answered Mar 31 '17 at 14:25

3

Hint:

if $s=s^2$ than $s(s-1)=0$ and $s$ is a zero divisor.

answered Mar 31 '17 at 14:25

Emilio Novati

62,675

How is $s$ a zero divisor? If $s=0$ then it is not a zero divisor. – Sun Apr 09 '20 at 03:36

If $S$ is a ring with the property that $s=s^2$ for each $s \in S$, then is $S$ commutative?

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