Let $R$ be a ring in which $r^2=r$ for all $r \in R$. Show that $r=-r$ for all $r \in R$ and that $R$ is commutative.

Question

Let $R$ be a ring in which $r^2=r$ for all $r \in R$. Show that $r=-r$ for all $r \in R$ and that $R$ is commutative.

I was given a hint to start with $(r+r)^2$. I got that $$(r+r)^2=r^2+2r^2+r^2=4r^2.$$ So $4r^2=4r \implies 4(r^2-r)=0 \implies r^2-r=0$ but from here I only get back to the assumption. Is there something else I should consider?

$4(r^2−r)=0⟹r^2−r=0$ only works if the ring has not characteristic 2 or 4. — Wuestenfux, Nov 15 '21 at 13:30

score 3 · Accepted Answer · answered Nov 15 '21 at 13:31

3

For every $r\in R$, $$r+r = (r+r)^2 = (r+r)(r+r) = r^2+r^2+r^2+r^2 = r+r+r+r$$ so we get $r+r=0$ giving $r = -r$.
For any $a,b\in R$, $$a+b = (a+b)^2 = a^2 + ab+ba+b^2 = a+ab+ba+b$$ giving $ab + ba = 0$. From the above property, $ba = -ba$, so we have $ab = ba$, i.e. $R$ is a commutative ring.

answered Nov 15 '21 at 13:31

stoic-santiago

14,877

Why not simply $-r=(-r)^2=r^2=r$? – Desperado Nov 15 '21 at 13:36
Can you justify why $(-r)^2 = r^2$? @Desperado – stoic-santiago Nov 15 '21 at 13:39

Let $R$ be a ring in which $r^2=r$ for all $r \in R$. Show that $r=-r$ for all $r \in R$ and that $R$ is commutative.

1 Answers1