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I had this question on a final exam. I was wondering if anyone knows a proof for it.

Let R be a ring not necessarily with unity ($1 \neq 0$ and $x \in R$ so that $x*1=x=1*x$) and let R have the property that every element of R is idempotent that is $\forall x \in R\quad x*x=x$. Prove that $R$ is commutative.

Not to put my whole answer up because I know I was wrong. I will put the part where I had trouble going to the next step.

My argument went like this: let $a,b \in R$ then $ab\in R$ cause $R$ is a ring. Therefore $abab=ab$.

Now $aabb=ab$ as well, so $$abab=aabb\ \Rightarrow abab-aabb=0 \Rightarrow\ a(ba-ab)b=0\ \because\text{ distributive property of $R$}.$$

Now because R is not a division ring then the element $a$ could be a divisor of zero or $b$ could be as well so you cannot just assume $ba-ab=0$. This is where I am stuck.

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marked as duplicate by Jack Schmidt, Julian Kuelshammer, O.L., Lord_Farin, Charles Jun 18 '13 at 19:37

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1 Answer 1

Note that $(x+y)=(x+y)^2=x^2+xy+yx+y^2$. Now, using the property that $x^2=x$ and $y^2=y$, we have $xy+yx=0$ so that $xy=-(yx)$. Also, since $x=x^2=(-x)^2=-x$, each element is its own additive inverse, i.e., $xy=yx$ so the ring is commutative.

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So (as shown above by Clayton) $x+x=0$ for any $x\in R$. –  juniven Dec 17 '12 at 3:12
    
@Clayton thanks for the help it kind of makes me angry that I missed the question on the test considering the answer is so simple :). –  drew Dec 17 '12 at 3:21
    
@drew, no problem! I'm glad you understand it; that is key. –  Clayton Dec 17 '12 at 3:39
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