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Let $\mathcal{R}$ be a unital ring with the property that $\forall p \not= 0_{R}$, $\exists$ unique $q$ such that $p = pqp$. Prove that $\mathcal{R}$ has no zero divisors.

These are some potentially helpful results I have managed to obtain:

  1. $p$ is the unique element such that $q = qpq$

  2. $\forall n > 0: \space pq = (pq)^{n}$

  3. $\forall n > 0: \space qp = (qp)^{n}$

I found this problem in an old exam paper when I was studying for an exam some time ago, literally every other question in the paper was straightforward, so I assumed I'd figure it out later and it would be good practice to do so... No such luck.

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1 Answer 1

up vote 6 down vote accepted

Suppose $p \neq 0$, $pr = 0$, and let $q$ be the unique element of $R$ such that $p = pqp$. Then $p(q-r)p = pqp - prp = p$. So $q-r = q$ by uniqueness of $q$, i.e. $r = 0$.

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2  
As a bonus, to show uniqueness of $q$ was necessary for the result, $R = \mathbb{Z}/6\mathbb{Z}$ has the property that for all $0 \neq p \in R$ there exists $q \in R$ such that $pqp = p$, but $R$ does have zero divisors. –  user73985 Dec 19 '13 at 15:42
    
If $pqp=pq'p$ then $p(q-q')p=0$ so $q \neq q'$ implies that $p=0$ or $p$ is a zero-divisor.(+1) –  drhab Dec 19 '13 at 15:52
2  
Corollary: such a ring is a division ring. Proof: $pq^2=pq=e$ is an idempotent, but if $e\notin\{0,1\}$, then $e(1-e)=0$ yields zero divisors. Thus $pq=1$. Similarly $qp=1$. –  rschwieb Dec 19 '13 at 17:25

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