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Suppose there are non-trivial idempotents in the ring without unity. Is it right that all of them are zero divisors?

If we're given unitary ring with unity $e$ and $a$ is non-trivial idempotent then $e-a \neq 0$. But $a(e-a) = 0$ so $a$ is zero divisor.

But i'm not sure about the case when ring doesn't have unity.

EDIT In the non-commutative case i wonder whether each non-zero idempotent must be SOME kind of zero divisors, i.e. left or right, or perhaps both.

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    $\begingroup$ What do you mean by non-trivial? Usually one demands $\neq 0,1$, but we cannot even formulate $\neq 1$ here! And of course it isn't true when we only demand $\neq 0$. $\endgroup$ – Martin Brandenburg May 21 '13 at 23:58
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    $\begingroup$ consider the ring which is the ideal generated by $x$ and $y$ in the quotient of the free algebra $k\langle x,y\rangle$ by the ideal generated by $x^2-x$ and $y^2-y$. Does that work? $\endgroup$ – Mariano Suárez-Álvarez May 22 '13 at 0:06
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A partial answer:

If the ring is commutative and finite, then the answer is yes. Take $e\in R$ an idempotent and define $f:R\to R$ by $f(a)=ea$. If $e$ is not a zerodivisor then $f$ is injective and therefore bijective. Now prove that $e$ is the unity of $R$: if $b\in R$ then there exists $a\in R$ such that $b=ea$ and we have $eb=e(ea)=ea=b$.

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  • $\begingroup$ If you mean the ideal $3\mathbb{Z}_6=\{0,3\}$, this has $3$ has identity element. If you meant something else, sorry. But what do you mean, then? $\endgroup$ – Julien May 21 '13 at 22:30
  • $\begingroup$ @julien Right. I've proved now that finite commutative rings can't provide a counterexample. $\endgroup$ – user26857 May 21 '13 at 23:50
  • $\begingroup$ Right, that was a good point make, +1. The argument works in the noncommutative case as well. You get $b=ce$ whence $be=b$. $\endgroup$ – Julien May 22 '13 at 1:04
  • $\begingroup$ @julien I know that this can work in the noncommutative setting as well, but what then means a zerodivisor: left, right, left and right? $\endgroup$ – user26857 May 22 '13 at 6:51
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    $\begingroup$ Being the identity follows from injectivity of multiplication by $e$ alone, so surjectivity and finiteness conditions are not necessary. $\endgroup$ – rschwieb Dec 28 '14 at 12:01
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Suppose $e$ is idempotent and not a right or left zero divisor. Since multiplication on either side by $e$ is injective, $(xe-x)e=0$ and $e(ex-x)=0$ for all x implies $e$ is the identity of the ring.

So you can conclude this:

In a ring without identity, a nonzero idempotent $e$ must be a zero divisor on one side or the other. If instead the ring has an identity other than $e$, $e$ is a zero divisor on both sides.

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