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Let $R$ be a finite ring that satisfies the following conditions:

(1) For any $x\in R$, if $x\ne 0$ then $x^2\ne 0$.

(2) There exists at most one nonzero element $y\in R$ that satisfies $y^2=y$.

How can we show that $R$ has no zero divisors?

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    If the ring is commutative the question ammounts to prove the ring is, in fact, a field. Knowing this perhaps can give some insight, yet the answer Mariano wrote below gives a general solution without assuming commutativity.2012-12-09

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From what you write, it is impossible to know what you know, so this may be incomprehensible to you, but here it goes. I will assume the ring has a unit.

Since $R$ is finite, it is Artinian. Its Jacobson radical $J(R)$ is therefore nilpotent, so in particular its elements are themselves nilpotent: in view of your first hypothesis, $J(R)=0$ and the ring is in fact semisimple. The theorem of Artin-Wedderburn tells us that $R$ is a direct product of matrix rings over division rings. Since there is only one idempotent, the product can at most contain one factor, and the size of the matrices in that factor must be one. We thus see that $R$ itself is a division ring.

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    Ah, of course. Extra characters.2012-12-09