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I know that in a finite ring, an element must be either a unit or a zero divisor. However, the proof that I have seems to also work for infinite ring.

My proof is as such: Assume that a $\in$ R is a unit and ab = 0 for some b $\in$ R. Then $b = a^{-1}(ab) = a^{-1}0 = 0$, so a is not a zero divisor. Similarly, if a is a zero divisor, then a is not a unit.

However, if I am not wrong, in $\mathbb{Z}$, all the non-zero elements are both not unit and zero divisor. What am I misunderstanding here?

user10024395
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    You seem to have proven that no element can be both a unit and a zero divisor, which is a little different from the statement that everything is either a unit or a zero divisor. – Platehead Jan 14 '15 at 12:43
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    In $Z$, $-1,1$ are units. All other non-zero elements are non-units and none are zero divisors – John McGee Jan 14 '15 at 12:44
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    In fact $\mathbb Z$ is a counter example of your claim – Anupam Jan 14 '15 at 12:50
  • You seem to be checking $p \to q$ and then the equivalent $\neg q \to \neg p$. – orangeskid Jan 14 '15 at 12:53
  • I agree with @Platehead. Btw, if it is proved that a unit is not a zero divisors then it is redundant to prove also that a zero divisor is not a unit. The statements are the same. – drhab Jan 14 '15 at 12:59

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You have Proved the following statement:

An element $r\in R$ cannot be both zero-divisor and a unit simultaneously.

But to prove that

An element must either be a zero divisor or unit in a commutative ring $R$ requires the ring $R$ to be finite.

Please look for the following link:

Every nonzero element in a finite ring is either a unit or a zero divisor

Kumar
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