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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?

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    $\begingroup$ 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. $\endgroup$ – Platehead Jan 14 '15 at 12:43
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    $\begingroup$ In $Z$, $-1,1$ are units. All other non-zero elements are non-units and none are zero divisors $\endgroup$ – John McGee Jan 14 '15 at 12:44
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    $\begingroup$ In fact $\mathbb Z$ is a counter example of your claim $\endgroup$ – Anupam Jan 14 '15 at 12:50
  • $\begingroup$ You seem to be checking $p \to q$ and then the equivalent $\neg q \to \neg p$. $\endgroup$ – Orest Bucicovschi Jan 14 '15 at 12:53
  • $\begingroup$ 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. $\endgroup$ – drhab Jan 14 '15 at 12:59

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