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Suppose a and b belong to a commutative ring and ab is a zero-divisor. Show that either a or b is a zero-divisor.

My answer goes like this:
If ab is a zero-divisor, then there exists a nonzero element c such that (ab)c = 0.
Assume b is not a zero divisor. Then bc is nonzero. By using associativity, we can write (ab)c = 0 as a(bc) = 0.
This means that a is a zero-divisor (because bc is assumed to be nonzero).
Is it really that easy? Seems like there should be more to it.

frierfly
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    Your answer is correct, if you find it easy look for harder, because there is always more difficult ones! – Elaqqad Apr 05 '15 at 23:19
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    Yes, every part of math is so easy once you fully get there.. – Berci Apr 05 '15 at 23:37
  • I think your proof glossed over an important logical fact, namely the use of the law of excluded middle , e.g. '$b$ is a zero divisor or $b$ is not a zero divisor'. In the former case the proof is done; in the latter case your proof continues. Also after "Assume $b$ is not a (non-zero) zero divisor" , you write "Then $bc$ is nonzero" , but this does not follow. It is possible that $b=0$. True, if $b=0$ then $ab =0$ which contradicts $ab$ being a (non-zero) zero divisor. But you should state this. Also it is worth mentioning $bc$ is nonzero, otherwise $b$ would be a zero divisor. – john Mar 04 '20 at 17:27
  • You should not use "either-or" here. This phrase is often understood in the sense of an exclusive or (exactly one of both is true). But admittedly there does not seem to exist a commonly accepted interpretation of "either-or" in mathematics. See https://math.stackexchange.com/q/2130327. – Paul Frost Mar 27 '23 at 10:10

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Let a and b belong to a commutative ring and ab be a zero-divisor. Then there is a c ≠ 0 such that (ab)c = 0. Observe that ab ≠ 0 implies that a ≠ 0 and b ≠ 0. Suppose that a is not a zero divisor. Then a(bc) = 0 implies that bc = 0 but this means that b is a zero-divisor. Hence, either a or b is a zero-divisor.

Rahul
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