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.
Best Answer
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.