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.
Given \(a, b \in R\) where \(R\) is a commutative ring and \(ab \neq 0\). By definition of zero divisor \(\exists c \in R\) such that \(c \neq 0\) and \((ab)c = 0\) (meaning \(c\) is also a zero divisor. This is because zero divisor has a mutual property). By associativity and commutativity \((ab)c = 0 = a(bc) = b(ca)\).
From \(a(bc) = 0\) we get \(a = 0\) or \(bc = 0\). \(a\) can’t equal \(0\) because that would make \(ab = 0\) which contradicts our given that \(ab\) is a non-zero element of \(R\) by definition of being a zero divisor. If \(bc = 0\) then \(b\) is a zero divisor or \(b = 0\), but \(b\) can’t equal \(0\) because that would make \(ab = 0\). Thefore \(b\) is a zero divisor.
We can do the same for \(b(ca) = 0\) to show \(a\) is also a zero divisor. This completes our proof.
I think there is a bit of redundancy in this proof but it makes the proof clearer in some ways. I’m using Gallian’s Contemporary Abstract Algebra for definitions. If you want to know more about proofs in general I recommend Velleman’s How to Prove It: A Structured Approach.