Verification of proof that for a,b in ring R, assuming ab is a zero divisor at least one of a and b is zero divisor

Question

I'm not so sure if this is correct but here's what I have so far:

ab is a zero divisor iff there is a c$\neq$0 s.t. (ab)c=c(ab)=0 given ab$\neq$0 and c$\neq$0. Then we have (ab)c-c(ab)=0=(ab)c+c-c-c(ab)=0 which breaks down into (ab+1)c-c(1+ab)=0. We already know c cannot be 0 so for 1+ab=0 ab must equal -1.

Is this what the proof is asking for? If not, how would I correct it?

Answer 1

No, that is not what the proof is asking for. You seem to have the wrong definition of zero divisor.

Let's assume that the ring $R$ is commutative for simplicity.

$x\in R$ is a zero divisor if there is $y\in R$, $y\ne0$, such that $xy=0$.

Thus, if $ab$ is a zero divisor, then there is $c\in R$, $c\ne0$, such that $(ab)c=0$.

Then, $0=(ab)c=a(bc)$. If $bc=0$, then $b$ is a zero divisor. If $bc\ne0$, then $a$ is a zero divisor.