Let $A$ be a commutative ring with identity. Is it always true that the sum of a nilpotent element and a zero divisor is a zero divisor? I tried to construct a counterexample but found it was true on the ring $\mathbb{Z}/n\mathbb{Z}$. Can anyone please give a specific example to this problem?
Some of my attempt: For a nilpotent element $x$ and a zero divisor $z$ the sum $x+z$ satisfies $z'(x+z)=z'x$ for some $z' \neq 0$, and since $x$ is nilpotent $z'^n(x+z)^n = 0$ for some $n$. I thought we can find a ring $A$ and some $n$, $z$ where $z'^n(x+z)^{n-1}$ becomes $0$ but it doesn't mean there are no other non-zero elements that annihilates $x+z$.
Best Answer
Suppose $zw = 0$ with $w \ne 0$, but $x+z$ is not a zero-divisor. Then $xw = (x+z)w \ne 0$, $x^2 w = (x+z) xw \ne 0$, and by induction $x^n w \ne 0$ for all positive integers $n$, implying $x^n \ne 0$ and $x$ is not nilpotent.