# Sum of a nilpotent element and a zero divisor in a commutative ring $A$.

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 $x+z$.

• @Suzet The point is not that $x$ can be a zero divisor, it is that in the remaining case, the non-zero factor that kills $x$ is of a very special form $z'x^k$ for some $k$. Then $z'x^k(x+z)=0$. Ah! comment deleted. Let me add what was there. – user574889 Jul 19 '18 at 1:47
• @cactus Yes, I agree. In fact, I just noticed how trivial my comment was so I had to erase it, it wasn't relevant whatsoever. – Suzet Jul 19 '18 at 1:48
• The whole argument: $(x+z)z'x^{n-1}=0$. Then either $z'x^{n-1}\neq0$ and $(x+z)$ is a zero divisor, or $z'x^{n-1}=0$. In the latter case $z'x^{k}\neq0$ for some $k$ maximum (take into account that for $k=0$, $z'x^0=z'\neq0$). Then $(x+z)z'x^k=0$ showing that $(x+z)$ is a zero divisor. – user574889 Jul 19 '18 at 1:51
• @Suzet What are you saying? It was the beginning to a complete proof. – user574889 Jul 19 '18 at 1:52

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.