Every ideal $J \supsetneq \sqrt{0}$ contains a non-zero-divisor

Question

I'm looking for an example of a commutative ring with unity, that has elements in each of the following three classes:

• Non-zero nilpotents
• Non-nilpotent zero-divisors
• Non-zero-divisors

and furthermore satisfies the following condition:

• Any ideal $J$ which strictly contains the nilradical $\sqrt{0}$ also contains a non-zero-divisor.

Any comments or partial answers are appreciated.

Answer 1

Such a ring doesn't exist because the second and last conditions are incompatible.

Let $x$ be a non-nilpotent element. By the last condition, the ideal $\big( x,\sqrt{0} \big) \supsetneq \sqrt{0}$ contains a non-zero-divisor, i.e. there exist elements $a, b$ with $ax + b$ a non-zero-divisor, and $b^n = 0$ for some positive integer $n$.

But then $(ax + b)^n$ is a non-zero-divisor, since if $(ax+b)^n k = 0$, then $(ax+b)(ax+b)^{n-1} k = 0$, contradicting that $ax+b$ is not a zero-divisor. Furthermore, expanding $(ax + b)^n$ shows it is of the form $cx$ for some element $c$, so $x$ is also a non-zero-divisor.