The problem is to prove that, in a commutative ring with identity, the set of ideals in which every element is a zero divisor has a maximal element with respect the order of inclusion, and that every maximal element is prime. But I´m thinking* that considering the set of all zero-divisors, this set as I see it´s an ideal, and thus must be the only maximal element.
An ideal is defined as a subset J of the ring R , such that for every $ x\in R$ we have $
xJ \subset J
$
I think that I´m wrong, only because it´s rare.
An extra question but related, there exist rings with element x, such that x is not a zero divisor nor a unit?
Best Answer
Your definition of an ideal seems nonstandard: for most folks, an ideal must also be a subgroup of the additive group of the ring. So, for counterexample to your conjecture, in the ring $\mathbf Z/6\mathbf Z$ there are zero-divisors $2$ and $3$, the sum of which is a unit.
To the second question: yes! As Alex notes in the comments, $2 \in \mathbf Z$ is an example. Any integral domain which is not a field also provides examples. But it's a good pigeonhole exercise to show that you cannot find such an element in a finite ring.