TITLE: Do the zero divisors form an ideal?
QUESTION [11 upvotes]: 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?

REPLY [8 votes]: It is not an ideal, as the next example show.
Consider the ring $\mathbb R \times \mathbb R$. The elements $(1,0)$ and $(0,1)$ are zero-divisors but their sum is not. Thus, the set of zero-divisors is not closed under sums.
For your second question, consider the polynomial ring in one variable $\mathbb R[x]$, clearly $x$ is neither a zero-divisor or a unit.