Is $Z(R)$ a maximal ideal? If $p$ and $q$ are two maximal ideals in the set of zero-divisors in a ring $R$ with non-zero intersection between $p$ and $q$.
does the set of all zero-divisors are a maximal ideal and equal the union of $p$ and $q$?  

Different phrasing:
Let $Z(R)$ be the set of all zero-divisors of $R$. Let $p$ and $q$ be maximal ideals contained in $Z(R)$ with $p\cap q\neq\{0\}$. Show that $Z(R)=p\cup q$. 
 A: Take any field $F$ and consider the ring $R=F^3$. It has three maximal ideals, corresponding to sets of elements where are zero on some fixed coordinate. Let's call them $I_1, I_2,I_3$ depending on if they contain elements that are zero on coordinate 1,2,3 respectively. Clearly all three consist of zero divisors. $I_1\cap I_2\neq \{0\}$ since it is the set of elements zero on both coordinates 1 and 2 (but they are nonzero on 3, often.) 
Now every element of $I_1\cup I_2$ is zero on either coordinate 1 or coordinate 2. But there are elements of $I_3$ (necessarily zero divisors) which do not satisfy this. Thus $I_1\cup I_2\neq ZD(R)$. 
In fact, $ZD(R)=I_1\cup I_2\cup I_3$. In any commutative Artinian ring, the zero divisors are the union of all the maximal ideals.
As for the title question/second part of your question about the zero divisors being an ideal, this is only the case in a commutative Artinian ring has exactly 1 maximal ideal. 
A: Let B= be the union of all prime ideals which consisting elements of zero-divisors. Clearly, B⊆Z(R). 
Now let a∈Z(R). Then (a)⊆Z(R), and (a)∩S=∅.
where S is a multiplicative closed set in R. So there exists a prime ideal P such that(a)⊆P and P∩S=∅. This implies a∈P⊆B.
 Hence B=Z(R). #                                                         
