Show that there exists a prime ideal contained in $J = \{a \in R : ab = 0 \text{ for some } b\neq 0\}$ Let $R$ be a commutative ring with unity. Let
$$J = \{a \in R : ab = 0 \text{ for some } b\neq 0\}$$
Show that there exists a prime ideal contained in $J$.
Attempt: 
I can see that this set contains all zero divisors along with zero itself.
Considering the contrapositive of the definition of prime ideal we need to prove that if $x,y \in J^c \implies xy \in J^c$.
Proof: Assume if possible $xy \in J$ then there exists a non-zero element $z$ such that $(xy)z = 0$, using the associative property of multiplication $x(yz) = 0$, now $yz \neq 0$ as $y \in J^c$ hence we can conclude $x(yz) \neq 0$. Which gives us the required contradiction.
There are two questions I want to ask:
Does the proof furnished above answer the question asked ?
But what I don't understand is why is the phrase "there exists a prime ideal contained in J" being used, why not use show that "J is the prime ideal"?
 A: $J$ is the set of zero divisors. You argued correctly that $x, y \in J^c \Rightarrow xy \in J^c$. However, $J$ is not an ideal in general, as my answer in the comments shows.
The question now is to show that $J$ contains a prime ideal $\mathfrak{p}$, i.e. an ideal $\mathfrak{p}$ consisting of only zero divisors ($\mathfrak{p} \subseteq J$) and such that $x, y \in \mathfrak{p}^c \Rightarrow xy \in \mathfrak{p}^c$. It's not clear from what you said yet that such an ideal exists.
Let's consider the set $\Omega$ of ideals contained in $J$. Clearly $\Omega \neq \emptyset$ as $\lbrace 0 \rbrace \in \Omega$. Using Zorn's Lemma one sees that this set $\Omega$ contains a maximal ideal $\mathfrak{m}$. Use the part of the argument you already had to show that $\mathfrak{m}$ is a prime ideal, and then you're done.
A: That is because the set of zero-divisors, in general, is not even an ideal.
Counter-example: in $\mathbf Z/6\mathbf Z$ the zero-divisors are the congruence classes of $2, 3, 4$. These are not an ideal, since $2+3=5$, for instance, is a unit in $\mathbf Z/6\mathbf Z$.
Hint:
Consider the set of ideals contained in $J$, ordered by inclusion, and apply Zorn's lemma to obtain an ideal contained in $J$ which is maximal w.r.t. inclusion. Then show this ideal is prime .
