Let $R$ be a ring and let $I$ be an ideal of $R$. Show that if $J$ is an ideal of $R$ that is maximal in the set of ideals of $R$ that annihilate elements of $R/I$, then $J$ is a prime ideal of $R$.
I've tried to show that J is prime directly by supposing that $ab\in J$. Without loss of generality suppose that $b\notin J$. Now let $\overline{i}\in R/I$. Then $ab\overline{i}=0$ in $R/I$... not sure where to go from here... need maximality of $J$.
Assuming commutative rings with identity, since so much is up in the air without it
Suppose $J=ann(x+I)$ is maximal among other point annihilators of elements in $R/I$. Suppose $ab\in J$ but $b\notin J$.
It follows that $b$ does not annihilate $x+I$, for if it did, $bR+J$ would also annihilate $x+I$ and would properly contain $J$ and contradict its maximality.
But $a\in ann(bx+I)$, since $abx\in I$.
Now note that $ann(x+I)\subseteq ann(bx+I)$. Can you finish from here?