An ideal whose radical is maximal is primary 
I've got to prove that an ideal $Q$ whose radical is a maximal ideal is a primary ideal. That is, I want to prove that if $xy\in Q$, then $x\in Q$ or $y^n\in Q$ for some $n>0$. 

I've been trying for a while and I'm not sure where to begin. All I've got is that if $\text{Rad}(Q)$ is maximal and by definition it's the intersection of all prime ideals $P_i$ containing $Q$, then it must be equal to each of these $P_i$. So there is only 1 prime ideal containing $Q$, namely $\text{Rad}(Q)$.
Could anyone point me in the right direction? Thanks for any replies.

Here is a different proof.  Assume $xy \in Q$ but $x \notin \sqrt{Q}$.  Since no maximal ideal contains both $Q$ and $(x)$, we have $Q + (x) = R$.  Thus $(y) = y(Q + (x)) \subseteq Q + (xy) = Q$.
