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.
 A: Hints: 
Reflect the problem in the quotient ring $R/Q$. Prove (if you don't already believe) that the definition of $Q$ being primary is equivalent to $R/Q$ satisfying "$ab=0$ implies $a=0$ or $b$ is nilpotent."
The condition that $\sqrt{Q}$ is maximal implies that $R/Q$ has exactly one maximal ideal, which is $\sqrt{Q}$. Thus it's a local ring.
Let $ab=0$ in $R/Q$ and suppose $b\notin \sqrt{Q}$. What can you say that is special about $b$ if it is outside the maximal ideal of a local ring? Conclude that $a=0$.
A: 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 $(y) = y(Q + (x)) \subseteq Q + (xy) = Q$.
A: Assume that $R$ is a commutative ring and $I⊆R$ is an ideal, such that the radical $r(I)$ of I is a maximal ideal. Then $I$ is a primary ideal. Proof. We will show, that every zero divisor in $R/I$ is nilpotent First of all, recall that $r(I)$ is an intersection of all prime ideals containing $I$. Since $r(I)$ is maximal, it follows that there is exactly one prime ideal $P=r(I)$ such that $I⊆P$. In particular the ring $R/I$ has only one prime ideal (because there is one-to-one correspondence between prime ideals in $R/I$ and prime ideals in $R$ containing $I$). Thus, in $R/I$ an ideal $r(0)$ is prime.Now assume that $α∈R/I$ is a zero divisor. In particular $α≠0+I$ and for some $β≠0+I∈R/I$ we have $αβ=0+I.$ But $0+I∈r(0)$ and $r(0)$ is prime. This shows, that either $α∈r(0)$ or $β∈r(0)$. Obviously $α∈r(0)$ (and $β∈r(0)$), because $r(0)$ is the only maximal ideal in $R/I$ (the ring $R/I$ is local). Therefore elements not belonging to $r(0)$ are invertible, but $α$ cannot be invertible , because it is a zero divisor. On the other hand $r(0)=\{x+I∈R/I: (x+I)^n=0 \ \ for  \ some \ \ n∈\mathbb{N}\}$. Therefore $α$ is nilpotent and this completes the proof.
A: There's not too much to the proof: 
Let $\sqrt{Q}$ be a maximal ideal $\mathfrak{m}$. Then $Q$ is $\mathfrak{m}$-primary. 
$\it{Proof}$: Suppose that $\alpha \beta \in Q$ and $\beta$ is not in $\sqrt{Q}=\mathfrak{m}$. Then by the maximality of $\mathfrak{m}$,  it follows that $\mathfrak{m} +R\beta=R$ where $R$ is a ring. Then, for some $m \in \mathfrak{m}$ and $r \in R$ we have $m+r\beta=1$. Now $m \in \mathfrak{m}= \sqrt{Q}$, hence $m^{n} \in Q$ for $n \geq 1$. Thus, $1=1^{n}=(m+r\beta)^{n}=m^{n} +s\beta$ for some $s \in R$. Multiply by $\alpha$ to get $\alpha=\alpha m^{n}+s\alpha \beta \in Q$.
