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.
Best Answer
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 $\alpha=\alpha m^{n}+s\alpha \beta \in Q$.