$Q$ is primary if and only if $Ass(R/Q) = \big\{\sqrt Q\big\}$
Definition $Q$ is primary if $ab\in Q$ and $a\not\in Q$, then $b^n \in Q$ for some $n\in \mathbb{N}$.
Definition A prime ideal $P\subset R$ is said to be associated to $R/Q$ if there is a nonzero $\tilde r\in R/Q$ such that $P = Ann(\tilde r)$, and the collection of associated primes is denoted by $Ass(R/Q)$.
Definition $\sqrt Q = \{r\in R : r^n \in Q \text{ for some } n\in \mathbb N\} = \bigcap_{Q \subset M \text{ maximal}} M$.
To show $(\Leftarrow )$, suppose $Ass(R/Q) = \big\{\sqrt Q\big\}$, then for each $\tilde r \neq \tilde0$ and $s\tilde r = \tilde 0$, we have $sr\in Q$ and $s\in \sqrt Q$, then $s^n \in Q$, this shows $Q$ is primary.
Now for $(\Rightarrow)$ if $Q$ is primary, for each $\tilde r \neq \tilde 0$, if $s\tilde r = \tilde 0$ then $sr\in Q$, since $Q$ is primary we have $s^n \in Q$, thus $s\in \sqrt Q.$ But here how can I show that $ann(\tilde r) \supset \sqrt Q$?
And there is no Noetherian assumption used.
Is my proof correct? it is a lot shorter than the ones posted here and here.