Primary ideals in Noetherian rings

For an $R$-module $M$ I have the following definition for a submodule $N\subset M$ to be $\mathfrak{p}$-primary: this is the case when $\text{Ass}(M/N) = \{\mathfrak{p}\}$, that is, $M/N$ is coprimary or $(0)$ is a primary submodule.

Now for an ideal $\mathfrak{q} \subset R$ this definition gives that $\mathfrak{q}$ is $\mathfrak{p}$-primary iff $\text{Ass}(R/\mathfrak{q}) = \{\mathfrak{p}\}$. It can be shown that this is equivalent with $\mathfrak{p}^n \subset \mathfrak{q}$ for some $n\in \mathbb{N}$ and the condition \begin{align*} \forall r,s \in R: rs \in \mathfrak{q}, r\not\in \mathfrak{p}\Rightarrow s \in \mathfrak{q}. \end{align*}

Now I suspect that if $R$ is Noetherian, and if $\mathfrak{q}$ is $\mathfrak{p}$-primary, then we have $\sqrt{\mathfrak{q}} = \mathfrak{p}$. I have shown this for PID's, but does this generaly hold?

And if so, is every ideal $\mathfrak{q}\subset R$ then $\sqrt{\mathfrak{q}}$-primary?

Many thanks.

Definition. An associated prime of an $R$-module $M$ is a minimal prime over $\operatorname{Ann}(x)$ for some $x\in M$, $x\ne 0$.

We have,

$\operatorname{Ass}(R/\mathfrak{q}) = \{\mathfrak{p}\}$ if and only if $\mathfrak p=\sqrt{\mathfrak q}$ and $\forall r,s \in R: rs \in \mathfrak{q}, s\not\in \mathfrak{q}\Rightarrow r \in \mathfrak{p}.$

"$\Rightarrow$" Let $\hat y\in R/\mathfrak q$, $\hat y\ne \hat 0$. A minimal prime over $\operatorname{Ann}(\hat y)$ is, by definition, associated to $R/\mathfrak q$, and therefore equals $\mathfrak p$. This shows that $\sqrt{\operatorname{Ann}(\hat y)}=\mathfrak p$. In particular, for $y=1$ we get $\sqrt{\mathfrak q}=\mathfrak p$.
If $rs \in \mathfrak{q}$, and $s\notin \mathfrak q$ consider a minimal prime over $\operatorname{Ann}(\hat s)$ which necessarily equals $\mathfrak p$ and since $r\hat s=\hat 0$ we get $r\in \mathfrak p$.

"$\Leftarrow$" Now let $\mathfrak p'\in\operatorname{Ass}(R/\mathfrak{q})$. (Note that $\operatorname{Ass}(R/\mathfrak{q})\ne\emptyset$ since $R/\mathfrak q\ne0$.) Then there is a non-zero element $\hat s\in R/\mathfrak q$ such that $\mathfrak p'$ is minimal over $\operatorname{Ann}(\hat s)$. But $\mathfrak q\subseteq\operatorname{Ann}(\hat s)\subseteq\mathfrak p$ and therefore $\sqrt{\operatorname{Ann}(\hat s)}=\mathfrak p$. This gives us $\mathfrak p'=\mathfrak p$.

Remark. I don't think you can prove that $\mathfrak p^n\subseteq \mathfrak q$ unless $R$ is noetherian.