In the setting of Chapter 4 of Atiyah and Macdonald's Commutative algebra, an ideal $\mathfrak{q}$ in a commutative ring is primary if, $xy \in \mathfrak{q}$ implies either $x \in \mathfrak{q}$ or $y^n \in \mathfrak{q}$ for some $n >0$. If $\mathfrak{q}$ is primary, then its radical $r(\mathfrak{q})$ is prime, and is moreover the smallest prime ideal containing $\mathfrak{q}$. (Conversely, if the radical of $r(\mathfrak{a})$ is maximal, then $\mathfrak{a}$ is primary.)
If $r(\mathfrak{q})=\mathfrak{p}$, we say that $\mathfrak{q}$ is $\mathfrak{p}$-primary. The authors show by way of example that we need not have $\mathfrak{q}=\mathfrak{p}^n$, and provide an example in the polynomial ring $k[x,y]$ of a case in which $\mathfrak{p}^2 \subsetneq \mathfrak{q} \subsetneq \mathfrak{p}$.
My question is:
If $\mathfrak{q}$ is $\mathfrak{p}$-primary, is it always the case that $\mathfrak{q}$ sits between two consecutive powers of $\mathfrak{p}$, i.e. we have inclusions $\mathfrak{p}^{n} \subset \mathfrak{q} \subset \mathfrak{p}^{n-1}$ for some $n$? If not, can anything be said in general about the relationship between $\mathfrak{q}$ and powers of $\mathfrak{p}$?
edit: This is quite a bad question, and its quick to come up with quite extreme counterexamples.
Fact: an ideal whose radical is a maximal ideal $\mathfrak{m}$ is $\mathfrak{m}$-primary.
You can take the ring $k[x_1,x_2,\dots]$ of polynomials in infinitely many variables and $\mathfrak{m}=(x_1,x_2,\dots)$. The ideal $(x_1,x_2^2,x_3^3,\dots,x_i^i,\dots)$ has $\mathfrak{m}$ as radical, hence is $\mathfrak{m}$-primary. However, it contains no power of $\mathfrak{m}$.
In a Noetherian ring you have the primary decomposition. Can you conclude anything?