 Solution

An ideal $\mathfrak{q}$ is $\mathfrak{p}$-primary if it is a proper primary ideal whose radical of an ideal is $\mathfrak{p}$. Explicitly,Equivalently, $ab\in \mathfrak{q}$ and $b\notin \mathfrak{p}$ imply $a\in \mathfrak{q}$: multiplication by every element outside $\mathfrak{p}$ is injective on the quotient ring $R/\mathfrak{q}$.