= 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,
$$
\boxed{\sqrt{\mathfrak q}=\mathfrak p,\qquad ab\in\mathfrak q,\ a\notin\mathfrak q\ \Longrightarrow\ b^n\in\mathfrak q\text{ for some }n\geq1.}
$$
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> $R/\mathfrak q$.
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