 Maximal-primary ideal criterion

In a Noetherian ring, an ideal $Q$ is $\mathfrak{m}$-primary for a maximal ideal $\mathfrak{m}$ if and only if ${\mathfrak{m}}^n\subseteq Q\subseteq \mathfrak{m}$ for some $n\ge 1$. For necessity, choose generators $x_1,\dots ,x_d$ of $\mathfrak{m}$ and exponents $e_i\ge 1$ with $x_i^{e_i}\in Q$. Every product of $1+{\sum }_i(e_i-1)$ generators lies in $Q$. If $\mathfrak{m}=0$, take $n=1$. Conversely the inclusions imply $\sqrt{Q}=\mathfrak{m}$, so $R/Q$ is a local ring with maximal ideal $\mathfrak{m}/Q$. Every element outside this maximal ideal is a unit, while every element inside is nilpotent. Therefore every zero divisor is nilpotent, which is the primary ideal condition. The reverse implication works for any commutative ring.