Primary ideal

Remarks

The radical $P = \sqrt{Q}$ of a primary ideal is a prime ideal, and one says that $Q$ is $P$-primary. Example: for a prime $p$ and $n \ge 1$, the ideal $(p^n)$ of $\Z$ is primary; if $p^n \mid ab$ and $p^n \nmid a$, then $p \mid b$, so $p^n \mid b^n$. Non-example: $(6)$ in $\Z$ fails clause 2, because $2 \cdot 3 \in (6)$ and $2 \notin (6)$, but no power of $3$ is divisible by $6$. Every prime ideal is primary, and every power of a maximal ideal is primary. A power of a prime ideal need not be primary: in $R = k[x, y, z]/(xy - z^2)$ over a field $k$, with $P = (x, z)$, one has $xy = z^2 \in P^2$, $x \notin P^2$ and no power of $y$ in $P^2$.