Primary ideals for a regular principal prime ($Q=(r^k)$)

If a prime ideal $P=(r)$ is generated by a non-zero-divisor, its $P$-primary ideals are exactly $(r^k)$ for $k\ge 1$, in any commutative ring. To prove $(r^k)$ is primary, if $ab\in (r^k)$ and $a\notin (r)$, the prime ideal condition gives $b=r{b}_1$; cancelling $r$ and inducting gives $b\in (r^k)$. Also $\sqrt{(r^k)}=(r)$. Conversely, for a $P$-primary ideal $Q$, choose the least $k$ with $r^k\in Q$. If $x\in Q\setminus (r^k)$, factor $x=r^{j}u$ as far as possible with $0\le j<k$. Then $u\notin (r)$, so the primary ideal condition forces $r^j\in Q$, a contradiction. Thus $Q=(r^k)$. The powers are strictly decreasing: equality $(r^k)=(r^{k+1})$ would let cancellation give $1=rc$. This also proves uniqueness of $k$.