= Primary ideals for a regular principal prime
{title2=$Q=(r^k)$}
If a <prime ideal> $P=(r)$ is generated by a <non-zero-divisor>, its $P$-<primary ideals> are exactly $(r^k)$ for $k\geq1$, in any <commutative ring>. To prove $(r^k)$ is <primary>, if $ab\in(r^k)$ and $a\notin(r)$, the <prime ideal> condition gives $b=rb_1$; 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 $x=r^ju$ as far as possible with $0\leq j<k$. 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 $k$.
\Comment[[
Exact target: finite-generation-clears-localization-kernels
Existing parent: Localization of a ring
Intended file: /home/ciro/bak/git/codex-wiki/commutative-algebra.bigb
]]
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