Past exam of the commutative algebra course of the University of Oxford / 2023 / 1 / e

Because $(r)$ is a proper prime ideal, each $(r^k)$ is proper and its radical of an ideal isIndeed, $(r^k)\subseteq (r)$ gives one inclusion of radicals of ideals, while $(rc{)}^k\in (r^k)$ gives the other.

We prove the saturation condition for a primary ideal: if $ab\in (r^k)$ and $a\notin \mathfrak{p}$, then $b\in (r^k)$. For $k=1$ this is the defining property of a prime ideal. For $k>1$, write $ab=r^{k}c$. Since $a\notin (r)$, the prime ideal condition gives $b=r{b}_1$. The element $r$ is a non-zero-divisor, so cancellation givesInduction gives $b_1\in (r^{k-1})$, hence $b\in (r^k)$. Together with the computed radical of an ideal, this proves

Since the radical of an ideal $\sqrt{\mathfrak{q}}$ is $(r)$, some positive power of $r$ belongs to $\mathfrak{q}$. Choose the least such exponent $k$. Then $(r^k)\subseteq \mathfrak{q}$ and $r^j\notin \mathfrak{q}$ for $0\le j<k$, with $r^0=1$ excluded because the primary ideal $\mathfrak{q}$ is proper.

Take $x\in \mathfrak{q}$. If $x\notin (r^k)$, choose the largest $j\in \{0,\dots ,k-1\}$ for which $x\in (r^j)$, and write $x=r^{j}u$. Maximality of $j$ implies $u\notin (r)=\sqrt{\mathfrak{q}}$: otherwise $x\in (r^{j+1})$, a contradiction, including when $j=k-1$. The primary ideal condition applied to $r^{j}u\in \mathfrak{q}$ therefore forces $r^j\in \mathfrak{q}$, contrary to the choice of $k$.

Consequently every $x\in \mathfrak{q}$ lies in $(r^k)$, andTogether with i, this classifies the primary ideals for a regular principal prime. The exponent is unique: if $(r^k)=(r^{k+1})$, then $r^k=r^{k+1}c$, and cancellation of the non-zero-divisor $r^k$ would give $1=rc$, contradicting that $(r)$ is proper. No Noetherian hypothesis is needed.