 Solution

First let $I={\mathfrak{p}}^e$ for a prime ideal $\mathfrak{p}$ and $e\ge 1$. If $\alpha \beta \in {\mathfrak{p}}^e$ but $\alpha \notin {\mathfrak{p}}^e$, then in the prime-ideal factorization valuationsThus $v_{\mathfrak{p}}(\beta )>0$, so $\beta \in \mathfrak{p}$ and ${\beta }^e\in {\mathfrak{p}}^e=I$. Hence every prime-ideal power is primary.

Conversely, the radical of any primary ideal is prime. To see this, suppose $ab\in \sqrt{I}$ and $a\notin \sqrt{I}$. For some $m$, $a^m{b}^m\in I$, while $a^m\notin I$. Primaryness gives $(b^m{)}^k\in I$ for some $k$, and hence $b\in \sqrt{I}$.

For nonzero $I$, unique factorization of ideals givessoThis radical is prime only when $r=1$. Therefore a nonzero primary ideal is ${\mathfrak{p}}^e$. The zero ideal is itself prime because ${\mathcal{O}}_K$ is a domain, so it is its own first power. Consequently