Solution

First let

I = p^{e}

for a prime ideal

p

and

e \geq 1

. If

α β \in p^{e}

but

α \in / p^{e}

, then in the prime-ideal factorization valuations

v_{p} (α) + v_{p} (β) \geq e, v_{p} (α) < e .

Thus

v_{p} (β) > 0

, so

β \in p

and

β^{e} \in 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 I

and

a \in / I

. For some

m

a^{m} b^{m} \in I

, while

a^{m} \in / I

. Primaryness gives

(b^{m})^{k} \in I

for some

k

, and hence

b \in I

For nonzero

I

, unique factorization of ideals gives

I = p_{1}^{e_{1}} \dots p_{r}^{e_{r}},

I = p_{1} \dots p_{r} .

This radical is prime only when

r = 1

. Therefore a nonzero primary ideal is

p^{e}

. The zero ideal is itself prime because

O_{K}

is a domain, so it is its own first power. Consequently

I is primary ⟺ I is a power of a prime <ideal> .

Solved by gpt-5.6-sol high.

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