Solution

Distinct prime-power ideals are pairwise comaximal. For comaximal ideals, product equals intersection, so induction gives

I = P_{1}^{m_{1}} \dots P_{k}^{m_{k}} = P_{1}^{m_{1}} \cap \dots \cap P_{k}^{m_{k}} .

The Chinese remainder theorem therefore gives the ring isomorphism

O_{K} / I ⟶ \prod_{i} O_{K} / P_{i}^{m_{i}} .

The ideal approximation theorem, proved by applying this Chinese-remainder map one prime power deeper, supplies

α \in I, α \in / P_{i} I (1 \leq i \leq k) .

Equivalently,

v_{P_{i}} (α) = m_{i}

at every prime dividing

I

Since

(α) \subseteq I

I^{'} = (α) I^{- 1}

I I^{'} = (α)

is a principal ideal. At every

P_{i} ∣ I

v_{P_{i}} (I^{'}) = v_{P_{i}} (α) - m_{i} = 0

, so

I + I^{'} = O_{K}

Choose

β \in I

and

γ \in I^{'}

with

β + γ = 1

. Then

α, β \in I

. For any

x \in I

x = x β + x γ \in (β) + I I^{'} = (β) + (α) .

Thus

I = (α, β),

so every ideal of

O_{K}

has a generating set of two elements.

Solved by gpt-5.6-sol high.

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