Let $K$ be a number field, and let $𝒪_K$ be its ring of integers.

Let $A$ be any non-zero proper ideal of $𝒪_K$. Show that there exist prime ideals $P_1, P_2, …, P_r$ such that $A=P_1 P_2 ⋯ P_r$ and that this factorisation is unique up to the order of factors.

Assume not every ideal $A$ has a prime factorisation. Let $A$ be such an ideal with $N(A)$ is minimal. There exists a maximal (hence prime) ideal $P_1$ containing $A$, and so $P_1 ∣ A$; hence there is an ideal $C$ with $A=P_1 C$.
If $A=C$ then $P_1 C=C=C 𝒪_K$ and $P_1=𝒪_K$, by cancellation. This is clearly impossible. Hence $A ⊂ C$, and by the definition of the norm we have $N(A)=N(C)[C: A]>N(C)$. Hence, by our minimality assumption for $A$, one can factor $C$ into prime ideals as $C=P_2 … P_r$ (or $C=𝒪_K$ and $A=P_1$). Therefore $A=P_1 … P_2$, a contradiction. Hence every proper non-zero ideal has a prime factorisation. Suppose \[ A=P_1 P_2 … P_r=Q_1 Q_2 … Q_s . \] Now $P_1 ∣ Q_1 … Q_s$. Let $k$ be minimal such that $P_1 ∣ Q_1 … Q_k$. If $k=1$ then $P_1 ∣ Q_1$. If $k>1$ then $P_1 ∣(Q_1 … Q_{k-1}) Q_k$, but $P_1$ does not divide $Q_1 … Q_{k-1}$. Since $P_1$ is prime, we must have $P_1 ∣ Q_k$. We therefore have $P_1 ∣ Q_k$ in either case. Since $Q_k$ is maximal this implies that $P_1=Q_k$. Without loss of generality we take $k=1$ and then, by the cancellation lemma, we have $P_2 … P_r=Q_2 … Q_s$. We may now repeat the process until every $P_i$ has been shown to equal some $Q_j$.

Let $P$ be a nonzero prime ideal of $𝒪_K$.

Let $R$ be any integral domain.
If $I$ is an ideal of $R$ then any ideal of $R / I$ is of the form $J / I$, where $J$ is an ideal of $R$ which contains $I$.
(Chinese Remainder Theorem) If $I_1, I_2$ are coprime ideals of $R$ then $R / I_1 I_2$ is isomorphic to $R / I_1 × R / I_2$.