# Why ideal containment in the proof of unique ideal factorization?

I want to understand the proof that the number fields (which are dedekind domains) have unique factorization of ideals. I am trying to read this proof here IDEAL FACTORIZATION - KEITH CONRAD but..

I couldn't understand the reason for the inclusions in this proof:

Corollary 3.5: For any nonzero ideal $\mathfrak b$ and nonzero prime ideal $\mathfrak p$, $\mathfrak {pb} \subseteq \mathfrak b$ with strict inclusion.

Proof. Easily $\mathfrak {pb} \subseteq \mathfrak b$. If $\mathfrak {pb} = \mathfrak b$ then for all $k\ge1$, $\mathfrak b = \mathfrak p^k \mathfrak b$ therefore $\mathfrak b \subseteq \mathfrak p^k$. rest of proof omitted

First of all these are all ideals of $\mathbb O_K$ the ring of integers of a number field $K$.

How do we know $\mathfrak {pb} \subseteq \mathfrak b$?

How do we know $\mathfrak {b} \subseteq \mathfrak p^k$?

Thank you very much.

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Hint $\,$ For ideals in a ring $\rm\,R,\,$ divides $\Rightarrow$ contains: $\rm\:I\mid J\:\Rightarrow\: I\supseteq J\:$ since $\rm\:J = I\,K \subseteq I\,R \subseteq I.\:$

(The converse implication is a characteristic property of Dedekind domains).

So, for example, $\:\mathfrak p^k \mathfrak b = \mathfrak b\:\Rightarrow\: \mathfrak p^k\mid \mathfrak b\:\Rightarrow\: \mathfrak p^k \supseteq \mathfrak b$

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I see! Thank you very much. – sperners lemma Nov 3 '12 at 20:23

Since $\mathfrak b$ is an ideal of $\mathbb O_K$ we have $\mathbb O_K \mathfrak b \subseteq \mathfrak b$, since $\mathfrak p$ is a subset of $\mathbb O_K$ we have $\mathfrak p \mathfrak b \subseteq \mathfrak b$.

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