Every ideal in ring of integers has a natural number

Let $O$ be the ring of integers of some number field and $I$ any nonzero ideal of $O$. Prove that there is some number $n \in \mathbb{Z}_+$ that is in ideal $I$. I suppose I should use that $O$ is Dedekind domain, so every ideal can be written as product of prime ideals, but I don't know how to use that. Any help is appreciated.

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Let $\mathbb K$ be a number field of degree $n$ and $O$ the ring of integers in $K$. Let $I$ be an ideal in $O$ and $x\in I$. Note that all $n$ conjugates $x=x_1, x_2, ..., x_n$ of $x$ are also algebraic integers, since they have the same minimum polynomial. Therefore, their product $x.x_2 ... x_n$ is an algebraic integer, and belongs to the ideal $I$. Finally, note that the product $x.x_2 ... x_n$ equals the norm $N(x)$, so it is a positive rational integer.

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Note that $x_2, \cdots, x_n$ do not need to be elements of $O$, or even $K$, so a priori it's not obvious that their product is an element of $I$. However, you can use the minimum polynomial to show $N(x)$ is a multiple of $x$. – Hurkyl Jan 9 '13 at 23:14
This really helps me.Thank you very much – Tirlas Jan 10 '13 at 16:59

Show that $O/I$ is finite for any nonzero ideal $I.$

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