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Suppose that K is a finite extension over $\mathbb{Q}$. Let $p$ be a prime in $\mathbb{Q}$, let $p\mathcal{O}_K=\mathfrak{P}_1\ldots\mathfrak{P}_n$ be its prime decomposition in $\mathcal{O}_K$, then is it true that any prime ideal in $\mathcal{O}_K$ above $p$ is a factor in the decomposition of $p$?

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In a word, yes. – Robin Chapman Nov 18 '10 at 11:28

Yes. I suppose that the definition of a prime ideal $\mathcal{P}$ being "above $p$" you are using is $\mathcal{P} \cap \mathbb{Z} = (p)$. From this it is clear that if $\mathcal{P}$ lies above $p$, then it contains $p\mathbb{Z}$, and -- since it is an ideal -- hence also $p \mathcal{O}_K$. But recall that in a Dedekind domain "to contain is to divide", hence $\mathcal{P}$ divides $p\mathcal{O}_K$ and appears in the (unique!) prime factorization of $\mathcal{O}_K$.

Note that this does not require that $K/\mathbb{Q}$ be Galois (as appears in the title but not the body of the question), and it has nothing to do with algebraic number fields per se: in general, for a finite extension of Dedekind domains $S/R$, the prime ideals of $S$ lying over a given prime $\mathfrak{p}$ of $R$ are exactly the ones appearing in the factorization of $\mathfrak{p}S$.

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Moreover, a finite field extension is not necessarily a Galois extension. – awllower Feb 27 '11 at 14:56
And we can in fact find an ideal such that the product of the ideal $\mathfrak{P}$ and that ideal is exactly the ideal $\mathfrak{p}$ if the inclusion holds to be true. – awllower Feb 27 '11 at 14:59

HINT $\rm\ \ P \supset \mathfrak{P}_1\ldots\mathfrak{P}_n\ \Rightarrow\ P \supset \mathfrak P_i\ $ for some $\rm\:i\:.\ $ This is true for any prime ideal $\rm\ P\ $ in any ring. Further $\rm\ P = \mathfrak P_i\ $ when the $\ \mathfrak P_i\ $ are maximal, e.g. in one dimensional domains, i.e. domains where nonzero prime ideals are maximal.

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