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I remember hearing somewhere that if $p^n$ is the largest power of a prime $p$ dividing the class number $h_K$ of a number field $K$, then there is unique factorization of the $p^n$-th powers of elements in $\mathcal{O}_K$ into $p^n$-th powers of irreducibles. I can see that there can't be any $p$-torsion for a $p^n$-th power of an ideal class (because $p^n$ is the largest power dividing $h_K$), but I'm confused as to how to translate this into a statement about factorization of elements, and why $p$ is arbitrary. Am I remembering this result correctly (if not, what is the right formulation)? Can someone provide a proof or reference?

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    Ehr... in a number field, any nonzero ideal of $\mathcal{O}_K$ can be factored uniquely into products of prime ideals; in particular, the $p^n$th power can be factored uniquely into $p^n$th powers of prime ideals, regardless of what $p$ and $n$ are. Perhaps you are aiming for something along the lines of the fact that the $h_K$th power of any ideal is principal, but with $p$th powers?2010-10-23
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    Gah, of course. I misstated the result I'm probably mis-remembering anyway. I'll edit.2010-10-23
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    You also need $n\gt 0$: otherwise, picking a prime that does not divide $h_K$, your assertion would be that the $p^0$th power of every nonzero element (that is, every nonzero element) can be factored uniquely into a product of irreducibles, i.e., unique factorization. Are you sure you don't mean that the $h_K$-th power of any element can be written uniquely as a product of irreducibles?2010-10-23
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    Good point with the $n>0$. Hmmm... I really thought there was something about prime powers in the statement, but I certainly could be wrong. At any rate, the result you mentioned sounds very close. Intuitively it seems true, but I'm having trouble proving it: take any $a\in\mathcal{O}_K$, factor $(a)=P_1^{e_1}\cdots P_n^{e_n}$, we get $(a^{h_K})=(a)^{h_K}=(P_1^{h_K})^{e_1}\cdots(P_n^{h_K})^{e_n}$, and each $P_i^{h_K}$ is principal, say $P_i^{h_K}=(x_i)$. Thus $a^{h_K}=x_1\cdots x_n$. But how do we get uniqueness from here?2010-10-23
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    Ehr... never mind. You also don't have the $x_i$ are irreducible. Think about $\mathbb{Q}(\sqrt{-5})$ with $h_K=2$. Since $6 = (1+\sqrt{-5})(1-\sqrt{-5}) = (2)(3)$, and each of $1+\sqrt{-5}$, $1-\sqrt{-5}$, $2$, and $3$ are irreducible (look at the norm function), you have $6^2 = (1+\sqrt{-5})^2(1-\sqrt{-5})^2= (2^2)(3^2)$, two different expressions as products of (squares of) irreducibles. Note that in a number field you can *always* factor a nonzero element into a product of irreducibles (induction on the norm); it's uniqueness that may fail (and will fail if $h_k\neq 1$).2010-10-23

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