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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... 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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