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Let $K$ be a finite field extension of the rational numbers and let $\mathcal{O}_K$ denote its ring of integers. If a rational integer $n$ factors into two distinct ways into irreducible elements in $\mathcal{O}_K$, that is,

$$n = \prod{a_j} = \prod{b_j},$$

where $a_j, b_j$ are irreducible and no $a_j$ is associate to any $b_j$, then $n^2 = \prod{a_j}^2 = \prod{b_j}^2 = \prod{a_j}\prod{b_j}$ has at least three distinct factorizations; by taking powers of $n$ one thus can see that the number of distinct factorizations of rational integers is unbounded. Is the number of distinct "primitive" factorizations of rational integers over $\mathcal{O}_K$ bounded, that is, factorizations that do not arise from a construction as above (that is, cannot be split into different "sub-factorizations")? If yes, can this bound be given explicitly in terms of the size of the class group and $[K : \mathbb{Q}]$?

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You might be interested in this question: math.stackexchange.com/questions/538959/… –  John M Mar 17 '14 at 14:38
The question has been answered here: mathoverflow.net/questions/162916/… –  streetcar277 Jun 19 '14 at 11:42

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