Let $K=\mathbb{Q}(\sqrt{d})$ for square-free integer $d$ and $\mathcal{O}_K$ be the integral closure of $\mathbb{Z}$ in $K$. Assume that $(a)=P_1\ldots P_n$ for prime ideals $P_i$ such that $N(P_i)\leqslant m$ for $m>0$. Prove that $a$ is a product of some elements $\pi_i\in \mathcal{O}_K$ with $|N(\pi_i)|\leqslant m^{h_K}$, where $h_K$ is a class number of $K$.
Factorization of element in quadratic integer ring
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number-theory
ring-theory
algebraic-number-theory
ideal-class-group
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0Let $J$ be the ideal class group, and let $P_i^{-1} \in J$ such that $P_i P_i^{-1} = (c_i)$. Then $\prod_i P_i^{-1}= (b)$ is principal because it is inverse of $(a)$ in $J$. And $(ab) = \prod_i P_i P_i^{-1}= \prod_i c_i$. – 2017-01-26
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0Where do you get $m$ from? Are there any requirements on $\pi_i$? – 2017-01-27
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0$m$ is given constant. No, $\pi_i$ are arbitrary elements of $\mathcal{O}_K$. – 2017-01-27