Factor (10) in QQ(sqrt(-6))

Find two distinct factorizations of 10 in \mathbb{Q}(\sqrt{-6}). Factor 10 as a product of four ideals in \mathbb{Q}(\sqrt{-6}).

Evidently 10 = 2 \cdot 5 = (2 + \sqrt{-6})(2 - \sqrt{-6}); moreover, these factorizations are distinct since, as the only units in \mathbb{Q}(\sqrt{-6}) are \pm 1, 2 is not associate to either of 2 \pm \sqrt{-6}.

We claim that (10) = (2,2+\sqrt{-6})(2,2-\sqrt{-6})(5,2+\sqrt{-6})(5,2-2\sqrt{-6}). To see the (\supseteq) direction, note that this ideal product is generated by all 2^4 possible selections of one generator from each factor. Name the generators \alpha_1,\alpha_2,\beta_1,\beta_2 such that \alpha_1\alpha_2 = \beta_1\beta_2 = 10. Now any selection which includes \alpha_1 (without loss of generality) and does not include \alpha_2 must include both \beta_1 and \beta_2, and so this ideal product is in (10).

Note also that 5 \cdot 5 \cdot (2 + \sqrt{-6}) \cdot (2 - \sqrt{-6}) - 6 \cdot 2 \cdot 2 \cdot (2 + \sqrt{-6}) \cdot (2 - \sqrt{-6}) = 10.

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