## 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 $5 \cdot 5 \cdot (2 + \sqrt{-6}) \cdot (2 - \sqrt{-6}) - 6 \cdot 2 \cdot 2 \cdot (2 + \sqrt{-6}) \cdot (2 - \sqrt{-6}) = 10$.