## Factor a given ideal in an algebraic integer ring

Let $K = \mathbb{Q}(\sqrt{-6})$ and let $\mathcal{O} = \mathbb{Z}[\sqrt{-6}]$ be the ring of integers in $K$. Factor the ideals $(2)$ and $(5)$ in $\mathcal{O}$.

We claim that $(2) = (2,\sqrt{-6})^2$. Indeed, $2 = -\sqrt{-6}^2 - 2^2$, so that $(2) \subseteq (2,\sqrt{-6})^2$. The reverse inclusion is clear.

Now we claim that $(2,\sqrt{-6})$ is maximal. By Corollary 9.11, $N((2)) = 4$, and by Theorem 9.14, we have $N((2,\sqrt{-6}))^2 = 4$, so that $N((2,\sqrt{-6})) = 2$. By Corollary 9.15, $(2,\sqrt{-6})$ is a prime ideal.

Thus $(2) = (2,\sqrt{-6})^2$ is the prime factorization of $(2)$ in $\mathcal{O}$.

Now we claim that $(5) = (5,1+2\sqrt{-6})(5,1-2\sqrt{-6})$. Indeed, we have $(5,1+2\sqrt{-6})(5,1-2\sqrt{-6}) = (25,5+10\sqrt{-6}, 5-10\sqrt{-6}, 25)$ $= (25, 5+10\sqrt{-6}, 10)$ $= (5)$.

Next we claim that $Q_1 = (5,1+2\sqrt{-6})$ and $Q_2 = (5,1-2\sqrt{-6})$ are proper. Suppose to the contrary that $1 \in Q_1$; then we have $1 = 5(a+b\sqrt{-6}) + (1+2\sqrt{-6})(h+k\sqrt{-6})$ for some $a,b,h,k \in \mathbb{Z}$. Comparing coefficients, $5a+h-12k = 1$ and $5b+k+2h = 0$, which yields a contradiction mod 5. So $Q_1$ is proper. Likewise, $Q_2$ is proper. In particular, neither $Q_1$ nor $Q_2$ have norm 1 as ideals. Now $25 = N((5)) = N(Q_1)N(Q_2)$, and neither factor is 1, so that $N(Q_1) = N(Q_2) = 5$. By Corollary 9.15, $Q_1$ and $Q_2$ are prime ideals.

Thus $(5) = (5,1+2\sqrt{-6})(5,1-2\sqrt{-6})$ is the prime factorization of $(5)$ in $\mathcal{O}$.