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}.

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