Solve a linear equation in the set of ideals of an algebraic integer ring

Let \mathcal{O} be the ring of integers in K = \mathbb{Q}(\sqrt{-5}), and let A = (3, 1+2\sqrt{-5}) and B = (1+2\sqrt{-5}). Show that A|B and exhibit an ideal C such that AC = B.


We will follow the algorithm given in Theorem 8.13 and Corollary 8.15 of TAN.

Note that the (ordered) conjugates of 3 and 1+2\sqrt{-5} for K are \alpha_1^1 = \alpha_1^2 = 3, \alpha_2^1 = 1+2\sqrt{-5} = \theta, and \alpha_2^2 = 1-2\sqrt{-5} = \overline{\theta}. Let g_{i,j}(x) = \alpha_1^ix + \alpha_2^jx^2. Now g_{1,1}g_{1,2}g_{2,1}g_{2,2} = x^4(81+108x+414x^2+252x^3+441x^4), and \mathsf{gcd}(81,108,414,252,441) = 9, and g_{1,2}g_{2,1}g_{2,2} = x^3(27+ (18+9\overline{\theta})x + (63+6\overline{\theta})x^2 + 21\overline{\theta}x^3). Let D = (27, 18+9\overline{\theta}, 63+6\overline{\theta}, 21\overline{\theta}). We simplify D as follows.

D  =  (27, 18+9\overline{\theta}, 63+6\overline{\theta}, 21\overline{\theta})
 =  (27, 18+9\overline{\theta}, 9+6\overline{\theta}, 21\overline{\theta})
 =  (27, -9+9\overline{\theta}, 9+6\overline{\theta}, 21\overline{\theta})
 =  (27, 15\overline{\theta}, 9+6\overline{\theta}, 21\overline{\theta})
 =  (27, 15\overline{\theta}, 9+6\overline{\theta}, 6\overline{\theta})
 =  (27, 3\overline{\theta}, 9+6\overline{\theta}, 6\overline{\theta})
 =  (27, 3\overline{\theta}, 9+6\overline{\theta}, 6\overline{\theta})
 =  (27, 3\overline{\theta}, 9)
 =  (3\overline{\theta}, 9)

By Theorem 8.13, AD = (9). Indeed, we can verify this since AD = (9\overline{\theta}, 27, 63, 9\theta) \subseteq (9) and 9 = 3\theta \cdot \overline{\theta} - 2\cdot 3 \cdot 9.

Now BD = (\theta)(3\overline{\theta}, 9) = (3\theta\overline{\theta}, 9\theta) = (63,9\theta) = (9)(7,\theta) = AD(7,\theta), so that B = A(7,\theta).

Indeed, we can verify that (3,\theta)(7,\theta) = (\theta\overline{\theta}, 3\theta, 7\theta, \theta^2) \subseteq (\theta) while \theta = 7\cdot \theta - 2 \cdot 3 \cdot \theta.

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