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 $\theta = 7\cdot \theta - 2 \cdot 3 \cdot \theta$.