Detect equivalences among ideals of an algebraic integer ring

Determine which of the following ideals of \mathbb{Z}[\sqrt{-6}] are equivalent: A = (2,\sqrt{-6}), B = 5-2\sqrt{-6},2+2\sqrt{-6}), C = (2+\sqrt{-6}), and D = (5,1+2\sqrt{-6}).

First we argue that B is principal. Indeed, B = (5-2\sqrt{-6}, 2+2\sqrt{-6}) = (7,2+2\sqrt{-6}) = (1+\sqrt{-6})(1-\sqrt{-6},2). Now (1-\sqrt{-6})(1+\sqrt{-6}) - 3 \cdot 2 = 1 \in (1-\sqrt{-6},2), so that B = (1+\sqrt{-6})(1) = (1+\sqrt{-6}). Indeed, 5-2\sqrt{-6} = (1+\sqrt{-6})(-1-\sqrt{-6}) and 2+2\sqrt{-6} = 2(1+\sqrt{-6}), and 1+\sqrt{-6} = (1-\sqrt{-6})(5-21\sqrt{-6}) + 4(2+2\sqrt{-6}).

In particular, B \sim C \sim (1).

Now we claim that AD and A^2 are both principal. Indeed, AD = (2,\sqrt{-6})(5,1+2\sqrt{-6}) = (10,2+4\sqrt{-6},5\sqrt{-6},-12+\sqrt{-6}) = (10,2+4\sqrt{-6},5\sqrt{-6},-10) = (10,2-\sqrt{-6},5\sqrt{-6}). Since N(2-\sqrt{-6}) = 10 and (2-\sqrt{-6})(-3+\sqrt{-6}) = 5\sqrt{-6}, we have AD = (2-\sqrt{-6}). So AD \sim (1). Similarly, A^2 = (s,\sqrt{-6})^2 = (4,2\sqrt{-6},-6) = (2), so that A^2 \sim (1). Now AD \sim A^2, so that A \sim D.

Finally, we claim that A is not principal. If (2,\sqrt{-6}) = (\alpha), then \alpha|2 and \alpha|\sqrt{-6}, so that N(\alpha) divides both 4 and 6. So N(\alpha) \in \{1,2\}. Note, however, that no element of \mathbb{Z}[\sqrt{-6}] has norm 2 since the equation a^2 + 6b^2 = 2 has no solutions in \mathbb{Z}. If N(\alpha) = 1, then we have 1 \in D. But then 1 = 2(h+k\sqrt{-6}) + \sqrt{-6}(u+v\sqrt{-6}), so that 1=2h-6v, which is impossible mod 2. So D is not principal.

In summary, we have B \sim C \sim (1), A \sim D, and (1) \not\sim D.

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