## 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 $(1) \not\sim D$.