Show that $\langle 2,1+\sqrt{-5}\rangle \langle3,1+\sqrt{-5} \rangle = \langle1-\sqrt{-5} \rangle$ In $\mathbb Z[\sqrt{-5}]$ I compute
$\langle 2,1+\sqrt{-5} \rangle \langle 3,1+\sqrt{-5} \rangle  = \langle 6,2+2\sqrt{-5},3+3\sqrt{-5},-4+2\sqrt{-5} \rangle $
So there must be $a+b\sqrt{-5}\in\mathbb Z[\sqrt{-5}]$ that $(a+b\sqrt{-5})(1-\sqrt{-5})=2+2\sqrt{-5}$ but then we have $b=\frac{2}{3}$ that it's not possible. Am I doing it wrong? How can I prove this?
 A: As you've showed, \begin{align} \langle 2,1+\sqrt{-5} \rangle \langle 3,1+\sqrt{-5} \rangle  &= \langle 6,\color{blue}{2+2\sqrt{-5}},\color{blue}{3+3\sqrt{-5}},-4+2\sqrt{-5} \rangle \\ \tag{1}&= \langle \color{blue}{6},1+\sqrt{-5}, \color{blue}{-4+2\sqrt{-5}} \rangle \\ \tag{2}&=\langle 6, \color{blue}{1+\sqrt{-5}},\color{blue}{2+2\sqrt{-5}} \rangle \\ \tag{3}& =\langle 6,1+\sqrt{-5} \rangle \\& \tag{4}= \langle \color{red}{1+ \sqrt{-5}} \rangle \end{align}

*

*Both $2 + 2 \sqrt{-5}$ and $3 + 3\sqrt{-5}$ are  multiples of $1 + \sqrt{-5}$, hence belong to $\langle 6, 1+ \sqrt{-5}, -4+2\sqrt{-5} \rangle$. On the other hand, $1+\sqrt{-5}$ = $(3 + 3\sqrt{-5}) - (2 + 2\sqrt{-5}) \in \langle 6, 2+ 2\sqrt{-5}, 3+ 3\sqrt{-5}, -4+2\sqrt{-5} \rangle$.

*$-4+2\sqrt{-5} = (2 + 2\sqrt{-5}) - 6 \in \langle 6, 1+\sqrt{-5}, 2+2\sqrt{-5} \rangle $. Conversely, $2+2\sqrt{-5} = (-4 + 2\sqrt{-5}) + 6 \in \langle 6, 1+\sqrt{-5}, -4+2\sqrt{-5} \rangle$.

*$2+2\sqrt{-5}$ is an element of $\langle 6, 1+\sqrt{-5} \rangle$ as it is a multiple of $1+\sqrt{-5}$.

*$ 6 = (1 + \sqrt{-5})(1-\sqrt{-5}) \in \langle 1 + \sqrt{-5} \rangle $.

The result $\langle 1+ \sqrt{-5} \rangle$  is not equal to $\langle 1 - \sqrt{-5} \rangle$, so your title presents wrong equality. To see this, suppose $\langle 1 + \sqrt{-5} \rangle = \langle 1 - \sqrt{-5} \rangle$. Then there exists $a + b\sqrt{-5} \in \mathbb{Z}[\sqrt{-5}]$ such that $1 + \sqrt{-5} = (a+b\sqrt{-5})(1-\sqrt{-5})$. Taking the norm of both sides yields $N(a+b\sqrt{-5}) = a^2 + 5b^2 =1$, whence $a=\pm 1$ and $b=0$. Equivalently, $1+\sqrt{-5} = \pm ( 1 - \sqrt{-5})$; contradiction.
A: More generally put  $\,a=1\!+\!\sqrt{-5},\ b = 2,\ c = 3\,$ below to get the correct product.
Lemma $\  I:=(a,\color{#c00}b)(a,\color{#c00}c) = (a)\ $ if $\,\color{#c00}{(b,c)\!=\!1},\, \color{#0a0}{a\mid bc}$
Proof $\ \ \  \ I= (aa,ab,ac,bc)\, =\, (a)(a,\,\color{#c00}{b,c},\,\color{#0a0}{bc/a}) = (a)$
