Confusion about multiplying two ideals.

Question

I am trying to solve this question:

Let $R$ be the quadratic integer ring $\mathbb Z[\sqrt{-5}].$ Define the ideals $I_2 = (2, 1 + \sqrt{-5}), I_3 = (3, 2 + \sqrt{-5}),$ and $I_3^{'} = (3, 2 - \sqrt{-5}).$ Prove similarly that $I_2 I_3 = (1 - \sqrt{-5})$ and $I_2 I_3^{'} = (1 + \sqrt{-5})$ are principal. Conclude that the principal ideal $(6)$ is the product of $4$ ideals: $I_2^2 I_3 I_3^{'}.$

But I am wondering why $I_2 I_3 = (1 - \sqrt{-5})$? I tried to multiply and I got the ideal $(6, 4 + 2\sqrt{-5}, 3(1 + \sqrt{-5}, 3(-1 + \sqrt{-5}))$ but I am unable to simplify it to the ideal $(1 - \sqrt{-5}).$ could anyone help me in this please?

Answer 1

Note that \begin{align*} 6 &= (1-\sqrt{-5})(1+\sqrt{-5})\\ 4+2\sqrt{-5} &= (1-\sqrt{-5})(-1+\sqrt{-5})\\ 3(1+\sqrt{-5}) &= (1-\sqrt{-5})(-2+\sqrt{-5})\\ 3(-1+\sqrt{-5}) &= -3(1-\sqrt{-5}). \end{align*}

Further, $(4+2\sqrt{-5}) - 3(-1+\sqrt{-5}) - 6 = 7-\sqrt{-5}-6= 1-\sqrt{-5}$.