I am working on an exercise that asks us to consider the ring $R = \mathbb{Z}[\sqrt{-3}]$ and the ideal $I = (2, 1 + \sqrt{-3})$ in $R$. Part (a) asks to show that $I^2 = (2)I$ but $I \neq (2)$, and part (b) asks to show that $I$ is the unique prime ideal in $R$ containing $(2)$, and to then conclude that $(2)$ does not factor into prime ideals in $R$.
A previous exercise asked to show that $(2, 1 + \sqrt{-5})(2, 1 - \sqrt{-5}) = (2)$ in $\mathbb{Z}[\sqrt{-5}]$. Now $(2, 1 + \sqrt{-5}) = (2, 1 - \sqrt{-5})$, so then $(2) = (2, 1 + \sqrt{-5})^2$. But this should work in $R$ too, so then $I^2 = (2)$. But $I$ is prime, so then $(2)$ does factor into prime ideals in $R$. Where did I go wrong? Thanks.
The equality $(2, 1 + \sqrt{-5})(2, 1 - \sqrt{-5}) = (2)$ that holds in $\Bbb Z[\sqrt{-5}]$ does not carry over into $R = \Bbb Z[\sqrt{-3}]$.
Here $(2, 1 + \sqrt{-3})(2, 1 - \sqrt{-3}) \ne (2)$. The inclusion $(2, 1 + \sqrt{-3})(2, 1 - \sqrt{-3}) \subset (2)$ still holds, but if you compute a generating set of $(2, 1 + \sqrt{-3})(2, 1 - \sqrt{-3})$, you only get
$$4 = 2\cdot 2 = (1+\sqrt{-3})\cdot (1-\sqrt{-3})$$ and $$2 \pm 2\sqrt{-3} = 2\cdot (1 \pm \sqrt{-3}).$$ Since $2 - 2\sqrt{-3} = 4 - (2 + 2\sqrt{-3})$, this generating set can be reduced to $(4, 2 + 2\sqrt{-3}) = 2I$. However, there is no way to express $2$ as a $\Bbb Z[\sqrt{-3}]$-linear combination of $4$ and $2 + 2\sqrt{-3}$.
The difference with the $\Bbb Z[\sqrt{-5}]$ case is that in that case you see that the corresponding products of generators include both $4$ and $6 = (1+\sqrt{-5})\cdot (1-\sqrt{-5})$, and thus $2 = 6-4$ lies in $(2, 1 + \sqrt{-5})(2, 1 - \sqrt{-5})$