Intuition behind Factoring $(6)$ in $\mathbb{Z}(\sqrt{-5})$

Question

It turns out that we have (ideal equality) $$(6)=(2,1+\sqrt{-5})^2(3,1+\sqrt{-5})(3,1-\sqrt{-5})$$ How does one get this from $(6)=(2)(3)$?

When factoring $(6)$ how do we know to continue to factor $(2), (3)$? Why do we have $(2)=(2,1+\sqrt{-5})^2$?

$2$ belongs to the left hand side, but it's not in the right hand side due to the square.

Answer 1

The equality $(2)=(2,1+\sqrt{-5})^2$ is one of ideals.

To see it, let us consider $(2,1+\sqrt{-5})^2$.

It is $$(2,1+\sqrt{-5})(2,1+\sqrt{-5}) = (4, 2( 1+ \sqrt{-5}), (1+\sqrt{-5})^2) = (4, 2 + 2\sqrt{-5}), -4+2\sqrt{-5}).$$ Since $2 = (2 + 2\sqrt{-5})) - (-4+2\sqrt{-5}))$ the right-hand side in fact contains.