Show $2 = (2, 1+\sqrt{-5})^2$ in $\mathbb{Z}[\sqrt{-5}]$

Question

Show $2 = (2, 1+\sqrt{-5})^2$ in $\mathbb{Z}[\sqrt{-5}]$.

Apologies if this may seem a trivial question, however, I am some difficulty showing this.

My Attempt:

$(2, 1+\sqrt{-5}) * (2, 1+\sqrt{-5}) = (4 + 1 + 2\sqrt{-5} - 5) = (2\sqrt{-5})$.

I am not sure how to proceed to show that $(2\sqrt{-5}) = (2)$

Answer 1

When multiplying ideals, you have to multiply each generator together to get generators of the new ideal.

So you actually have $(2,1+\sqrt{-5})\cdot(2,1+\sqrt{-5}) = (4, 2+2\sqrt{-5}, 2+2\sqrt{-5}, -4+2\sqrt{-5})$.

If you look at the four generators (three unique generators) of the new ideal, you should quickly see they have a gcd of $2$.

Answer 2

We have $$2=2\cdot 2^2-2(1+\sqrt{-5}) + (1+\sqrt{-5})^2, $$ so $(2) \supset (2,1+\sqrt{-5})^2$. The other way is easier.

Answer 3

Note $\,(a,b)^2 = (a^2,ab,b^2)\,$ so with $\,w=\sqrt{-5},\,$ using $\,(2,I) = (2,\, I\bmod 2)\,$ yields

$\ \ (2,1\!+\!w)^2 = 2(2,\underbrace{1\!+\!w,-2\!+\!w}_{\large I}) = (2)\ $ by $\bmod 2\!:\ I = (1\!+\!w,w) = (1,w) = (1)$