Prove that $\langle 2,1+\sqrt{-5} \rangle ^ 2 \subseteq \langle 2 \rangle$

Question

I've got an exercise whose goal I pressume is to show how $\mathbb{Z}[\sqrt{-5}]$ is not a PID. They define $I = \langle 2,1+\sqrt{-5} \rangle$ then they show that $I$ is a maximal ideal on $\mathbb{Z}[\sqrt{-5}]$. What I have doubts is about proving that: $I^2 \subseteq \langle 2 \rangle$. Apparently this should lead to the fact that $I$ is not a principal ideal.

But I ask you, how to show $I^2 \subseteq \langle 2 \rangle$ without going through endless algebraics equations of at least (currently in my approach) 6 parameters? Is there a simple way?

Answer 1

The ideal $I^2$ will be generated by all possible products of generators of $I$. There are four of these, though two are obviously the same: $$\displaylines{ 2^2=4\cr 2(1+\sqrt{-5})=(1+\sqrt{-5})2=2+2\sqrt{-5}\cr (1+\sqrt{-5})^2=-4+2\sqrt{-5}\ .\cr}$$ All of these are multiples of $2$, so $I\subseteq\langle2\rangle$.

Answer 2

$I^2$ is generated by elements of the form $$(2a + (1+\sqrt{-5})b)(2c + (1+\sqrt{-5})d).$$ (Here, $a,b,c,d$ are elements of $\mathbb Z[\sqrt{-5}]$.)

If we expand this out, the first three of four terms clearly have a factor of $2$ in them, and the fourth term is $(1+\sqrt{-5})^2 bd = (-4 + 2\sqrt{-5})bd$, which also has a factor of $2$.

So each such element is contained in $(2)$, and therefore $I^2 \subseteq (2)$.

Answer 3

\begin{align} (2,1+\sqrt{-5})^2 &=(2,1+\sqrt{-5})(2,1-\sqrt{-5})\\ &=(4,2(1+\sqrt{-5}),2(1-\sqrt{-5}),6)\\ &=(2)(2,1+\sqrt{-5},1-\sqrt{-5},3)\\ &=(2)(1)\\ &=(2) \end{align}