I want to show that $\mathbb{Z}\left[\sqrt{-5}\right]$ is not a principal ideal domain.
As a hint we should look at the $2$. I have shown that $2$ is irreducible in $\mathbb{Z}\left[\sqrt{-5}\right]$. If I can show that $2$ is not prime in $\mathbb{Z}\left[\sqrt{-5}\right]$, then I am finish.
So I need numbers $a\pm b\sqrt{-5}$ with $2\mid a^2+5b^2$ but $2\not \mid a\pm b\sqrt{-5}$.
Best Answer
One possible method to show $\mathbb{Z}[\sqrt{-5}]$ is not a PID is showing that it is not a Unique Factorization Domain(UFD), which contains Principal Ideal Domain.
Consider the element $6$. $6=2\times3=(1-\sqrt{-5})(1+\sqrt{-5})$, so $6$ is not uniquely factorized in this domain. Thus $\mathbb{Z}[\sqrt{-5}]$ is not a UFD, which leads to the fact that $\mathbb{Z}[\sqrt{-5}]$ is not a PID.
Hope my answer helped.