$\mathbb{Z}\left[\sqrt{-5}\right]$ is not a principal ideal domain

Question

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 $2\not \mid a\pm b\sqrt{-5}$.

Answer 1

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.

Answer 2

Notice that $2\mid a+b\sqrt{-5}$ if and only if both $a$ and $b$ are even.

So, any pair of odd $a,b$ will work.

Answer 3

You can also prove that $(2,\sqrt{-5})$ is not principal (prove it is a proper ideal, to begin with). Indeed, if $a+b\sqrt{-5}$ divides $2$, then $a^2+5b^2$ must divide $4$, so $b=0$ and $a=\pm2$. But $2+0\sqrt{-5}$ cannot divide $\sqrt{-5}$.