Why is Z[√−n],n≥3\mathbb{Z}[\sqrt{-n}], n\ge 3 not a UFD?

I’m considering the ring $\mathbb{Z}[\sqrt{-n}]$, where $n\ge 3$ and square free. I want to see why it’s not a UFD.

I defined a norm for the ring by $|a+b\sqrt{-n}|=a^2+nb^2$. Using this I was able to show that $2$, $\sqrt{-n}$ and $1+\sqrt{-n}$ are all irreducible. Is there someway to conclude that $\mathbb{Z}[\sqrt{-n}]$ is not a UFD based on this? Thanks.

Answer

If $n$ is even, then $2$ divides $\sqrt{-n}^2=-n$ but does not divide $\sqrt{-n}$, so $2$ is a nonprime irreducible. In a UFD, all irreducibles are prime, so this shows $\mathbb{Z}[\sqrt{-n}]$ is not a UFD.

Similarly, if $n$ is odd, then $2$ divides $(1+\sqrt{-n})(1-\sqrt{-n})=1+n$ without dividing either of the factors, so again $2$ is a nonprime irreducible.

This argument works equally well for $n=3$, but fails for $n=1,2$, and in fact $\mathbb{Z}[\sqrt{-1}]$ and $\mathbb{Z}[\sqrt{-2}]$ are UFDs.

Attribution
Source : Link , Question Author : Danielle Intal , Answer Author : Chris Eagle