# why its fail at n = $1$ and $2$?

i have some confusion in this answer

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

My attempt : we know that $$1$$ is odd and $$2$$ is even . So if $$n$$ is even then $$n=2$$ , then obviously $$2$$ divides $$\sqrt{-2}^2=-2$$ but does not divide $$\sqrt{-2}$$, so $$2$$ is a nonprime irreducible

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

But the user chris eagle said that it fail for $$n= 1, 2$$

why is fail for $$n =1 , 2$$??

• Doesn’t $2=(1+\sqrt{-1})(1-\sqrt{-1}) =-\sqrt{-2}^2$ show $2$ is reducible in $\Bbb Z[\sqrt{-1}$ and – J. W. Tanner Aug 25 at 16:04
• @J.W.Tanner $2$ does not divide $( 1+ \sqrt {-1})$ – jasmine Aug 25 at 16:06
• I meant $\Bbb Z[\sqrt{-1}]$ and $\Bbb Z[\sqrt{-2}]$, respectively? – J. W. Tanner Aug 25 at 16:10

The problem with your argument is that $$2$$ is actually reducible in $$\mathbb{Z}[\sqrt{-2}]$$, as $$2=(-\sqrt{-2})\sqrt{-2}$$. So there is no problem with it being not prime.
Similarly, in $$\mathbb{Z}[i]$$ we have $$2=(1+i)(1-i)$$, so again it is reducible.
• ....@Mark But $1+ i$ is an irreducible element of $\mathbb{Z}[i]$ How it is reducible ? – jasmine Aug 25 at 16:25
• $1+i$ is irreducible, but why is that important? I'm saying $2$ is reducible as it is a product of two non-invertible elements. – Mark Aug 25 at 16:29