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

Question

i have some confusion in this answer

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

My confusion is

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 '19 at 16:04
I meant $\Bbb Z[\sqrt{-1}]$ and $\Bbb Z[\sqrt{-2}]$, respectively? — J. W. Tanner, Aug 25 '19 at 16:10

score 3 · Accepted Answer · answered Aug 25 '19 at 16:05

3

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.

answered Aug 25 '19 at 16:05

Mark

39,605

....@Mark But $1+ i $ is an irreducible element of $\mathbb{Z}[i]$ How it is reducible ? – jasmine Aug 25 '19 at 16:25
1

$1+i$ 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 '19 at 16:29
oh soorry @Mark u – jasmine Aug 25 '19 at 16:30

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

1 Answers1