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i have some confusion in this answer

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

My confusion is
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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 $??

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

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  • $\begingroup$ ....@Mark But $1+ i $ is an irreducible element of $\mathbb{Z}[i]$ How it is reducible ? $\endgroup$ – jasmine Aug 25 at 16:25
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    $\begingroup$ $1+i$ is irreducible, but why is that important? I'm saying $2$ is reducible as it is a product of two non-invertible elements. $\endgroup$ – Mark Aug 25 at 16:29
  • $\begingroup$ oh soorry @Mark u $\endgroup$ – jasmine Aug 25 at 16:30

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