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

Are $\mathbb{Z}[\sqrt{-5}]$ and $\mathbb{Z}[\sqrt{-6}]$ UFDs?

I think that $\mathbb{Z}[\sqrt{-5}]$ is UFD because for example $6=2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5})$ but and $\mathbb{Z}[\sqrt{-6}]$?

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marked as duplicate by Gerry Myerson, Micah, Erick Wong, 5PM, Brandon Carter Jan 24 '13 at 5:35

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Do you mean $\mathbb{Z}[\sqrt{-5}]$ and $\mathbb{Z}[\sqrt{-6}]$ ? –  Alan Simonin Jan 23 '13 at 20:42
    
yes, thank for your comment –  Sophie Germain Jan 23 '13 at 20:48
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You wrote "I think that ... is UFD" and then gave an example of two different factorizations. Did you mean to write "isn't a UFD"? Then the second half of the question is "But and $\Bbb Z[\sqrt{-6}]$?" meaning "what about this other ring?" –  rschwieb Jan 23 '13 at 21:11
    
My question is: Is $\mathbb{Z}[\sqrt{-6}]$ a UFD? –  Sophie Germain Jan 23 '13 at 21:19
    
This was asked before for general $\Bbb{Z}[\sqrt{-n}]$: math.stackexchange.com/questions/70976/… –  Chris Eagle Jan 23 '13 at 23:32
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1 Answer 1

$\mathbb Z[\sqrt{-6}]$ is not an UFD since $6=2\cdot 3=-(\sqrt{-6})(\sqrt{-6})$.

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Ok, I see it. The answer is no. Lets consider the number 2. Looking for a factorization of 2 over $ \mathbb{Z}$ we have . Is this factorizartion unique ? If we are looking over another domain like $\mathbb{Z}[\sqrt{-6}]$ , then how easy is to find all the posible factorizations of 2 ? If we find one is it indeed unique? considering the norm of 2 it's easy. :) –  Sophie Germain Jan 23 '13 at 22:48
    
$2$ is irreducible in $\mathbb Z[\sqrt{-6}]$. –  user26857 Jan 23 '13 at 23:17
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Why do you add the $i$? Why not say $-6=-2\cdot 3=\sqrt{-6}^2$? –  Olivier Bégassat Jan 23 '13 at 23:19
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