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Let $ n \geq 3 $. By factorising $ n $ or $n + 1 $ (as appropriate), shat that $ \mathbb{Z}[\sqrt{-n}] $ is not a UFD.

My thoughts so far:

Define $ N(a + b \sqrt{-n}) = a^2 + n b^2 $.

Suppose $ n $ is odd. Then $ n + 1 $ is even, say $ n + 1 = 2k $. Now $ N(2) = 4 $, and the norm of an element in this ring can never be 2, so we have that 2 is irreducible. Now note that $ 1 + n = (1 + \sqrt{-n})(1 - \sqrt{-n}) $. Is $ 1 + \sqrt{-n} $ irreducible? Well, if $ 1 + \sqrt{-n} = z_1 z_2 $, then $ N(z_1)N(z_2) = 1 + n $. So $ N(z_i) \leq \frac{n+1}{2} < n $. But this means both $ z_i$ must be purely real, which clearly can't be the case. Similarly, $ 1 - \sqrt{-n} $ is irreducible. Neither of these factors are equal to 2, and so 2 appears in one factorisation but not another. Hence for $ n $ odd, we don't have a UFD.

What about $n$ even? How can I factorise $ n $ other than as $ 2k $ for some $k $?

Thanks

EDIT: I overlooked $ n = \sqrt{-n} ( -\sqrt{n}) $!

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But this is not always true, although in most cases it is. –  Adrián Barquero May 6 '11 at 22:53
    
Can you give me a counterexample? –  user938272 May 6 '11 at 22:56
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@Adrian, perhaps you are thinking of $n$ such that the ring of integers in ${\bf Q}(\sqrt{-n})$ is a UFD? But if those $n$ are $3$ mod $4$, then the ring of integers is not ${\bf Z}[\sqrt{-n}]$. –  Gerry Myerson May 6 '11 at 23:08
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@user938272: In your edit, you mean $n=-\sqrt{-n}\sqrt{-n}$. Note that $\sqrt{n}\notin\mathbb{Z}[\sqrt{-n}]$. –  Arturo Magidin May 7 '11 at 0:34
    
@Gerry: And if $n$ is not square free, you also get other things. If $n=4$,then $\mathbb{Z}[\sqrt{-4}]$ is not the ring of integers of $\mathbb{Q}(\sqrt{-4}) = \mathbb{Q}(i)$, either. –  Arturo Magidin May 7 '11 at 0:38

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This CW answer intends to remove the question from the unanswered queue


Your thoughts are correct and if you correct your edit according to this comment by Arturo Magidin you will find a similar argument for even $n$.

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