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

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
@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
@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

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 $n$.