3

Let $ n \geq 3 $. By factorising $ n $ or $n + 1 $ (as appropriate), show 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}) $!

user26857
  • 52,094
user938272
  • 923
  • But this is not always true, although in most cases it is. – Adrián Barquero May 06 '11 at 22:53
  • Can you give me a counterexample? – user938272 May 06 '11 at 22:56
  • 1
    @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 06 '11 at 23:08
  • 1
    @user938272: In your edit, you mean $n=-\sqrt{-n}\sqrt{-n}$. Note that $\sqrt{n}\notin\mathbb{Z}[\sqrt{-n}]$. – Arturo Magidin May 07 '11 at 00:34
  • @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 07 '11 at 00:38
  • @Gerry Yes, you're absolutely right. I was thinking of the rings of integers of the imaginary quadratic fields of class number 1. I also thought about what you said that in the case in which $-n \equiv 1 \pmod{4}$ then $\mathbb{Z}[\sqrt{-n}]$ is not the full ring of integers. But I was just leaving my office and didn't have a chance to look at them (I don't remember all of them =P) and now that I look what they are I realize that all of them have $-n \equiv 1 \pmod{4}$. My bad for leaving such a comment when I don't have time to actually look things up. But at least I learned something new =) – Adrián Barquero May 07 '11 at 01:26

1 Answers1

2

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 $n$.

Community
  • 1
Julian Kuelshammer
  • 9,670