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I'm considering the ring $\mathbb{Z}[\sqrt{-n}]$, where $n\ge 3$ and square free. I want to see why it's not an UFD.

I defined a norm for the ring by $|a+b\sqrt{-n}|=a^2+nb^2$. Using this I was able to show that $2$, $\sqrt{-n}$ and $1+\sqrt{-n}$ are all irreducible. Is there someway to conclude that $\mathbb{Z}[\sqrt{-n}]$ is not a UFD based on this? Thanks.

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1 Answer 1

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If $n$ is even, then $2$ divides $\sqrt{-n}^2=-n$ but does not divide $\sqrt{-n}$, so $2$ is a nonprime irreducible. In a UFD, all irreducibles are prime, so this shows $\mathbb{Z}[\sqrt{-n}]$ is not a UFD.

Similarly, if $n$ is odd, then $2$ divides $(1+\sqrt{-n})(1-\sqrt{-n})=1+n$ without dividing either of the factors, so again $2$ is a nonprime irreducible.

This argument works equally well for $n=3$, but fails for $n=1,2$, and in fact $\mathbb{Z}[\sqrt{-1}]$ and $\mathbb{Z}[\sqrt{-2}]$ are UFDs.

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Thanks very much, Chris! –  Danielle Intal Oct 9 '11 at 2:46
    
@ChrisEagle Why is $\mathbb{Z}[\sqrt{-2}]$ a UFD? I tried showing it's Euclidean using the same method as for the Gauss integers, but I found that the remainder could have norm equal to that of the divisor (and not strictly less). –  JessicaB Nov 23 '12 at 15:18
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@JessicaB: You're doing something wrong, then. Every point in the plane is distance at most $\sqrt{3}/2<1$ from a lattice point, so you can always get the remainder strictly smaller than the divisor. –  Chris Eagle Nov 23 '12 at 15:26
    
@ChrisEagle Quite right... Our lattice has edges 1 and $\sqrt{2}$ not $\sqrt{2}$ and $\sqrt{2}$. Oops. –  JessicaB Nov 23 '12 at 15:32
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@Knight The norm $|a + b\sqrt{-n}| = a^2 + nb^2$ is multiplicative, and $2$ has norm $4$. When $n \ge 4$, the only other element with norm dividing $4$ is $1 = 1 + 0\sqrt{-n}$. When $n = 3$, $\pm 1 \pm \sqrt{-3}$ also have norm $4$, but still do not divide $2$. –  Goos May 4 at 21:28

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