Let $k$ be an algebraically-closed field of characteristic not two. Then is the ring $$k[x,y,z]/(x^2+y^2-z^2)$$ a UFD? I admit that $k[x,y,z]/(xy-z^2)$ is not a UFD.
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3Aren't the two rings obviously isomorphic (due to $x^2+y^2=(x+iy)(x-iy)$) ? – darij grinberg Jun 07 '13 at 14:58
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Oh, didn't notice it! +1. – darij grinberg Jun 07 '13 at 17:14
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The problem is more difficult than it first appears. Easy enough it is to say "Oh in the quotient ring we just have $z^2 = (x+iy)(x-iy)$ and so it is not a unique factorization domain". Though, one needs to prove that $x+iy, x-iy$ and $z$ are really irreducible elements in the quotient ring.
However you can reduce your problem to what you already know about $k[x,y,z]/(xy - z^2)$: I claim that in fact your ring is isomorphic to this! Indeed define variables \begin{eqnarray*} u&:=& x+iy\\ v &:=& x-iy .\end{eqnarray*}
and note that $k[x,y,z] = k[u,v,z]$ by using that $x = \frac{u+v}{2} $ and $ y = \frac{u-v}{2i}$. Then $$k[x,y,z]/(x^2 + y^2 - z^2) \cong k[u,v,z]/(uv - z^2)$$
where the ring on the right hand side you already know is not a UFD, so you have reduced your problem to what you already know!
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No, consider $\mathbb{C}[x,y,z]/(x^2 + y^2 -z^2)$ (for instance). Then look at the factorization of $$z^2 = z \cdot z = (x + i y) (x-iy)$$ Where all of these products are taken in the quotient ring.
Andrew Maurer
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7Dear andybenji, I think the problem is harder than it appears. For example, how do you know that $z$, $x+iy$ and $x-iy$ are actually irreducible elements of this quotient ring? Regards, – Jun 07 '13 at 02:45
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@BenjaLim This was just to provide intuition, but it could be made more rigorous. Proving $z$ is irreducible is easy - show it's prime by taking the quotient by $(z)$, which is a domain. Thus it suffices to prove that $z$ does not divide $x + i y$ or $x - i y$, which is also obvious. So these are two different factorizations, rendering $\mathbb{C}[x,y,z]/(x^2 + y^2 - z^2)$ not a UFD. That being said, your solution above seems more elegant, so props on that. – Andrew Maurer Jun 07 '13 at 04:16
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