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As the title says,

I am curious as to whether $A =\mathbb{R}[x,y,z]/(x^2+y^2+z^2)$ is a UFD.

I believe the answer is yes.

A thought I had was to apply Nagata's criterion, say by localizing at $z,$ to get $A_z = \mathbb{R}[x,y,z,z^{-1}]/((x/z)^2+(y/z)^2+1).$ Further, from that, I was hoping to maybe show that $A_z$ is isomorphic to $\mathbb{R}[x',y',z,z^{-1}]/(x'^2+y'^2+1)$ by sending $x/z$ to $x'$ and $y/z$ to $y'.$ However, for this I would still need to show that $\mathbb{R}[x',y']/(x^2+y^2+1)$ is an UFD, something I'm not quite sure how to do.

Any help would be welcome.

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Tedar
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  • Interestingly, Hartshorne Exercise II.6.5(b) (together with Prop. II.6.2) seems to suggest that it's not. However, Hartshorne forgot to assume that $k$ is algebraically closed, and his method is certainly wrong. (However, that doesn't mean the result is necessarily false.) Very curious. – Remy Jan 10 '16 at 22:01
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    @Remy There is no surprise: the factoriality depends on the base field. In this case, if $\mathbb R$ is replaced by an algebraically closed field of characteristic $\ne2$, then the ring is not a UFD and its divisor class group is cyclic of order two. – user26857 Jan 10 '16 at 23:34

1 Answers1

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We have $$A_z\simeq\mathbb R[X,Y,Z,Z^{-1}]/\langle (XZ^{-1})^2+(YZ^{-1})^2+1\rangle.$$ Since $\mathbb R[X,Y,Z,Z^{-1}]=\mathbb R[XZ^{-1},YZ^{-1},Z,Z^{-1}]$, we get $$A_z\simeq\mathbb R[XZ^{-1},YZ^{-1},Z,Z^{-1}]/\langle (XZ^{-1})^2+(YZ^{-1})^2+1\rangle.$$ Now set $U=XZ^{-1}$, $V=YZ^{-1}$. Check that $U,V$ are algebraically independent over $\mathbb R$. Then $$A_z\simeq\mathbb R[U,V,Z,Z^{-1}]/\langle U^2+V^2+1\rangle.$$

Now use that $\mathbb R[U,V]/\langle U^2+V^2+1\rangle$ is a UFD; see here.

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