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.
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.