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I'm working through a proof that $R=\mathbb R[X,Y,Z]/ \left\langle X^2+Y^2+Z^2 -1 \right\rangle $ is a UFD. The idea is to localize at $1-x$ and show the result is a UFD. Since $R$ is atomic as a quotient of a Noetherian ring, and $1-x$ is prime, Nagata's lemma will imply the resut.

Let $x,y,z$ be the images of $X,Y,Z$ in the quotient.

By exactness of localization $R_{1-x}\cong \mathbb R[X,Y,Z]_{1-X}/ \left\langle X^2+Y^2+Z^2 -1 \right\rangle$. If I understand correctly, $\mathbb R[X,Y,Z]_{1-X}$ means we're deleting the plane $X=1$ from $\mathbb R^3$.

Take $T=(1-X)^{-1}$ and note $\mathbb R[X,Y,Z]_{1-X}\cong \mathbb R[X,Y,Z,T]$. Descending to $R_{1-x}$ as the quotient above, some manipulations enable me to show $\mathbb R[x,y,z,t]\cong \mathbb R[ty,tz,t^{-1}]$, and since the latter is the localization at $t$ of a UFD, Nagata ends the proof.

I would like to understand what exactly is going on geometrically here and I really have no clue where to start because already at the quotient $R_{1-x}\cong \mathbb R[X,Y,Z]_{1-X}/ \left\langle X^2+Y^2+Z^2 -1 \right\rangle$ I have no idea what I should be visualizing.

Arrow
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  • I'm not sure what exactly you want to visualize. The ring $\Bbb R[X,Y,Z]$ is basically all (polynomial) functions on $\Bbb R^3$. The quotient $R$ then identifies two such functions if they are the same on the unit sphere. The localization then just extends this to functions that are ratios of polynomials, where the denominator is not allowed to vanish on the sphere when $x=1$. – Gregory Grant Jul 17 '16 at 14:21
  • @GregoryGrant I would like to understand why the ring of such rational functions (identified if they're the same on the sphere) is a UFD geometrically, if possible. – Arrow Jul 17 '16 at 14:23
  • Well, $\Bbb R[X,Y]$ is a UFD and is a much simpler space. The UFD property there is just the fact that every curve is a unique union of irreducible curves (varieties). So for example $XY$ is a union of two lines each of which is irreducible. The analogous property in $\Bbb R[X,Y,Z]$ is that every surface is a unique union of irreducible surfaces. So somehow in $R$ the same property must hold, all objects are unions of irreducibles in the same way, even if you identify the unit sphere to zero. In fact I'm having trouble visualizing when the UFD property might fail. Do you have an example? – Gregory Grant Jul 17 '16 at 15:20
  • @GregoryGrant $\Bbb R[X,Y]/ \left\langle X^2+Y^2-1 \right\rangle $ is not a UFD. – Arrow Jul 17 '16 at 15:37
  • Okay but how do you visualize the non-UFD property of that ring? And doesn't it seem strange to be a UFD in 3 variables but not in 2? – Gregory Grant Jul 17 '16 at 21:18
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    @GregoryGrant yes it seems very strange to me - visualizing these things is exactly my problem! – Arrow Jul 18 '16 at 06:33
  • @Gregory, Arrow: I have tried to provide some "visualization" in my answer. – Georges Elencwajg Jul 18 '16 at 11:42

1 Answers1

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Given a noetherian integrally closed domain $A$ we have the equivalence $$ A \operatorname { is a } UFD \iff Cl(A)=0$$where $Cl(A)=0$ is the class group of $A$.
If your case $\operatorname {Spec}(R)$ is smooth so that $Cl(R)=Pic(R)$, the Picard group of $R$.
So the problem boils down to proving that the algebraic Picard group $Pic(S^2)$ of a sphere is trivial.
I don't know a way of showing that simpler than the calculation you made but intuitively this is not so surprizing since all continuous (or differentiable) line bundles on the sphere are trivial because $H^1(S^2,\mathbb Z/2\mathbb Z)=0$.
Contrastingly the ring $\mathcal O(S^1)=\mathbb R[X,Y]/\langle X^2+Y^2-1 \rangle$ is not a UFD since the circle $S^1$ possesses a non-trivial line-bundle, the notorious Möbius line bundle, which can be given an algebraic structure.

Georges Elencwajg
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