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This is exercise 14.2 N in Vakil, self-study. Similar questions have been asked a few times on this site, but none of the answers use a method that I believe Vakil intended: here, here, and here.

We are to show $\mathbb R[x, y]/(x^2 + y^2 -1)$ is not a UFD, but over $\mathbb C$, it is, using exercise 14.2 L, which says, among other things, that $\mathbb P^n - Y$ is not the spectrum of a UFD if $Y$ is a hypersurface of degree $d > 1$.

The issue is that I don't see how to write the unit circle as the complement of a degree $d > 1$ hypersurface in a projective space, if indeed that is what we are supposed to do, nor do I see how, supposing we have done this, the result will change over $\mathbb C$, since presumably 14.2 L still applies.

Johnny Apple
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2 Answers2

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This is quite a late answer but maybe better late than never. I just want to spell out your nice attempted answer which should basically be Vakil's intended idea.

We note that there is an isomorphism $(k[x,y]_{(x^2+y^2)})_0 \cong k[x,y]/(x^2+y^2-1)$ for $\operatorname{char}{k} \neq 2$.* (This could for instance be seen and motivated from the isomorphism between $\mathbb{P}^1$ and the projective circle.) Therefore, $$ \operatorname{Spec}k[x,y]/(x^2+y^2-1) = \mathbb{P}_k^1 \setminus V_+(x^2+y^2).$$ The difference between $k= \mathbb{R}$ and $k = \mathbb{C}$ lies in the distinction of whether $x^2+y^2$ is an irreducible polynomial.

  • If $k = \mathbb{R}$, then $x^2+y^2$ is irreducible (of degree 2). We have seen in 14.2.L that $\mathbb{P}_\mathbb{R}^1 \setminus V_+(x^2+y^2)$ has non-trivial class group.
  • If $k = \mathbb{C}$, then $x^2+y^2 = (x+iy)(x-iy)$ splits into two hyperplanes. Then, $$\mathbb{P}_{\mathbb{C}}^1 \setminus V_+(x^2+y^2) = \mathbb{P}_{\mathbb{C}}^1 - V_+(x+iy) - V_+(x-iy). $$ By 14.2.L (after changing coordinates) this is the spectrum of a UFD.

*The embedding of the projective unit circle is given by a map of graded rings (see MSE/4086633) via $$\mathbb{P}_k^1 \cong \operatorname{Proj}{k[x,y]^{(2)}} \overset{\varphi}{\hookrightarrow} \mathbb{P}_k^2.$$ The patch $\operatorname{Spec}{k[x_{02}, x_{12}]/(x_{02}^2+x_{12}^2 - 1)}$ is given by the preimage of $D_+(x_2)$ which is $D_+(\varphi^*x_2) = D_+(x^2 + y^2)$. On homogeneous localizations, there is an isomorphism between a ring and its Veronese subrings, so this pulls back to $D_+(x^2+y^2)$ through $\mathbb{P}_k^1 \cong \operatorname{Proj}{k[x,y]^{(2)}}$.

Арсений Кряжев
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Qi Zhu
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    Your isomorphism of rings $k[x,y]_{(x^2+y^2)}\cong k[x,y]/(x^2+y^2-1)$ looks a little funny to me, though maybe it's just late and I'm not seeing it. Could you explain a little more about what you're doing there? – Hank Scorpio Apr 07 '21 at 07:10
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    @HankScorpio: agreed, I'm similarly confused. Certainly, it seems like it cannot be true if $k$ has characteristic $2$, since then the RHS is not even a domain, but the left is. Even if $k$ has characteristic not $2$ though, I still don't follow... – Alex Wertheim Apr 07 '21 at 07:52
  • @HankScorpio It's indeed also the part that worries/worried me the most about this argument. I added an explanation, I'm sorry if I made a careless mistake. The map of graded rings does use $\operatorname{char}{k} \neq 2$ here. – Qi Zhu Apr 07 '21 at 07:54
  • @AlexWertheim You're right, in characteristic 2, there is a problem in the isomorphism between the projective unit circle and $\mathbb{P}^1$. I added an explanation for that isomorphism but please tell me about any mistakes I might have made. – Qi Zhu Apr 07 '21 at 07:57
  • Naively, the idea is that the isomorphism to the projective unit circle $\mathbb{P}k^1 \cong C$ pulls the affine patch $s^2+t^2 = 1$ back to $D+(x^2+y^2)$ which was also Johnny's idea. I tried to make that rigorous. – Qi Zhu Apr 07 '21 at 08:04
  • I think I see the motivation, but I still don't quite see the map of rings. Could you write down where $x$ and $y$ in $k[x,y]/(x^2+y2-1)$ go under you map? Or the other direction, if it's easier. – Hank Scorpio Apr 09 '21 at 07:32
  • @HankScorpio The map of rings is of the form $k[x_0, x_1, x_2] \to k[x,y]^{(2)}$ that corresponds to $\varphi$. It is given in the thread that I've linked. – Qi Zhu Apr 09 '21 at 07:49
  • Ahh, okay, it's $k[x,y]/(x^2+y^2-1)\to k[x,y]_{(x^2+y^2)}$ by $x\mapsto \frac{x^2-y^2}{x^2+y^2}$ and $y\mapsto \frac{2xy}{x^2+y^2}$. I guess I don't understand why you didn't just write that down somewhere, but whatever, this was a good answer. – Hank Scorpio Apr 09 '21 at 09:26
  • @HankScorpio Oh, that‘s what you meant. Nice! I guess I just didn‘t really see the explicit isomorphism and didn‘t bother working it out from the abstract argument... But your comment is good and closes the case :) – Qi Zhu Apr 09 '21 at 09:35
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Attempt at an answer:

Over $\mathbb C$, Euler's formula and parameterization of the unit circle give us that our ring is isomorphic to $\mathbb C[x]_x$, a UFD.

Over $\mathbb R$, the projectivized unit circle is isomorphic to $\mathbb P^1_{\mathbb R}$, and the image of the affine unit circle under this isomorphism is the complement of $x^2 + y^2 = 0$ (thinking of points on this curve as points in $\mathbb C$), so 14.2 L concludes.

Johnny Apple
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