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I read a prove that the projective space $\mathbb P_{R}^{n}$ is not affine (n>0): (Remark 3.14 p72 Algebraic Geometry I by Wedhorn,Gortz).

It said that the canonical ring homomorphism $R$ to $\Gamma(\mathbb P_{R}^{n}, \mathcal{O}_{\mathbb P_{R}^{n}})$ is an isomorphism. This implies that for n>0 the scheme $\mathbb P_{R}^{n}$ is not affine, since otherwise $\mathbb P_{R}^{n}=Spec R$.

I am not clear about why is it impossible to have $\mathbb P_{R}^{n}=Spec R$? And is what sense does the $=$ mean here?

3 Answers3

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Here is a proof that does not need dimension argument and Noetherian assumptions. Set $X$ to be $\mathbb{P}^n_R$. Then $X=Spec (R)$ means that the map $X\to Spec (R)$ by glueing $n+1$ maps $U_i=Spec (R[T_1,\dots,T_n])\to Spec (R)$ is isomorphic. This is contrary to the fact that the map $U_1 \to Spec (R)$ is surjective and $U_1\neq X.$ The map $U_1 \to Spec (R)$ is surjective since for each $\mathfrak{p}\in Spec (R)$, $(\mathfrak{p},T_1,\dots,T_n)\in U_1$ which is mapped to $\mathfrak{p}.$ And $U_1\neq X$ since $(\mathfrak{p},T_1,\dots,T_n)\in U_2-U_1$ for some prime ideal of $R.$

KReiser
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Nguyễn Xuân Bách
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  • This isn't important for the argument, but isn't U_i the spec of a polynomial ring over R with n variables? (and not n+1) – Chen Van Dam Feb 18 '22 at 15:29
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    Thanks. I fixed it. – Nguyễn Xuân Bách Feb 19 '22 at 16:05
  • For now, for me, I think that this is most standard proof for this statement. And can I ask more to make certainly sure? Why the "glued map $\psi : X \to \operatorname{Spec}(R)$ is isomorphims" leads to contradiction? If $\varphi_1 : U_1 \to \operatorname{Spec}(R)$ is surjective, then it necessiates $U_1 \neq X$? If so, how? How can we use $\varphi_1 = \psi|_{U_1}$ ? – Plantation Sep 27 '23 at 02:14
  • O.K. I understand why that leads to the contradiction. I edited your post a little bit to explain such step more in detail. Do you agree? – Plantation Sep 27 '23 at 02:45
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    @Plantation it is inappropriate to make significant changes to someone else's work like this. Fixing minor typos, grammar, or MathJax issues is fine, but nearly tripling the character count is too large of an edit to make unilaterally. – KReiser Sep 27 '23 at 04:38
  • @KReiser : Uhm. Anyway O.K. And in my personal opinion, I think that it is not significant change, what should I say, it is just a reinforce to fill in the leaped part. I want to leave it as a record. Can you allow? – Plantation Sep 27 '23 at 04:50
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    @Plantation It's not a question of me allowing this or not - this is the accepted convention here on MSE, which you can find at meta. If you wish, you could put your expanded argument in a new answer. – KReiser Sep 27 '23 at 05:00
  • @KReiser : O.K. I got it. Thanks. ~~ – Plantation Sep 27 '23 at 05:02
  • @KReiser : Btw, beacuse of wondering, perhaps is there someone who rollback my edited question? The author? P.s. I don't want to post my expanded argument as a new answer, beacause its essential(?) ideas are not changed and I don't want to copy it :) What should I do? – Plantation Sep 27 '23 at 05:10
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First, let us review the definition of an affine scheme. An affine scheme $X$ is a locally ringed space isomorphic to $\operatorname{Spec} A$ for some commutative ring $A$. This means that if one knows one has an affine scheme $X$, then all one has to do to recover $A$ such that $X=\operatorname{Spec} A$ is to take global sections of the structure sheaf, ie $A\cong\Gamma(X,\mathcal{O}_X)$.

In order to prove that $\mathbb{P}^n_R$ is not affine, it suffices to show that $\operatorname{Spec}(\Gamma(\mathbb{P}^n_R,\mathcal{O}_{\mathbb{P}^n_R}))\cong \operatorname{Spec} R$ is not isomorphic to $\mathbb{P}^n_R$. This is due to a dimension argument- assume $R$ is noetherian, and $\dim R=d$. Then $\dim\mathbb{P}^n_R=d+n$, as $\dim R[x_1,\cdots,x_n]=d+n$. Unless $n=0$, the two cannot be isomorphic.

KReiser
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  • Iam not clear why any map of $\mathbb{A}^{1}{R}$ into $\mathbb{P}^{n}{R}$ for n>0 misses at least one point? –  Sep 20 '13 at 23:11
  • If $n>1$, then it's by a dimension argument. If $n=1$, then such a map must first be one-to-one and therefore is equivalent to picking a coordinate patch on $P^1_R$- apply the standard transition functions to see that there is a point in the other patch that is missed. (Reference the definition of $P^n_R$ by gluing datum in the same section this remark is from). – KReiser Sep 20 '13 at 23:18
  • I still cannot understand this:"apply the standard transition functions to see that there is a point in the other patch that is missed." When proving a straight line cannot be homeomorphic to the circle, we use compactness, could there be a similar argument? Maybe I am missing some of your point, I am not clear why the image of the patch of $\mathbb{A}^1_R$ is one of the gluing datum? As in the topological case, the image of the line segment may not be one of the gluing datum, or maybe the two cases are different? Regards –  Sep 20 '13 at 23:57
  • I have updated my answer to only include a dimension argument- I had confused $A^1=Spec R[x]$ and $A^0=Spec R$ earlier- my mistake. Compactness isn't something that we can use to distinguish $A^1$ and $P^1$ since open sets in $A^1$ are the cofinite sets, and thus $A^1$ is also compact. – KReiser Sep 21 '13 at 00:14
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A partial answer: Scheme morphisms between affine schemes are in natural bijection with ring morphisms between the base rings. Hence, if the canonical homomorphism is an isomorphism and the projective scheme were affine, there would then be an isomorphism in the category of schemes. This is what the = means here. Now, an isomorphism of schemes carries with it an isomorphism of the spaces associated to the schemes. If $R$ is some field, then the space for $Spec R$ is a single point, while $P^n_R$ will have lots of points, since each homogeneous prime ideal is a point. So there can be no homeomorphism between the spaces.

If $R$ is not a field, I'm not really sure how to argue, since if it's something weird like $k[x_1,x_2,x_3,\ldots]$, then the rings $R[x_1,x_2,\ldots, x_n]$ are all isomorphic to $R$ and so maybe something weird could happen. Hopefully an expert can clear it up.

Zach L.
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