Projective space is not affine

Question

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?

Answer 1

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.

Answer 2

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 $R.$

Answer 3

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.