Projective $n$ - space is not affine over any ring $R$

Question

A similar question has been asked here already, but there was no final answer to the problem in the most general case. I wish to show that:

For $n>1$ and a ring $R$, the projective $n$ - space $\mathbb{P}_R^n$ is not affine unless $R=0$.

What I have so far: Assume $\mathbb{P}_R^n$ was affine, then $\mathbb{P}_R^n\simeq \operatorname{Spec}(R)$. Now by construction $\mathbb{P}_R^n$ contains the affine subspace $\mathbb{A}_R^n\simeq \operatorname{Spec}(R[\frac{t_1}{t_0},...,\frac{t_1}{t_0}])$ as an open subscheme. Hence the inclusion $\mathbb{A}_R^n\hookrightarrow\mathbb{P}_R^n$ induces some ring homomorphism $R\rightarrow R[\frac{t_1}{t_0},...,\frac{t_1}{t_0}]$. And that is about it...

The book I read (Bosch, Algebraic Geometry and Commutative Algebra) uses for the case that $R=K$ is a field an argument based on $K'$ - valued points $\mathbb{P}_K^n = \operatorname{Hom}_K(\operatorname{Spec}(K'),\mathbb{P}_K^n)$, where $K'$ is field extension of $K$. He shows that if $\mathbb{P}_K^n = \operatorname{Spec}(K)$ was affine, it would be a one - point space and then constructs a bijection $\mathbb{P}_K^n(K')\leftrightarrow \mathbb{P}^n(K')$, where the RHS is the ordinary projective $n$ - space over $K'$.

I was hoping to argue similarly, but I am lost at this point.

Answer 1

Take $R\neq 0$, assume $\Bbb P^n_R\to \operatorname{Spec} R$ is an isomorphism, and consider the composite map $\Bbb A^n_R \to \Bbb P^n_R \to \operatorname{Spec} R$, where the first map is the standard open immersion of $\Bbb A^n_R\to \Bbb P^n_R$ with image $D(x_0)$. Then both maps are injective on underlying topological spaces, so the composite map $\Bbb A^n_R \to \operatorname{Spec} R$ must be injective as well. On the other hand let $x$ be a closed point of $\operatorname{Spec} R$ with corresponding maximal ideal $\mathfrak{m}$ and residue field $k$. The fiber of $\Bbb A^n_R \to \operatorname{Spec} R$ over $x$ is exactly $\Bbb A^n_k \to \operatorname{Spec} k$, which has at least two points: $(0,\cdots,0)$ and $(1,0,\cdots,0)$. This is a contradiction, so it cannot have been the case that $\Bbb P^n_R\to\operatorname{Spec} R$ was an isomorphism and we're done.

Answer 2

The statement can be reduced to its version over fields, for if $\mathrm{Spec}(k)\to\mathrm{Spec}(R)$ is a closed point, $\mathbb P_R^n\times_R \mathrm{Spec}(k)= \mathbb P_k^n$ and $\mathrm{Spec}(R)\times_R \mathrm{Spec}(k) =\mathrm{Spec}(k)$. Thus, if $\mathbb P_R^n$ were isomorphic to $\mathrm{Spec}(R)$, then so were $\mathbb P_k^n$ and $\mathrm{Spec}(k)$. Assuming the statement is known for fields, it follows that either $n=0$ or $R=0$.