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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.

KReiser
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Teddyboer
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2 Answers2

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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 $R=0$.

Ben
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    I also like this proof because the restriction is based on basic facts about fiber and tensor producs. – Teddyboer Feb 21 '20 at 12:37
  • Why $\mathbb P_R^n\times_R \mathrm{Spec}(k)= \mathbb P_k^n$ ? – Alex Gomez Apr 13 '21 at 03:58
  • Dear @A.Gomez, I was assuming that this is standard knowledge in the context of the question. There are different ways to prove that. I think the most elementary way is to show more generally $\mathrm{Proj}(S\otimes_A B)=\mathrm{Proj}(S)\times_A\mathrm{Spec}(B)$ and apply this to $A=R$, $B=k$ and $S=k[x_0,\ldots,x_n]$. I’m assuming this question has been asked on MSE, so you could look it up if you need more help. – Ben Apr 13 '21 at 08:38
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    Dear @Ben, your comment was helpful enough. I am following Gortz' book, so the details that you mentioned I appear latter in the book. – Alex Gomez Apr 13 '21 at 08:57
  • Dear @Ben,this maybe obvious but why $\mathbb P_R^n$ is a $R$-scheme? – Alex Gomez Apr 13 '21 at 10:26
  • I’m sure this will be somewhere in your textbook. The structure morphism $\mathrm{Proj}(S)\to\mathrm{Spec}(A)$—for any graded $A$-algebra $f\colon A\to S$—takes a homogeneous prime $\mathfrak p \subset S$ to the contraction $f^{-1}\mathfrak p\subset A$. Note that, incidentally, if you know the $k$-scheme-structure of $\mathbb P^n_k$, then from the isomorphism of my previous comment, now taking $S=k[x_0,\dots,x_n]$, $A=k$ and $B=R$, you get $\mathbb P_R^n=\mathbb P^n_k\times_k\mathrm{Spec}(R)$ and the structure map is just the projection. – Ben Apr 13 '21 at 19:17
  • Can I ask now? : Perhaps, the existence of such a closed point $\operatorname{Spec}k = \operatorname{Spec} k(\mathfrak{m}) \to \operatorname{Spec}R$ is gauranteed by the existence of at least one maximal ideal $\mathfrak{m} \subseteq R$ ( $R \neq 0$ ) ? – Plantation Sep 27 '23 at 02:06
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    @Plantation exactly. – Ben Sep 27 '23 at 09:29
  • @Ben Thanks.~~~ – Plantation Sep 27 '23 at 09:35
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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.

KReiser
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  • Thank you for your detailed answer. I still have a question on a detail. What exactly do you mean by "The fiber of $\mathbb{A}_R^n\rightarrow Spec(R)$ over $X$ is exactly $\mathbb{A}_k^n\rightarrow Spec(k)$? – Teddyboer Feb 20 '20 at 12:21
  • I mean that the fiber over the point $x$ is the affine space $\Bbb A^n_k$ and the map from the fiber to the point is the standard map $\Bbb A^n_k\to \operatorname{Spec} k$. (Talking about fibers is the standard way to talk about preimages of morphisms of schemes, in case that's what's puzzling you.) – KReiser Feb 21 '20 at 00:56
  • It was. Thank you! – Teddyboer Feb 21 '20 at 11:09
  • I think my problem is that I, as a beginner learning about schemes, wish to see what every map "actually does" instead of throwing some (well founded) identifications at a problem such that it looks trivial. For example, I don't see why the fiber of the natural map $\mathbb{A}_R^n\rightarrow Spec(R)$ over the closed point $x$ is $\mathbb{A}_k^n$, although I assume it comes from the natural map $R \rightarrow R/\mathcal{m_x}=k$ – Teddyboer Feb 21 '20 at 11:40
  • You should look up the fiber product, then. The fiber product of the maps $i:{x}\to \operatorname{Spec} R$ and $\Bbb A^n_R\to \operatorname{Spec} R$ can be computed as Spec of the tensor product of the corresponding map on rings - ${x}\to \operatorname{Spec} R$ corresponds to the natural map $R\to k$, and $\Bbb A^n_R\to \operatorname{Spec} R$ corresponds to $R\to R[x_1,\cdots,x_n]$, so the tensor product is $k\otimes_R R[x_1,\cdots,x_n]\cong k[x_1,\cdots,x_n]$, which after taking Spec is exactly $\Bbb A^n_k$. (The other answer does the same thing but uses a slightly different ending.) – KReiser Feb 21 '20 at 18:48
  • Thank you. I figured out how the details of the second answer but in your argument I couldn't see the relation. Thank you for making that clear, I think I get it know – Teddyboer Feb 23 '20 at 20:39