Projective space is not affine 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?

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 $R.$
