# What kind of ring $R$ makes $\mathcal O(\mathrm{Proj}\,R[x_0,\cdots,x_n])=R$?

Consider the ring $\Bbb Z[x_0,\cdots,x_n]$ with the grading defined by the degrees of polynomials, then we have $\mathcal O(\mathrm{Proj}\,\Bbb Z[x_0,\cdots,x_n])=\Bbb Z$.

Let $R$ be a nonzero commutative ring and consider the ring $R[x_0,\cdots,x_n]$ with the grading defined by the degrees of polynomials. What kind of ring $R$ makes $\mathcal O(\mathrm{Proj}\,R[x_0,\cdots,x_n])=R$?

Consider the standard affine cover of $\Bbb P^n_R :=\operatorname{Proj} R[x_0,\cdots,x_n]$ given by $U_i=D(x_i)$, ie $U_i$ is the subscheme where $x_i\neq 0$. Each $U_i$ is affine and isomorphic to $\operatorname{Spec} R[\frac{x_0}{x_i}\cdots,\frac{x_n}{x_i}]$. Suppose $f$ is a global section of $\mathcal{O}_{\Bbb P^n_R}$. It must restrict to a section $f_i$ on each $U_i$ such that the sections agree on overlaps, ie $f_i|_{U_j}=f_j|_{U_i}$.
What does this really mean? It means that if $f_i=p(\frac{x_1}{x_i},\cdots,\frac{x_n}{x_i})$ and $f_j=q(\frac{x_1}{x_j},\cdots,\frac{x_n}{x_j})$, then $f_i=f_j$ where $x_i,x_j\neq 0$. Combining the observation that the degree of $p$ with respect to $\frac{x_j}{x_i}$ and $q$ with respect to $\frac{x_i}{x_j}$ must be non-negative with the fact that $f_i|_{U_j}=f_j|_{U_i}$, we see that after varying $j$, it must be the case that $p$ was actually constant. Varying $i$, we see that the restriction of a global section to any of the standard affine opens must be constant, which shows the result.