I want to prove that $\mathscr{O}_{\mathbb{P}^n_R}(\mathbb{P}^n_R) = R$ where $\mathbb{P}^n_R$ denotes the projective space over ring $R$ - the scheme glued out of schemes $U_i = \operatorname{Spec}(R[T_0/T_i,\dots,T_n/T_i])$ with given families of open sets required $U_{ij}=\operatorname{Spec}(R[T_0/T_i,\dots,T_n/T_i]_{T_j/T_i})$ and natural identifications.
By standard theorem we have $\Gamma(U_i,\mathscr{O}_{U_i})=R[T_0/T_i,\dots,T_n/T_i]$. To show that $\Gamma(\mathbb{P}^n_R,\mathscr{O}_{\mathbb{P}^n_R})=R$ it's sufficient to show $$\varprojlim \limits_{i=1,\dots,n} R[T_0/T_i,\dots,T_n/T_i] = R$$ But how neatly prove the above equality? Generally, are my mussings correct?
Thanks in advance.
Using your notations, we have the covering $\mathbb{P}_R^n = \bigcup_{i=0}^n U_i$. By the defining property of a sheaf, there is an exact sequence, $$0\rightarrow \Gamma(\mathbb{P}_R^n,\mathcal{O}_{\mathbb{P}_R^n}) \rightarrow \prod_i \Gamma(U_i,\mathcal{O}_{\mathbb{P}_R^n}) \rightarrow \prod_{i<j} \Gamma(U_i\cap U_j,\mathcal{O}_{\mathbb{P}_R^n})$$
Using the identifications $$\Gamma(U_i,\mathcal{O}_{\mathbb{P}_R^n}) \cong R[T_0/T_i,...,T_n/T_i], \Gamma(U_i\cap U_j,\mathcal{O}_{\mathbb{P}_R^n}) \cong R[T_0/T_i,...,T_n/T_i,T_i/T_j],$$ we have that $\Gamma(\mathbb{P}_R^n,\mathcal{O}_{\mathbb{P}_R^n})$ can be identified with the $n+1$ tuples of polynomials over $R$, for which the $i$'th component is a polynomial in $T_0/T_i,...,T_n/T_i$, and the $i$'th and $j$'th polynomials are the same when considered polynomials in $T_0/T_i,...,T_n/T_i,T_0/T_j,...,T_n/T_j$. But clearly an $n+1$-tuple has this property precisely if it is of the form $(a,...,a)$ for some $a\in R$. Thus $\Gamma(\mathbb{P}_R^n,\mathcal{O}_{\mathbb{P}_R^n})\cong R$.