Geometric meaning of the positive part of graded ring Let's say I have a complex projective variety $X\subseteq\mathbb P^n$ with homogeneous coordinate ring $S=\bigoplus_{d\ge 0} S_d$. The localization by some homogeneous $f\in S$ (of nonzero degree) yields a graded ring $S_f=:R=\bigoplus_{d\in\mathbb Z} R_d$. Now, instead of looking at the spectrum of the degree zero part, I could look at $R_{\ge 0}=\bigoplus_{d\ge 0} R_d$. I have an inclusion of graded $\mathbb C$-algebras $S\hookrightarrow R_{\ge 0}$ and both are positively graded, so I may very well consider the morphism 
$$Y:=\mathrm{Proj}(R_{\ge 0})\xrightarrow{\quad\textstyle\pi\quad}\mathrm{Proj}(S)=X.$$ 
Since $\mathrm{Spec}(R_0)=X_f$, the variety $Y$ is projective over $X_f$. However, what is the nature of $\pi$? Is it surjective? I am sure this must have interested someone before, but I cannot find anything in the 'standard' literature on algebraic geometry.
More generally, is there any deeper geometric meaning to taking the positive part of a graded ring?
 A: Up to replacing $S$ by the "Veronese subring" $S_{(e)} := \oplus_d S_{de}$, which does not change Proj, you may as well assume that $f$ has degree $1$.  Consider the graded $R_0$-algebra homomorphism $u:R_0[t]\to R_{\geq 0}$ by  $u(t) = f$.  
First, $u$ is surjective.  Indeed, for every homogeneous element $g$ in $R_d$, then $h=gf^{-d}$ is in $R_0$.  Thus $ht^d$ is an element in $R_0[t]_d$ such that $u(ht^d)$ equals $g$. 
Next, $u$ is injective.  To prove this, observe that, since $u$ is a graded $R_0$-algebra homomorphism, also $\text{Ker}(u)$ is a homogeneous ideal.  Every element in $R_0[t]_d$ is of the form $ht^d$ for some $h\in R_0$.  Also, $u(ht^d)$ equals $hf^d$. Since $f$ is not a zero divisor in $R$ (by construction), if $hf^d$ equals $0$, then $h$ equals $0$, hence $ht^d$ equals $0$.  Therefore the only homogeneous element in $\text{Ker}(u)$ is $0$, i.e., $u$ is injective.
Since $u$ is an isomorphism of graded $R_0$-algebras, also $\text{Proj}(R_{\geq 0})$ equals $\text{Proj}(R_0[t])$.  Of course $\text{Proj}(R_0[t])$ is just $\text{Spec}(R_0)$.  Thus $\text{Proj}(R_{\geq 0})$ is canonically isomorphic to $\text{Spec}(R_0)$.
