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?



Geometric meaning of the positive part of graded ring
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 $\text{Spec}(R\_0)$.


