(Reference) Does a ring's grading commute with the formation of fractions? Question:

Given a graded integral domain, when one forms the corresponding field of fractions, does the resulting field of fractions have a grading "compatible" with the original field of fractions?

If it is true, I imagine the proof is long and tedious, so please don't feel obligated to include or copy the proof here. A pointer to a reference containing the proof will be sufficient for an answer.
Example:
I am not sure how to state the result more precisely. Hopefully the following example clarifies.
Let $k$ be an algebraically closed field, consider a projective variety $V \subseteq \mathbb{P}^n(k)$, then $V$ corresponds to a homogeneous prime ideal $I$. Therefore the ring $k[x_0, \dots, x_n]/I$ is a graded integral domain. Moreover, the following isomorphism holds $$k[x_0, \dots, x_n]/I \cong \bigoplus\limits_{d=0}^{\infty} \frac{R_d}{I \cap R_d}$$ where $R_d$ is the additive subgroup of $k[x_0, \dots, x_n]$ consisting of homogeneous polynomials of degree $d$ (i.e. the $R_d$ form the grading of $k[x_0, \dots, x_n]$), and I believe it is meant to be understood that the direct sum is an external direct sum. 
Because the quotient ring given above is an integral domain (since $I$ is a prime ideal), we can form the field of fractions of this ring, denote it by $$\mathscr{F}\left( 
\bigoplus\limits_{i=0}^{\infty} \frac{R_d}{I \cap R_d} \right) \,.$$ If we denote the set (and, I believe, additive group) of all fractions whose numerator and denominator consist both of elements of $\frac{R_d}{I \cap R_d}$ by $\mathscr{F}(\frac{R_d}{I \cap R_d})$, then does one have the following: 

$$\mathscr{F}\left( 
\bigoplus\limits_{d=0}^{\infty} \frac{R_d}{I \cap R_d} \right) \cong \bigoplus\limits_{d=0}^{\infty} \mathscr{F}\left( \frac{R_d}{I \cap R_d} \right) \quad ? $$ 

The right-hand side is the definition which was given in my textbook for the field of functions of $V$, $\mathscr{K}_V$. A problem asked me to verify that $\mathscr{K}_V$ was a field. Since the fractions (localization?) of an integral domain is always a field, it is easy for me to verify that the left-hand side is a field, so that if the desired isomorphism actually holds, then I would have a quick and elegant argument that $\mathscr{K}_V$ is a field. As it stands, I was able to verify that it was a field by checking all of the axioms one by one, but I am still curious to know whether the more elegant approach could be justified.
 A: Fields almost never have interesting natural gradings.  In fact, any $\mathbb{Z}$-graded field must be concentrated in degree $0$.  To prove this, suppose $K$ is a field with a grading which is not concentrated in degree $0$ and let $x\in K$ be some homogeneous element of nonzero degree (say, degree $d>0$).  Write $(1+x)^{-1}=\sum_{i=m}^n y_i$ where $y_i$ is homogeneous of degree $i$ and $y_m$ and $y_n$ are nonzero.  Then since $(1+x)y=1$, all the homogeneous parts of $(1+x)y$ in nonzero degrees must be $0$.  But the homogeneous part of $(1+x)y$ in degree $m$ is just $1\cdot y_m=y_m\neq 0$, which is a contradiction.  (More generally, a similar argument shows any unit in a $\mathbb{Z}$-graded integral domain must be homogeneous.)
In your example, the equation $$\mathscr{F}\left( 
\bigoplus\limits_{d=0}^{\infty} \frac{R_d}{I \cap R_d} \right) \cong \bigoplus\limits_{d=0}^{\infty} \mathscr{F}\left( \frac{R_d}{I \cap R_d} \right)$$ is very false.  First of all, the sets on the right-hand side have nontrivial intersection, so you can't take an internal direct sum of them.  For instance, if $f\in R_d$, then $f/f\in \mathscr{F}\left( \frac{R_d}{I \cap R_d} \right)$ but also $f/f=1/1\in \mathscr{F}\left( \frac{R_0}{I \cap R_0} \right)$.  Second of all, the sets on the right-hand side may not even be subgroups, since typically a sum of two fractions of homogeneous elements of degree $d$ will be a fraction of homogeneous elements of degree $2d$, not of degree $d$.  Third, the sum of the sets on the right-hand side is typically not the entire fraction field.  For instance, if $I=0$, the fraction field contains $x_i$ for $i=0,\dots,d$ but the sum of the sets on the right hand side does not.
