Construction of Projective Varieties Question I am struggling in understanding Mumford's construction of Projective Varieties. In the image I uploaded here, Are we to understand each $R_n$ as $M_n/P_n$, where $M_n:=${homogeneous polynomials in $k[x_1,...,x_n]$ of degree n} and $P_n:=${homogeneous polynomials in $P$ of degree n}? In which case, how does one consider $K(X)$, as Mumford defines it, as a ring?

 A: 1) Yes, your idea is correct:
If $S=\bigoplus_n S_n$ is a graded ring and $I \subseteq S$ is a homoreneous ideal, it basically means that $I=\bigoplus_{n}I_n$, where each $I_n \subseteq S_n$ is a subgroup. Then we have
$$R=S/I=\left(\bigoplus_n S_n\right)/\left(\bigoplus_n I_n\right)\simeq \bigoplus_n S_n/I_n, $$
which shows that the quotient ring $R$ has a natural grading, i.e. $S_n/I_n$ is considered to be the group of homogeneous elements of degree $n$.
2) Now, since we had a graded prime ideal, the resulting quotient is a integral domain, and by localizing at the set $S$ of all homogeneous elements (notice that it is multiplicatively closed), one gets some ring $S^{-1}R$, not necessary a field. However, since all the denominators are homogeneous, it makes sense to define degree as follows: Whenever $f, g \in R$ are homogeneous, define $$\deg \left(\frac{f}{g}\right)=\deg f - \deg g.$$
You can check that this  defines grading on $S^{-1}R$. Now take the subring consisting of all these fractions of degree zero, and you end up precisely with the field Mumford is describing in the last line. 
3) By the way, observe that the result is indeed a field.
A: The abelian group $R_k$ is the image of $k[X_0,\dots,X_n]_k$ under  the canonical  ring morphism $$k[X_0,\dots,X_n]\to R=\frac {k[X_0,\dots,X_n]}{P}$$.
The notation $k[X_0,\dots,X_n]_k$ stands for the set of homogeneous polynomials of degree $k$ (I have modified Mumford's poor notation where $n$ stands for two unrelated numbers) .
Noether's standard isomorphism shows that $R_k$ is canonically isomorphic to $\frac {k[X_0,\dots,X_n]_k}{P\cap k[X_0,\dots,X_n]_k}$, just as you correctly supposed.
Anyway $k(X)$ is certainly a ring and actually even a field: the rational function field of $X$. 
