Construction of Projective Varieties Question

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? Mumford text

The Red Book of Varieties and Schemes, Second Expanded Edition — kfriend, Jul 13 '15 at 19:23

Pavel Čoupek · Answer 1 · 2015-07-13T19:21:23.923

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.

This makes perfect sense, but say we have $f_1/g_1, f_2/g_2 \in R$, say the degree of $f_1, g_1$ is $m$, and that of $f_2,g_2$ is $r$. Then say $m \leq r$ -- do we multiply $f_1$ and $g_1$ by a power of some variable before adding $f_1/g_1$ and $f_2/g_2$? — kfriend, Jul 13 '15 at 19:15
@kfriend: If $\deg f_1=\deg g_1=m$ and $\deg f_2=\deg g_2=r$, i.e. the fractions $f_i/g_i$ are of degree $0$, then $f_1/g_1+f_2/g_2=(f_1g_2+f_2g_1)/(g_1g_2)$ has still both numerator and denominator homogeneous and of the same degree, i.e. $n+r$ (provided that the numerator is nonzero, of course). — Pavel Čoupek, Jul 13 '15 at 19:19

score 1 · Answer 2 · answered Jul 13 '15 at 19:07

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 $X$.

Construction of Projective Varieties Question

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