Let $R$ be a graded ring. There are two ways to take the localization of $R$.
Let $\mathfrak{p}$ be a homogeneous prime ideal, $T$ be the set of all homogenous elements of $R\setminus \mathfrak{p}$. Then $R_{(\mathfrak{p})}$, the subring of $T^{-1}R$ consisting of all $\dfrac{f}{g}$ where $f$ and $g$ are homogeneous of the same degree, is called homogeneous localization of $R$.
Let $R$ be a graded ring, $S\subset R$ is a multiplicative closed subset of $R$. For any $f\in R, g\in S$ define the degree of $\dfrac{f}{g}$ to be $\deg f-\deg g$. Then it is not hard to check that this is well defined. The localization $S^{-1}R$ is a graded ring.
My question are the following:
-
What is the difference between these two localizations?
-
What are the applications of them in higher commutative algebra ?
Best Answer
To supplement the comments by QiL:
An arbitrary element of $R$ has no degree. Only the homogeneous elements have some degree. You seem to be confused with the notion of degree of polynomials. But here, there are no polynomials at all.
If $S \subseteq R$ is a homogeneous submonoid, we can grade the usual localization $S^{-1} R=R_S$ (i.e. of the underlying rings) as follows: The homogeneous elements of degree $d$ are those of the form $r/s$, where $r \in R$, $s \in S$ are homogeneous elements such that the degree of $r$ equals $d$ plus the degree of $s$. As always the homogeneous elements of degree $0$ constitute a subring. This is usually called $R_{(S)}$. When $\mathfrak{p}$ is some homogeneous prime ideal, then the set of homogeneous elements $s \in R$ such that $s \notin \mathfrak{p}$ is an example for $S$. Therefore, we have a graded ring $R_{\mathfrak{p}}$ and the subring of elements of degree $0$ is called $R_{(\mathfrak{p})}$.
So there is no real difference: Rather one construction is the special case of the other (properly phrased).