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Let $R$ be a graded ring. There are two ways to take the localization of $R$.

Let $\mathfrak{p}$ be a homogenous 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 homogenous of the same degree, is called homogenous 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 ?

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    I think in the second case, $S$ must contain only homogeneous elements and one defines the degree of $f/g$ only when $f$ is homogeneous. Otherwise you don't get a graded ring $S^{-1}R$ by you construction.2012-05-20
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    @Qil :No, I do not think so. We can think $\dfrac{f}{g}$ as a polynomial of degree degf-deg g.2012-05-20
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    Can you show that the set of elements of a given degree (including the element $0$) is stable by addition ?2012-05-20
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    The second one is more general than the first one. In fact, the first one picks special multiplicative set, i.e. the homogeneous elements in $R - \mathfrak{p}$, and also it only take the degree $0$ part of the graded ring $S^{-1}R$. Also two more remarks: $S^{-1}R$ is graded in the sense that it has both positive and negative degrees, unlike $R$ is usually just non-negatively graded; the homogeneous localization defined in the first way is in fact a local ring, not a graded ring.2013-10-12

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