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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. –  user18119 May 20 '12 at 6:46
@Qil :No, I do not think so. We can think $\dfrac{f}{g}$ as a polynomial of degree degf-deg g. –  Arsenaler May 20 '12 at 8:06
Can you show that the set of elements of a given degree (including the element $0$) is stable by addition ? –  user18119 May 20 '12 at 10:26
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. –  Secret Math Oct 12 '13 at 2:11

1 Answer 1

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).

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