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Let $R$ be a graded ring and $S$ a multiplicatively closed set of homogeneous elements of $R$. Prove that $R_S$ is a graded ring, where
$$(R_S )_n = \{ \frac{r}{s} ∈ R_S : r,s \textrm{ are homogeneous, and } \deg r − \deg s = n\}.$$

I tried to prove that $R_S$ is a direct sum of $(R_S)_n$ where $(R_S)_n$ is additive subgroup but I found problems, for example: when I try to prove it's additive subgroup I take $\frac{r}{s}$ and $\frac{r'}{s'}$ two elements in $(R_S)_n$ so we have $\frac{r}{s} + \frac{r'}{s'} = \frac{ar+br'}{u}$ where $u = as = bs'$

Now, $\deg \frac{ar+br'}{u} = \deg(ar) - \deg(as) = [\deg(a) \deg(r)] - [ \deg(a) \deg(s) ] $.

So how can I prove this equals $n$?

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    You should move this comment into the question which should address why it's on hold. As to the content, you may want to recheck what degree of a product is. – 2017-01-28

1 Answers 1

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$$\frac{r}{s} + \frac{r'}{s'} = \frac{(rs'+r's)}{ss'}.$$
By definition, $\deg r-\deg s=n=\deg r'-\deg s'$.

So $\deg rs'=\deg r+\deg s'=\deg r'+\deg s=\deg r's.$
Now $$\deg\frac{(rs'+r's)}{ss'}= \deg (rs'+r's)- \deg ss'= \deg rs' - \deg ss'=\\ \deg r+\deg s'-(\deg s+\deg s')= \deg r-deg s=n.$$

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    Thank u so much .. Ok what about the condition of graded $ (R_S)_n (R_S)_m \subseteq (R_S)_{nm} $ ?? Lock what I tried , Let $ \frac{r}{s} \in (R_S)_n $ and $ \frac{r'}{s'} \in (R_S)_m $ Now, $$ \deg\frac{rr'}{ss'} = \deg (rr') - \deg (ss') = \deg r + \deg r' - ( \deg s + \deg s' ) = ( \deg r - \deg s ) + ( \deg r' - \deg s' ) = n + m $$ this is hold if G is an additive group but what about the multiplicative group ?? Please correct my answer that I was wrong – 2017-01-29
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    **1-** Definition. Let G be an abelian group ( *written additively* ) and R a commutative ring. A G-grading for R is.... ---\\ **2-** By definition, u should prove : "$(R_S)_n (R_S)_m \subseteq (R_S)_{n+m}$", (not $(R_S)_n (R_S)_m \subseteq (R_S)_{nm}$.) – 2017-01-29
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    If (G , + ) is a group then the grading defined by : $ (R_S)_n (R_S)_m \subseteq (R_S)_{n+m} $ but if ( G , . ) is agroup then it replaced by $ (R_S)_n (R_S)_m \subseteq (R_S)_{n.m} $ it depends of the binary operation of G – 2017-01-29
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    Yes, but then you have two multiplication.... I suggest you think about the issue and read definitions carefully. then i welcome questions – 2017-01-29
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    This is E.C.Dad definition "fix a multiplicative group G with identity $1= 1_G$. A G-graded ring $R$ is a ring, also denoted by $R$, together with a direct sum decomposition: (1.1a) $R = \sum_{\sigma \in G } ^{.} R_{\sigma}$ (as additive groups) into additive subgroups $R_{\sigma} , \sigma \in G$, such that: $R_{\sigma} R_{\tau} \subseteq R_{\sigma \tau }$ " and Sorry for disturbing – 2017-01-29
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    you are welcome. with this definition (i.e. multiplicative group G), u should change all definitions Adapted with multiplication instead of addition, e.g. $(R_S )_n = \{ \frac{r}{s} ∈ R_S : r,s \textrm{ are homogeneous, and } \deg r . (\deg s)^{-1} = n\}.$ – 2017-01-29
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    Excuse me can u give me an example of $R_S$ ?? – 2017-02-02