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Is this a equivalent definition for a graded ring?

Let $R$ be a ring. We say that $R$ is graded if there are subgroups $R_n, n\in \mathbb{Z}$ of $R$ such that given $x\in R$, there are $x_n \in R_n$ such that $x=\sum x_n$ where the sum is finite, i.e., there are only a finite number of non-zero $x_n$ such that this expression is unique.

The definition I am familiar is:

$R$ is graded if $R=\bigoplus R_n$ as abelian groups and $R_nR_m\subset R_{m+n}$

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You have now changed your question, incorporating my answer and the numerous comments in the new version. You shouldn't do that because now the answer seems not to address the question. Moreover everything has been answered so you should try to understand on your own what you have been told and review the definition of direct sum. – Georges Elencwajg Mar 15 '12 at 19:10
Apologies, that was not what I meant, I'll left the question in original form. – Jr. Mar 15 '12 at 19:17
I just want to understand why some books use the first definition, I think that the function $\(x_n\)_{n \in \mathbb{Z}} \rightarrow \sum x_n$ is bijective – Jr. Mar 15 '12 at 19:23
Thanks, Jr: I have deleted the edit mentioning your former change in my answer. – Georges Elencwajg Mar 15 '12 at 19:50
up vote 4 down vote accepted

No, your definition is not equivalent to the usual one since it says nothing about the compatibility of the grading with the multiplicative structure of the ring.

For example let $k$ be a field and write $R=k[X]=\oplus R_n$ with $R_n=kX^{2n}\oplus kX^{2n+1}$.
Then $R$ is graded in your sense but not in the usual sense because $X^3\in R_1$ and yet $X^3\cdot X^3 \notin R_2$

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and what about I if add the hipothesis: $R_nR_m\subset R_{n+m}$, in my definition,is it equivalent now? – Jr. Mar 15 '12 at 16:45
@Jr. Yes $ \\\\\ $ – Georges Elencwajg Mar 15 '12 at 17:18
why?How do I show that? – Jr. Mar 15 '12 at 17:38
Your definition was equivalent to just $R=\bigoplus R_n$ – Georges Elencwajg Mar 15 '12 at 17:45
the elements of $\bigoplus R_n$ are of the form $\(x_n\)_{n \in \mathbb{Z}}$, aren't they? – Jr. Mar 15 '12 at 18:08

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