Let M be a R graded module with $M= \oplus M_i$. If M is noetherian then $M_i=0 $ for i << 0. My question is this, isn't $M_i = 0$ for all i >> 0 as well? If $(M_{n_i})_{i} \neq 0, n_i > 0$ then $M_{n_1} \nsubseteq M_{n_1} \oplus. .. M_{n_2} \nsubseteq M_{n_1} \oplus. .. M_{n_2} \oplus. .. M_{n_3} \nsubseteq. ..$ isn't a contradiction with ACC rule?
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9$\begingroup$ Is $R$ a $\mathbb{N}$-graded ring? Then is $M$ a graded $R$-module, compatible with $R$'s grading? If that's the case, then the various $M_i$ you are writing are not even $R$-modules, they are $R_0$-modules. $\endgroup$– Karl SchwedeCommented Jun 17, 2011 at 20:01
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$\begingroup$ K you are right. My mistake is that they are not R modules. $\endgroup$– markCommented Jun 17, 2011 at 20:50
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$\begingroup$ The mathoverflow bot just bumped this to the front page. Maybe you should just accept Mariano's answer below to prevent this from happening again. Really, the comment above answers the question, but mathoverflow wants you to accept an answer before it considers this question resolved. $\endgroup$– David WhiteCommented Jul 2, 2011 at 16:03
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$\begingroup$ Mathoverflow bot just bumped this up to the front page again. Perhaps we should close the question? $\endgroup$– David WhiteCommented Jul 30, 2011 at 14:20
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Pick your favorite noetherian graded ring $R$, and consider the free module $M=R$. Are you saying it must vanish in high degree?
answered Jun 17, 2011 at 20:06
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1$\begingroup$ This should have been a comment... $\endgroup$ Commented Jun 17, 2011 at 20:12