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Hello!

This is a very short question:

Given a local graded Noetherian ring $R_{\bullet}$, is it true that any graded projective module over $R_{\bullet}$ is free?

In the ungraded case, this is true, but I do not know where the graded case is considered. Are there any references?

Thank you!

Hanno

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Martin- Graded local rings aren't local, so, no. –  Ben Webster Nov 17 '10 at 13:55
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2 Answers

up vote 8 down vote accepted

There is a proof of the ungraded result on page 10 of Matsumura's "Commutative Ring Theory", and you can insert gradings everywhere in a straightforward way to prove the graded result.

The main reason why you cannot just appeal to the ungraded result is as follows: a local graded ring has (by definition) precisely one homogeneous ideal that is maximal among proper homogeneous ideals, but typically there will be many maximal inhomogeneous ideals, so the underlying ungraded ring will not be local.

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Dear Neil, the OP's hypothesis that the graded ring be Noetherian is not necessary, is it? –  Georges Elencwajg Nov 17 '10 at 14:52
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@Georges: you are correct. (In stable homotopy theory we care about the case $R_*=BP_*=\mathbb{Z}_{(p)}[v_1,v_2,\dotsc]$ with $|v_i|=2(p^i-1)$, and of course that one is not Noetherian.) –  Neil Strickland Nov 17 '10 at 15:26
    
Thank you for the example, Neil. Unfortunately, I don't know what stable homotopy theory is. But the ring you mention reminds me of the cohomology of infinite projective space . Does the $B$ in $BP_\ast$ stand for "classifying space" ? –  Georges Elencwajg Nov 17 '10 at 21:48
    
@Georges: no, BP stands for Brown-Peterson here. See for example this book: goo.gl/snmbm or Wikipedia: en.wikipedia.org/wiki/Brown%E2%80%93Peterson_cohomology –  Neil Strickland Nov 17 '10 at 22:04
    
Ah, I was really off base with my suggestion! Thanks for the bibliography. –  Georges Elencwajg Nov 17 '10 at 22:24
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Neil's answer is certainly fine, but let me give another reference which says this explicitly.

Proposition 1.5.15(d) (on page 36) of Bruns and Herzog, "Cohen-Macaulay Rings".

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