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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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2 Answers 2

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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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    $\begingroup$ Dear Neil, the OP's hypothesis that the graded ring be Noetherian is not necessary, is it? $\endgroup$
    – Georges Elencwajg
    Commented Nov 17, 2010 at 14:52
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    $\begingroup$ @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.) $\endgroup$
    – Neil Strickland
    Commented Nov 17, 2010 at 15:26
  • $\begingroup$ 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" ? $\endgroup$
    – Georges Elencwajg
    Commented Nov 17, 2010 at 21:48
  • $\begingroup$ @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 $\endgroup$
    – Neil Strickland
    Commented Nov 17, 2010 at 22:04
  • $\begingroup$ Ah, I was really off base with my suggestion! Thanks for the bibliography. $\endgroup$
    – Georges Elencwajg
    Commented Nov 17, 2010 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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