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In the graded context, if $R = K[x_1,...,x_n]$ where $K$ is a field, is a projective R-module a free R-module?

Eric Wofsey
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  • See also the related https://en.wikipedia.org/wiki/Quillen%E2%80%93Suslin_theorem – morrowmh Jan 31 '24 at 01:49
  • Does the "graded context" make this question different from the proposed duplicate? I can't see how... – rschwieb Jan 31 '24 at 16:21
  • @rschwieb: Yes it does, since a priori a graded module could be free as an (ungraded) module but not as a graded module (i.e., it has no basis consisting of homogeneous elements). – Eric Wofsey Jan 31 '24 at 16:23
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    This can't happen for graded modules that are bounded below (see here), but I don't know about the general case for modules over a polynomial ring. – Eric Wofsey Jan 31 '24 at 16:28
  • @EricWofsey I see. So maybe all one can say is that all projective graded modules are free as ungraded modules, but there's a gap after that point concluding that they are free as graded modules. Thank you for reopening it. I was about to based on your comment. – rschwieb Jan 31 '24 at 16:33
  • I know there are theorems that under very mild hypotheses any non finitely generated projective module is free. I don’t know the proofs of these off the top of my head though so I don’t know whether they still work in a graded context. – Eric Wofsey Jan 31 '24 at 16:45
  • Maybe Kaplansky's Theorem 2 (any projective module over a local ring is free) in his paper "Projective Modules" (Ann. Math. 68 (1958), 372-377) works in the graded case. – John Palmieri Jan 31 '24 at 18:30
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    By the way, you should specify the grading on the ring $R$. If you set $\deg x_j = 0$ for all $j$, then the "graded context" is not the same as if $\deg x_j = 1$. – John Palmieri Jan 31 '24 at 18:31

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