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Let $R$ be an $\mathbb{N}^m$-graded ring and let $M$ be an $\mathbb{N}^m$-graded module over $R$. Supposing that $M$ is free as an $R$-module, does there necessarily exist an $R$-basis homogeneous with respect to the $\mathbb{N}^m$-grading?

I am happy to assume that $R$ is noetherian and $M$ is finitely generated if that matters.

It seems to me intuitively that there should, and I've read a few papers that seem to treat this as a standard fact, at least in some specific noetherian / f.g. contexts, but it's not obvious to me how to prove it, and I can't find it by googling. Looking forward to your thoughts.

Ben Blum-Smith
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    See http://math.stackexchange.com/questions/557402/graded-free-is-stronger-than-graded-and-free. The answers there discuss the case $m=1$, but the exact same arguments work for any $m$. – Eric Wofsey Feb 17 '17 at 05:49
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    In particular, you do need one additional hypothesis, which is that every projective $R_0$-module is free. Indeed, if $P$ is a projective but not free $R_0$-module and $R=R_0$, you can find $Q$ such that $P\oplus Q$ is free and then take $M=P\oplus Q$ graded so that $P$ and $Q$ are in different degrees. – Eric Wofsey Feb 17 '17 at 05:54
  • @EricWofsey - thanks. I think that constitutes an answer. (I.e. if you put it in an answer, I can accept.) Aside: once you linked that question of course I remembered it! I'm all up in the comments! – Ben Blum-Smith Feb 17 '17 at 06:19

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This is not always true. For instance, if $R=R_0$ and $P$ is a projective $R$-module that is not free, let $Q$ be such that $P\oplus Q$ is free. If you take $M=P\oplus Q$, graded so that $P$ and $Q$ are in different degrees, then $M$ has no multihomogeneous basis.

On the other hand, if every projective $R_0$-module is free, then the answer is yes. To sketch the proof, $M/R_+M$ will be a free $R_0$-module, and hence each graded piece of it is free since every projective $R_0$-module is free. Lifting a multihomogeneous basis of $M/R_+M$ to multihomogeneous elements $M$ then gives a multihomogeneous basis of $M$: they generate $M$ by the graded Nakayama lemma, and there can be no relations between them since there are no relations mod $R_+$ and the submodule of relations is a direct summand since $M$ is a free $R$-module. For more details (as well as an alternate argument that works if $M$ is finitely generated), see the answers to "Graded free" is stronger than "graded and free"?.

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Eric Wofsey
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