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.
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"?.