If $R$ is a PID, then we know that any projective module over $R$ is free. Is there any graded version of this? That is, let us call a graded commutative ring $R=\oplus _{g \in G} R_g$ a graded PID if every homogeneous ideal (i.e. graded ideal) of $R$ is generated by a homogeneous element (I don't know if this is a standard definition. If there is a standard definition, please let me know). If $M=\oplus_{g\in G} M_g$ is a projective graded module over $R$ (note that gradedprojective is equivalent to graded and projective), then obviously $M$ is free; my question would be: Is $M$ gradedfree, i.e., does there exist a basis for $M$ over $R$ consisting of homogeneous elements? If this is not true in general, does it hold with some extra condition on $M$ or $R$, like $M$ being finitely generated, or something else?
1 Answer
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This is true. More generally, we have the following:
Let $\psi\colon G\rightarrow H$ be an epimorphism of abelian groups. We denote by $\bullet_{[\psi]}$ the $\psi$coarsening functor from the category of $G$graded $R$modules to the category of $H$graded $R_{[\psi]}$modules. Let $R$ be a $G$graded ring, and let $M$ be a $G$graded $R$module. If $M$ is free, then so is $M_{[\psi]}$. If $R$ is principal, then the converse holds.
Note that the "free" and "principal" are understood in the graded sense.

$\begingroup$ Can you please add a proof or give a reference of a proof ? And are you using the same notion of "gradedprincipal" as I defined (is that the standard ? ) ? $\endgroup$– user111524Nov 1, 2017 at 14:16

$\begingroup$ I use the same definition as you; it is probably standard, and in any case reasonable. I do not know of a published proof (nor a published statement), but I will add some hints later on. $\endgroup$ Nov 1, 2017 at 15:34

$\begingroup$ Okay, please do add some elaborations later on . Thanks a lot. $\endgroup$– user111524Nov 1, 2017 at 15:36

$\begingroup$ Dear @users, if you disclose your identity, then I can send you some old notes of mine about these things by email. I do not have enough time now to write this down here (and it will probably take a very long time until I manage to bring it (in a far bigger framework) into publishable form...). $\endgroup$ Nov 1, 2017 at 18:58