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Let $R$ be a commutative ring and let $G$ be a group (not necessarily finite). Let $\mathcal A$ be the category of (left) $R[G]$-modules that are finitely generated over $R$; one might call this the category of finite dimensional representations of $G$ over $R$. Consider the following statement $P(R,G)$:

  • For all $M\in\mathcal A$, there is an epimorphism $N\to M$ in $\mathcal A$ where $N$ is free as an $R$-module.

For what conditions on $R$ and $G$ is $P(R,G)$ true?

For instance, if $R$ is a field, then $P(R,G)$ is true for all $G$, since any $M\in \mathcal A$ is already free as an $R$-module. Similarly, if $G$ is finite, then $P(R,G)$ is true for all $R$, since any $M\in\mathcal A$ will be a quotient of a finitely generated free $R[G]$-module, which is in turn finitely generated and free as an $R$-module (if I'm not messing anything up in my head...).

Thus, the question is only interesting in the hard case: representations of infinite groups over a ring that is not a field.

Partial progress on a question I asked recently indicates that $P(R,G)$ is true if $R$ is Noetherian and $G$ is free; at the very least, it seems to be true (I haven't checked the details) when $G=\mathbb Z$ and $R$ is Noetherian local, and when $G$ is free and $R$ is a PID.

Andrea B.
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  • It should be true if $R$ is Noetherian and $G$ is finitely generated by other other question, what do you think? My gut feeling is that it doesn't hold in general. – Martin Brandenburg Sep 13 '24 at 19:56
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    Much more general question: Let $\mathcal{C}$ be a Grothendieck abelian category. Let $\mathcal{D} \subseteq \mathcal{C}$ be a full generating additive subcategory consisting of projective objects (in your case, $\mathcal{C}$ is a module category, $\mathcal{D}$ consists of free modules). For which index categories $\mathcal{I}$ is the full subcategory $[\mathcal{I},\mathcal{C}] \subseteq [\mathcal{I},\mathcal{D}]$ also a generator? (In your case, $\mathcal{I}$ is a one-object groupoid, but I don't know if this restriction is useful, hence my comment.) – Martin Brandenburg Sep 14 '24 at 00:35

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