Given a field $k$ with characteristic $p$ and a finite cyclic $p$-group $G$ of order $p^a$, it is well-known that all the indecomposable representations of $kG$ are given by mapping a generator $x$ of $G$ to the Jordan matrix $J_s\in M_s(k)$ with all eigenvalues one for $1\leq s\leq p^a$. If we replace $k$ by a commutative ring $R$ with characteristic $p$, then what are the indecomposable representations of $RG$? Is it the same as in the situation of $kG$?
Indecomposable representations for group ring $RG$ over commutative ring $R$ with characteristic $p$
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1$\begingroup$ You may want to add some conditions on $R$. Otherwise all the weird complications $R$-modules can have, will potentially be present in $RG$-modules as well. For example: If you have non-trivial idempotents in $R$, you have "unexpected" central idempotents ins $RG$ that further decompose every module that you may have expected to be indecomposable. $\endgroup$– Johannes HahnCommented Aug 25, 2021 at 14:53
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3$\begingroup$ Your question is at least as hard as classifying the indecomposable modules of an arbitrary commutative ring $R$ but also, since $R[G]$ is a commutative ring, is a special case of classifying the indecomposable modules of an arbitrary ring $R$. So this is more a question about rings than groups. Even if you put conditions on $R$, as Johannes suggests, note that $R[G]=R[u]/(u^{p^n}-1)= R[x]/ ( (1+x)^{p^n}-1 ) = R[x]/ (x^{p^n})$ so $R[G]$ is fairly similar to $R$, so you will need the conditions to be very stringent for this to be more of a question about $G$ than a question about $R$. $\endgroup$– Will SawinCommented Aug 25, 2021 at 14:59
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$\begingroup$ @Will Sawin Maybe I should first think the case that $R$ is UFD with characteristic $p$. $\endgroup$– Master GangCommented Aug 26, 2021 at 0:19
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$\begingroup$ @MasterGang Maybe a PID or a Dedekind domain would be a better start. I think Mare's example shows $R= K[x,y]$ is impossible, so UFDs of dimension at least two are hopeless. $\endgroup$– Will SawinCommented Aug 26, 2021 at 0:29
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One might ask whether one can classify all indecomposable $RG$-modules when one knows all indecomposable $R$-modules but the example $R=K[x,y]/(x^2,y^2)$ shows that this is not possible.
The answer will in general be that one can not classify the indecomposable representations as those algebras are most often of "wild" representation type (see for example https://www.tandfonline.com/doi/abs/10.1080/00927879108824178 ). If $G$ is a non-trivial cyclic group and $R$ is a representation-infinite finite dimensional $K$-algebra (for example $R=K[x,y]/(x^2,y^2)$) then $RG \cong R \otimes_K K[x]/(x^n)$ for some $n$ and this will have wild representation type since the quiver of $RG$ will have at least three loops.
In the example of $R=K[x,y]/(x^2,y^2)$ one can classify all indecomposable $R$-modules but for $RG$ this is a wild problem already.
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$\begingroup$ Thanks. You have shown me that it is really impossible to classify all indecomposable representations under my hypothesis. $\endgroup$ Commented Aug 26, 2021 at 0:15