$\newcommand{\Z}{\mathbb{Z}}$ I'm interested in a classification of indecomposable modules over the $\Z/p^2\Z$-group-ring $R$ of the cyclic group $C_p$ of order $p$: $$ R = \Z/p^2\Z[C_p] = \Z/p^2\Z[T]/(T^p - 1) $$
(I actually want to replace $p^2$ with $p^n$, but I figured this would be a good place to start.)
I was hoping it would be the case that all indecomposables over $R$ were cyclic $R$-modules, which is true when we replace $\Z/p^2\Z$ with the field $\Z/p\Z$, but I don't think this is the case. Consider the $R$-module $M$ with underlying abelian group $\Z/p\Z \times \Z/p^2\Z$ where $T$ acts by the matrix \begin{equation*}\begin{pmatrix} 1 & 0 \\ p & 1+p \end{pmatrix}.\end{equation*}
By direct computation, one can verify that the cyclic submodule generated by every* nonzero element has order $p^2$. (*The element $(0,p)$ generates a submodule of order $p$, but $p\Z/p^2\Z$ is not a direct summand of $M$ even as an abelian group.) In particular, no single element generates $M$ (so it is not cyclic), but no two elements can be used to break $M$ up into a direct sum, since any direct sum must involve a submodule of order $p$.
I spoke with someone who gave me the following argument for the cyclicity of indecomposables (which I now think is false, but I don't know in which of the first two steps problem is as I'm not comfortable thinking about projective covers over this ring):
- $M$ has a projective cover as an $R$-module.
- The projective cover of an indecomposable module is indecomposable.
- $R$ is local so projectives are free.
- Free + indecomposable implies rank 1.
- Thus every indecomposable is a quotient of $R$.
Does anyone know how I might modify this argument to something correct (but necessarily weaker)? Or maybe someone can point me toward the classification I'm looking for or a reference that will help me find it?