Let $R$ be a commutative ring (with unit) and let $R'$ be the $R$-algebra $R[x]/(x^n)$. Let $r$ be a positive integer. Let $V$ be a finitely generated $R'$-module which is free of finite rank over $R$.
Does $V$ decomposes as a direct sum $\bigoplus_{i=1}^s(x^{e_i})/(x^n)$ for integers $0\leq e_0 \leq e_1\leq ...\leq e_s<n$? If it exists, is this decomposition unique?
I believe that, if true, one shall use $x$ as a nilpotent $R$-linear morphism on $V$, but I am unable to complete the argument.
Many thanks!
Even the modified question is not true. Here is a counterexample: let $k$ be a field and consider $R = k[\varepsilon]$ with the relation $\varepsilon^2=0$. Let $R' = R[x]/(x^2)$, and consider the $R'$-module $V = R$, where $x$ acts by multiplication with $\varepsilon$.
If $V\cong \bigoplus_{i=1}^s(x^{e_i})/(x^2)$, for $0\le e_i \le 1$, then necessarily $s=1$ and $e_1 = 1$, because $V$ has rank $1$ as an $R$-module. But $x$ acts trivially on $(x)/(x^2)$, whereas it acts non-trivially on $V$. Hence $V$ does not admit such a decomposition.