version of Nakayama's lemma for group rings I have the following rather complicated setting and I would like to know if something that resembles Nakayama's lemma can be proved in this setting, but I can make no progress with it:
Let $G$ be a finite group and let $K$ be a field of characteristic $0$. Assume that $M$ is a module over the polynomial ring $K[X]$, that $M$ is finitely generated as a $K$-module and also that there exists an action of the group $G$ on $M$, so that $M$ becomes a module the group ring $(K[X])[G]$. 
I know in addition that the annihilator of $M$ in $K[X]$ is of the form $(X-c)^d$ for some $c \in K$, $d>0$ and that the set of elements in $M$ annihilated by $(X-c)$ is a cyclic $K[G]$-module (also the action of $X$ and $G$ commute). Does it follow that $M$ is cyclic as a $(K[X])[G]$-module?
 A: Maybe there's a more elegant way to prove this, but one way to do it is to just classify all such modules.  First, we may assume $c=0$.  By Artin-Wedderburn, $K[G]$ is isomorphic to a product $\prod M_{n_i}(D_i)$ of matrix rings over division rings, and then $K[G][X]\cong \prod M_{n_i}(D_i[X])$.   Any $K[G][X]$-module $M$ then splits as a product $\prod M_i$ of modules over each $M_{n_i}(D_i[X])$, and $M$ is cyclic iff each $M_i$ is cyclic.
So, it suffices to prove the following.  Suppose $D$ is a division ring and $M$ is a finitely generated $M_n(D[X])$-module that is annihilated by some power of $X$ such $N=\{a\in M:Xa=0\}$ is a cyclic $M_n(D)$-module.  Then $M$ is cyclic.
To prove this, note first that any finitely generated $D[X]$-module which is annihilated by some power of $X$ is isomorphic to a direct sum of modules of the form $D[X]/(X^i)$.  (There several ways to prove this; some commonly used proofs of this fact over fields (which amounts to Jordan normal form for nilpotent matrices) work just as well over division rings.  Another approach is to show that $D[X]/(X^i)$ is injective as a module over itself using Baer's criterion (the only ideals in $D[X]/(X^i)$ are those generated by powers of $X$) and then use this to repeatedly peel of direct summands of the form $D[X]/(X^i)$ where $i$ is minimal such that $X^i$ annihilates the module.)  By the Morita equivalence of $D[X]$ and $M_n(D[X])$, similarly any finitely generated $M_n(D(X))$-module which is annihilated by some power of $X$ is a direct sum of modules of the form $D^n\otimes_D D[X]/(X^i)$.
So, $M$ is a direct sum of modules of the form $D^n\otimes_D D[X]/(X^i)$, and then $N$ will be a direct sum of copies of $D^n$, one for each summand in $M$.  An $M_n(D)$-module is cyclic iff it is a direct sum of at most $n$ copies of $D^n$, so if $N$ is cyclic, there are at most $n$ summands in $M$.  But then $M$ is a quotient of a direct sum of at most $n$ copies of $D^n\otimes_D D[X]$, and so $M$ is a cyclic $M_n(D[X])$-module.
