I have met this recently in my abstract algebra course dealing with modules over PIDs and we are dealing with cyclic modules at the moment, the problem I am having difficulty with is as follows:
Let V be a finite dimensional vector space over a field F, we define a linear operator $ T : V \to V $ and the ring of polynomials $ D = \mathbb{F}[x] $ and we define M to be the D module which T induces on V i.e. $ (p(x) \cdot v = p(T)v ) $. We assume also that T is invertible and that M is a cyclic D-module. We define $ M_1 $ to be the D module induced by $ T^{-1} $ (the inverse of T) and we are asked to show that $ M_1 $ is also a cyclic D-module.
My problem is here I know the ring is a principal ideal domain and that M is obviously finitely generated as a cyclic D module, but I cannot seem to show that the module induced by the inverse $ T^{-1} $ is also cyclic and I do not know how to even proceed, I appreciate all help on this
$M$ is a cyclic $D$-module is equivalent to saying that there exists $x\in M$ such that $(x,T(x),...T^m(x)$ is a basis of $M$. This implies that for every $y\in M$, there exists $c_0,...,c_n\in F$ such that $T^n(y)=c_0x+c_1T(x)+...+c_nT^n(x)$. We deduce that $y=T^{-n}(T^n(y))=T^{-n}(c_0x+c_1T(x)+...+c_nT^n(x))=c_nx+c_{n-1}T^{-1}(x)+...+c_0T^{-n}(x)$. This is equivalent to saying that $M_1$ is generated by $x$ thus is a cyclic $D$-module.