Vector space as simple $K[x]$-module I am trying to solve the problem:
Let $V$ be a vector space and $T$ a linear transformation $T:V \to V$. Let $(V,T)$ be a $K[x]$-module. 

Show that $(V,T)$ is simple if and only if $V$ is finite dimensional and $\mathcal X_T=m_T$ is irreducible (where $\mathcal X_T$ is the characteristic polynomial and $m_T$  is the minimal).

I am pretty lost with the exercise, I would appreciate any suggestions on how could I show both implications.
 A: If $V$ is finite dimensional then $V$ is a finitely generated torsion $K[x]$-module. Now apply the structure theorem and find that $V$ is a direct sum of cyclic $K[x]$-modules. Since $\chi_T$ is the product of all invariant factors and $m_T$ is the last of them, we can conclude that $V$ has only one invariant factor equal to $\chi_T=m_T$, so $V\simeq K[x]/(m_T)$. By hypothesis $m_T$ is irreducible, hence $V$ is a simple $k[x]$-module.
Conversely, if $V$ is a simple $K[x]$-module we get $V\simeq K[x]/(f)$, where $f$ is irreducible. Then $V$ is finitely dimensional (of dimension $\deg f$) and by the uniqueness of the structure theorem we get $f=\chi_T=m_T$.
A: Here's a basic argument without using any structure theorem (though I shall use that $K[x]$ is a PID).
First off, $K[x]$ does not have any simple modules that are infinite dimensional over $K$. Any element$~v\neq0$ of a simple module generates the whole module, and if $P.v=0$ for some nonzero $P\in K[x]$, the generated module would be of dimension $\deg P<\infty$, so an infinite dimensional simple module would have to be free of rank$~1$; however since $K[x]$ has nonzero proper ideals (is not a field), the free rank$~1$ module is not simple.
So henceforth suppose $\dim V=n<\infty$. For every $v\in V$ there is a lowest degree monic$~P_v\in K[x]$ with $P_v.v=0$, and $d=\deg P_v$ is the smallest value such that $x^d.v=T^d.v$ is a linear combination of $x^i.v$ for $0\leq i<d$, which values then span a $d$-dimensional $T$-stable subspace (a submodule). Also we know from the Cayley-Hamilton theorem that $\def\XT{\mathcal X_T}\XT.v=0$ for all$~v$, from which it follows that $P_v$ divides $\XT$.
Suppose $(V,T)$ is a simple module. Then for all $v\neq0$ one has $d=n$ whence $P_v=\XT$, and this clearly also is the minimal polynomial $m_T$ of$~T$. Moreover $m_T$ is irreducible: if one had $m_T=Q_1Q_2$ for non-constant polynomials, then $m_T[T]=0=Q_1[T]\circ Q_2[T]$ while (by minimality of $m_T$) neither of $Q_1[T],Q_2[T]$ is the zero endomorphism shows that $Q_2[T](V)\subseteq\ker(Q_1[T])$ is a nonzero proper submodule, which contradict simplicity.
Conversely suppose $m_T$ is irreducible of degree$~n$. Since every $P_v$ divides $m_T$ this forces $P_v=m_T$ whenever $v\neq0$, so $(V,T)$ is a simple module.
