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.

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 $f=\chi_T=m_T$.
