No nontrivial subspaces implies irreducibility of characteristic polynomial Let $k$ be a field and let $V$ be a $k$-vector space. Let $T:V\to V$ be a linear transformation, and suppose $T$ has no invariant subspaces in $V$ other than $0$ and $V$ itself. Does it follow that the characteristic polynomial of $T$ is irreducible?
It's easy to show that the minimal polynomial $\mu_T$ must be irreducible: if $\mu_T=fg$ Is a nontrivial factorization then $W=ker(f(T))$ is a nontrivial subspace of $V$. Indeed if $W=0$ then $\mu_T$ divides $g$, and if $W=V$ then the minimal polynomial divides $f$, a contradiction.
 A: I think I got it. The linear transformation $T$ makes $V$ into a $k[x]$-module. This is a PID so it has a decomposition into invariant factors. The matrix of $T$ with respect to this decomposition will be in block diagonal form, where the blocks are the companion matrices for the invariant factor polynomials. The characteristic polynomial is then the product of the invariant factors. If there are no nontrivial invariant subspaces in $V$, the characteristic polynomial is therefore equal to the minimal polynomial, which we know to be irreducible. 
A: I did not read your answer; yet , in fact, it's an equivalence.
i) If there is a proper $T$-invariant space, then there is a basis s.t. the matrix of $T$ is in the form $\begin{pmatrix}A&B\\0&C\end{pmatrix}$, and therefore, its characteristic polynomial $\chi_T=\chi_A\chi_C$ is reducible.
ii) Conversely, assume that $\chi_T$ is reducible; then, there is a vector $v\in V\setminus \{0\}$ that is not $T$-cyclic (I wrote a post on this site about that, but I cannot find it). Then $span(v,Tv,T^2v,\cdots)$ is a proper $T$-invariant space.
