$K[X]$ module and characteristic polynomial Let $(V, \phi)$ a $K[X]$-module with $V$ a finite-dimensional vector space.
Do you think that $ V \cong \dfrac{K[X]}{(\chi_{\phi})}$ with $\chi_{\phi}$ characteristic polynomial of $\phi$?
How to prove it? (I know the invariant factor decomposition if needed).
Thanks and regards.
 A: Presumably, $K$ is a field, $\phi\in \operatorname{End}_K(V)$, and the action of $K[X]$ on $V$ is given by
$$f(X)\cdot v=f(\phi)(v)$$
for all $f(X)\in K[X]$ and $v\in V$.  The answer is no.  Take $V=K^2$ and $\phi=0$.  Then
$$\chi_\phi(X)=X^2$$
so 
$$\frac{K[X]}{\big(\chi_\phi(X)\big)}=\frac{K[X]}{(X^2)}$$
is a $2$-dimensional indecomposable $K[X]$-module, but
$$V\cong K\oplus K$$
is decomposable (where each copy of $K$ is the $K[X]$-module on which $X$ acts by $0$).  
However, if each irreducible factor of $\chi_\phi(X)$ occurs with multiplicity $1$, then the claim is true.  More generally, let $\Sigma$ be the set of all monic irreducible polynomials in $K[X]$.  For each $p(X)\in \Sigma$, let
$$V^p=\big\{v\in V\big|\exists k\in\Bbb N,\ \big(p(X)\big)^k\cdot v=0\big\}.$$
Then, $V^p$ is a $K[X]$-submodule of $V$ and
$$V=\bigoplus_{p(X)\in \Sigma}V^p.$$
It follows that $V\cong \frac{K[X]}{\big(\chi_\phi(X)\big)}$ if and only if $V^p$ is an indecomposable $K[X]$-module for every irreducible factor $p(X)$ of $\chi_\phi(X)$, or equivalently, the minimal polynomial $\mu_\phi(X)$ of $\phi$ satisfies $$\mu_\phi=\chi_\phi.$$
