Irreducible characteristic polynomial and invariant subspaces

Question

Let $K$ be a field and let $V$ be a finite-dimensional vector space over $K$. Let $\alpha$ be an endomorphism of $V$, with irreducible characteristic polynomial. I'm trying to show that there is no proper non-zero subspace $W \subset V$ with $\alpha(W) \subset W$.

So far, I've consider $V$ as a $K[X]$-module by $$ K[X] \times V \to V : (\sum \lambda_i X^i,v) \mapsto \sum \lambda_i \alpha^i(v) $$ And then noticed that the $\alpha$-invariant subspaces of $V$ are exactly the $K[X]$-submodules of $V$. Furthermore, since $K[X]$ is a PID and $V$ finite dimensional, we have by the structure theorem $$ V \cong K[X]/p_1^{e_1} \oplus \ldots \oplus K[X]/p_k^{e_k}, $$ with $p_1,\ldots,p_k \in K[X]$ irreducible. Also since the characteristic polynomial of $\alpha$, $p_\alpha$ is irreducible, it is equal to the minimum polynomial, and we have that $K[X]/(p_\alpha)$ is a field, which is expect we need to use somewhere.

Any hints on how to proceed?

Answer 1

This is very simple. Suppose $W$ is an invariant subspace for$~\alpha$, then one can restrict $\alpha$ to$~W$ to obtain an endomorphism $\alpha|_W$ of$~W$. Then the characteristic polynomial of $\alpha|_W$ divides the characteristic polynomial of $\alpha$, contradicting (if $0\subset W\subset V$) the assumption that the latter is irreducible.

The fact that the characteristic polynomial of the restriction to an invariant subspace$~W$ divides that of the full endomorphism is immediate from the definition of the characteristic polynomial, using a basis of $V$ that extends a basis of$~W$ to get a matrix for the endomorphism: this matrix is block upper triangular, with a matrix for the restricted endomorphism as upper left diagonal block.