Let $E$ be an $n$-dimensional vector space on a field $\mathbb K$, $u$ an endomorphism of $E$, and $\pi_u$ its minimal polynomial of degree $d$.
I have this exercise :
Show that $u$ is diagonalisable if and only if $\mathbb K [u]$ is isomorphic (as a $\mathbb K$-algebra) to $\mathbb K^d$.
If $u$ is diagonalisable, then its minimal polynomial can be written $\pi_u(X) = (X-\lambda_1) \dots (X- \lambda_d)$ with distinct $\lambda_i \in \mathbb K$.
And then :
$$ \mathbb K [u] \simeq \mathbb K[X]/(\pi_u) \simeq \mathbb K[X]/(X- \lambda_1) \times \dots \times \mathbb K[X]/(X- \lambda_d) \simeq \mathbb K^d $$
I'm struggling to prove the other part of the assertion. If $\pi_u$ is not square-free then there would a nilpotent element $\mathbb K [u]$, but what if $\pi_u$ does not split in $\mathbb K$ ?
Hint. If $\pi$ doesn't split then $\mathbb{K}[u]$ contains a proper field extension of $\mathbb{K}$. OTOH, if an element of $\mathbb{K}^d$ is algebraic over $\mathbb{K}$ then all its components are roots of its minimal polynomial.