Let $F$ be an endomorphism of $V$ over some field $\mathbb{k}$ and characteristic polynomial of $F$ is the same as its minimal polynomial. Is that true that $Z_F \overset{\operatorname{def}}{=} \{G \in End(V) | FG = GF\}$ is the same as $\mathbb{k}[F]$?
I know that $V $ is the same as $\mathbb{k}[t]$-module $\mathbb{k}[t]/(f_1) \oplus \dots \oplus \mathbb{k}[t]/(f_k)$ with operation of multiplication by $[t]$ being operator F, and because $\mu_F = \chi_F$, all $f_i$ are mutually prime.
Multiplication by some polynomial $P$ commutes with multiplication by $[t]$, so $\mathbb{k}[t] \subseteq Z_F$
How do i proove that there are none operators that commute with $F$ and do not represent multiplication by some polynomial in $\mathbb{k}[t]$-module? I need to somehow use that $F$ is diagonazable, and can't think of a way to use it.