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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.

cawabynga
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    No, it's not true, take e.g. the identity. It is true if $F$ has distinct eigenvalues, or equivalently if the minimal polynomial of $F$ has degree $\dim V$. – Qiaochu Yuan Jan 29 '23 at 20:26
  • @QiaochuYuan, i made a mistake when writing a question, and initially i meant to ask about the case when $\mu_F = \chi_F$. Can you give me a hint on how it is proved? Tx in advance – cawabynga Jan 29 '23 at 21:38
  • If $charpoly(F)$ is separable then it is easy: given $GF=FG$, for simplicity assume that $k$ is algebraically closed, replacing $G$ by $G+cI$ we can assume that $\det(G)\ne 0$, for $Fv_j=\lambda_jv_j$ we get that $F Gv_j = G Fv_j= \lambda_j G v_j $ so $G v_j$ is an eigenvector with eigenvalue $\lambda_j$, ie. $G v_j = c_j v_j$.

    Taking $P\in k[X]$ such that $P(\lambda_j) = c_j$ we get that $G=P(F)$.

     – reuns Jan 29 '23 at 21:52
  • Not sure what tools you have at your disposal, but in that case $V$ is cyclic with respect to $F$, and it can be proved that the commutant of $F$ equals the bicommutant of $F$, which by a standard result equals the set of polynomials in $F$. – blargoner Jan 30 '23 at 02:58
  • @caw It is not necessarily true that $F$ is diagonalizable; this only holds if $\mu_F = \chi_F$ has no repeated factors. – Ben Grossmann Jan 30 '23 at 16:46
  • @caw Yes, it is true. For one proof of this fact using Jordan normal form, see Matrix Analysis by Horn and Johnson, Theorem 3.2.4.2 – Ben Grossmann Jan 30 '23 at 16:47
  • @caw See this post – Ben Grossmann Jan 30 '23 at 16:59

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