Let $F$ be a field and let $p(x)\in F[x]$ be a monic irreducible of degree $\deg p>1$. Let $r$ be any one root of $p$. Then $F(r)\simeq F[x]/p(x)F[x]$ and $[F(r):F]=\deg p$. Now $F(r)$ is a $\deg p$-dimensional $F$-vector space, and its left action on itself makes sense of the embedding $F(r)\subset \mathrm{gl}(F(r),F)$. I think $F(r)$ is self-centralizing in $\mathrm{gl}(F(r),F)$, i.e., $[T,F(r)]=0$ implies $T\in F(r)$ for all $T\in \mathrm{gl}(F(r),F)$.
Question: Is it true that $F(r)$ is even (Lie) self-normalizing in $\mathrm{gl}(F(r),F)$, i.e., does $[T,F(r)]\subset F(r)$ imply $T\in F(r)$ for all $T\in\mathrm{gl}(F(r),F)$?
Brute force calculations seem to confirm this for $F=\mathbb{R}$.
Edit: In view of my own answer below, I have to rephrase my question. Can anyone, please, provide a reference where this fact can be found? I don't want to prove things that are well known. Thank you.
$[T,F(r)]\subset F(r)$ implies in particular $Tr=rT+s$ for some $s\in F(r)$. Now $$ Tr^2=(rT+s)r=r(rT+s)+sr=r^2T+2rs. $$ One can prove by induction that $Tr^m=r^mT+mr^{m-1}s$ for all $m\in\mathbb{N}$. This in turn implies that $TP(r)=P(r)T+P'(r)s$ for every $P(x)\in F[x]$. Take now $Tp(r)=p(r)T+p'(r)s$. Then $p(r)=0$ forces $p'(r)s=0$. But $p'(r)\neq0$ because $p$ is the minimal polynomial for $r$, thus $s=0$. This means, $[T,r]=0$ and therefore $[T,F(r)]=0$, which by the self-centralizing property gives $T\in F(r)$.