Let $q$ be a prime power. Let $\mathbb{F}_q$ be the finite field with $q$ elements. Then $\mathbb{F}_{q^n}$ is a field extension of $\mathbb{F}_q$ of degree $n$ and can be considered as an $n$-dimensional vector space $V$ over $\mathbb{F}_q$. Now consider the action of $GL(V)$ on $V$. Any element $x$ in $\mathbb{F}_{q^n}$ acts on itself by multiplication and hence defines an element in $GL(V)$; we denote this element as $g_x$. Suppose now an element $h\in GL(V)$ commutes with $g_x$ for some $x\neq0,1\in \mathbb{F}_{q^n}$. Does it imply that $h=g_y$ for some $y\in \mathbb{F}_{q^n}$ as well? Thanks.
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$\begingroup$ If $\Bbb{F}_{q}(x)$ is smaller than $\Bbb{F}_{q^n}$ then not necessarily (try with $x=1$). If $\Bbb{F}_{q}(x)=\Bbb{F}_{q^n}$ then yes if the minimal polynomial of $h$ is irreducible ? $\endgroup$– reunsCommented Mar 3, 2020 at 6:16
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$\begingroup$ Sorry I should have said x not equal to 1 or 0. $\endgroup$– DapsCommented Mar 3, 2020 at 6:57
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$\begingroup$ I see. The answer should be false in general. But when would it be true? $\endgroup$– DapsCommented Mar 3, 2020 at 7:01
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2 Answers
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Let $F$ be any field and $F<E$ a finite field extension. Fix $x\in E^*$ and consider the multiplication operator $g_x\in \text{GL}_F(E)$, the group of invertible $F$-linear transformations of $E$. The centralizer of $g_x$ could be naturally identified with the subgroup $\text{GL}_{F[x]}(E)<\text{GL}_F(E)$, where $F[x]<E$ is the subfield of $E$ generated by $x$ over $F$. We thus have $$ E^* \simeq \text{GL}_E(E)<\text{GL}_{F[x]}(E)<\text{GL}_F(E).$$ The inclusion $\text{GL}_E(E)<\text{GL}_{F[x]}(E)$ is an equation iff $F[x]=E$. To see the "only if" part it is enough to recall that $\text{GL}_n$ is never commutative for $n\geq 2$. Thus every element in the centralizer of $g_x$ is of the form $g_y$ for $y\in E^*$ iff $E=F[x]$, that is $x$ generates $E$ over $F$.
Specializing to $F=\mathbb{F}_q$ and $E=\mathbb{F}_{q^n}$, we get that every element in the centralizer of $g_x$ is of the form $g_y$ for $y\in \mathbb{F}_{q^n}^*$ iff $x$ generates $\mathbb{F}_{q^n}$ over $\mathbb{F}_q$.
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$\begingroup$ Uri is right the result is immediate because "$h\in End_{\Bbb{F}_q}(\Bbb{F}_{q^n})$ commutes with $g_x$" is the definition of "$h$ is $\Bbb{F}_q[x]$ linear" thus it is defined by its action on a $\Bbb{F}_q[x]$ basis of $\Bbb{F}_{q^n}$. $\endgroup$– reunsCommented Mar 3, 2020 at 8:56
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Let $k=\Bbb{F}_{q},K=\Bbb{F}_{q^n}$, $x\in K$ and $g_x\in M_n(k)\cong End_k(K)$ the matrix of the multiplication by $x$.
If $k(x)$ is smaller than $K$ then every $k(x)$-linear endomorphism of $K$ commutes with $x$, most of them are not of the form $g_x$.
Assume that $K=k(x)$
$g_x$ has an eigenvector $v\in K^n$ with eigenvalue $x$. Since $g_x\in M_n(k)$ then $$g_x(v^{q^l}) =(g_x(v))^{q^l}=(xv)^{q^l}=x^{q^l}v^{q^l}$$ Since the $x^{q^l}$ are distinct the $v^{q^l}$ must be distinct and we have diagonalized $g_x$ $$g_x(\sum_{l=1}^n c_l v^{q^l})=\sum_{l=1}^n c_l x^{q^l} v^{q^l}$$
If $h\in M_n(k)$ commutes with $g_x$ then $h$ has an eigenvector which is an eigenvector of $g_x$, wlog we can assume it is $v$ so that $h(v)=av$ for some $a \in F$ where $F$ is the splitting field of $h$'s minimal polynomial. As before we obtain the diagonalization from $$h(v)=av \implies h(v^{q^l}) = a^{q^l} v^{q^l}, \qquad h(\sum_{l=1}^n c_l v^{q^l})=\sum_{l=1}^n c_l a^{q^l} v^{q^l}$$ $v^{q^n}= v$ implies that $a^{q^n}=a$ thus $a\in K$. Whence $a = f(x)$ for some $f\in k[T]$ and $$h = f(g_x)=g_a$$
