Let $\sigma_p:\Bbb F_q \to \Bbb F_q$ be the Frobenius automorphism $\sigma_p(x)=x^p$ where $q=p^n$. Now viewing $V=\Bbb F_q$ as a vector space over $\Bbb F_p$ of dimension $n$.
Now how to prove that the matrix we get corresponding to Frobenius automorphism
- this matrix is diagonalizable over $\Bbb F_p$ iff $n|(p-1)$
- this matrix is diagonalizable over algebraic closure of $\Bbb F_p$ iff $(n,p)=1$
My try: I think the matrix of the linear transformation as the basis be $\{a,a^p,\cdots, a^{p^{n-1}}\}$. Then $$ \sigma_p(a)=a^p, \sigma_p(a^p)=a^{p^2}, \cdots, \sigma_p(a^{p^{n-1}})=a^{p^{n}}=a$$ where $V=\Bbb F_q=\Bbb F_p(a)$
$$(b_{ij})= \begin{pmatrix} 0 & 0 & \cdots & 0 & 1 \\ 1 & 0 & \cdots & 0 & 0\\ 0 & 1 & \cdots & 0 & 0\\ \vdots & \vdots& \ddots & \vdots &\vdots\\ 0 & 0 & \cdots & 1 & 0\\ \end{pmatrix}$$ is the matrix.
I can also calculate the characteristic polynomial of $(b_{ij})=x^n+(-1)^{2n-1}=X^n-1$= minimal polynomial [the minimal polynomial has to divide X^n -1 It cannot be a proper divisor cause otherwise, min polynomial would have deg less than $n$ Then the extension deg would be less than $n$]
Diagonalizable over F_p iff it splits into distinct linear factors over F_p iff F_p has an element of multiplicative order $n$ iff n divides the order of the cyclic multiplicative group, which is p-1.
How to prove that 2) "this matrix is diagonalizable over algebraic closure of $\Bbb F_p$ iff $(n,p)=1$"?
For 2). Since $B$ is a Frobenius block, each of its eigenspaces has dimension $1$.
Thus $B$ is diag. over $\overline{F_p}$ iff each of the (non-zero) roots $x_i$ of $x^n-1$ has multiplicity $1$ iff $n{x_i}^{n-1}\not= 0$ iff $(n,p)=1$.