Canonical forms of the Fröbenius Endomorphism (13.6.11-12)

Dummit and Foote Abstract Algebra, section 13.6, exercises 11-12:

11-12. Let $φ$ denote the Fröbenius map $$φ:\mathbb{F}_{p^n}→\mathbb{F}_{p^n}$$$$x↦x^p$$Find the rational and Jordan (when it exists) canonical form of $φ$.

Proof: (11) We saw $φ^n=1$, so since $[\mathbb{F}_{p^n}~:~\mathbb{F}_p]=n$ we note $φ$ is an $n \times n$ matrix and the invariant factors of $φ$ all divide $x^n-1$. Assume there are $m > 1$ invariant factors, so $h_1(x)~|~h_2(x)$ are invariant factors and write $h_2(x)=f(x)h_1(x)$. Observe elements of the form $\bigoplus_{k=1}^m a_k$ where $f(x)~|~a_k$ when $k=2$, and $a_k=0$ when $k > 2$. These are all in the kernel of $h_1(x)$, and yet there are more than $p^a$ of them, implying there are more than $p^a$ solutions to the polynomial of degree $p^a$ represented by the linear transformation $h_1(x)$, contradiction. Therefore the sole invariant factor is one of degree $n$ dividing $x^n-1$, necessarily $x^n-1$ itself and the Jordan canonical form is the matrix with $1$s along the subdiagonal and a $1$ in $1,n$.

(12) Let $n=p^km$ for $p \not | m$. We see $x^n-1=(x^m-1)^{p^k}$ in $\mathbb{F}_{p^n}[x]$, so that by previous investigations the invariant factors are some powers of $(x-\alpha)$ where $\alpha$ is a power of an $n^{th}$ primitive root of unity, and moreover there are $m$ such distinct roots. By the same reason as above, and since the degree of the product of the polynomials must be $n$, the Jordan canonical form (when it exists, i.e. when there is an $m^{th}$ primitive root in $\mathbb{F}_{p^n}$, iff $m~|~p^n-1$) of $φ$ is the matrix with Jordan blocks $(x-\zeta_{m})^{p^k}$.