Frobenius Automorphism as a linear map Let $\phi(x) = x^p$ be the Frobenius automorphism on $\mathbb F_{p^n}$. We can view $\mathbb F_{p^n}$ as an $n$-dimensional vector space over $\mathbb F_p$. In this case, $\phi$ is a linear transformation. What are its characteristic and minimal polynomials? What are its Jordan and rational canonical forms? I know that $\phi$ satisfies $x^n - 1$, so its minimal and characteristic polynomials must divide this one. But, I can't seem to make much progress beyond this. Any suggestions?
 A: If the minimal polynomial $m(T)=\sum_{i=0}^k a_iT^i\in F_p[T]$ of the Frobenius automorphism were of degree $k<n$, then all the $p^n$ elements of $F_{p^n}$ should be zeros of the polynomial
$$
f(x)=\sum_{i=0}^ka_ix^{p^i}
$$
of degree $p^k$. But a polynomial of degree $p^k$ can have at most $p^k$ zeros in a field.
Thus the minimal polynomial and the characteristic polynomial are both equal to $T^n-1$.
Edit: Sorry, I missed the question about the canonical forms. The Jordan canonical form is somewhat interesting. Write $n=mp^t$, where $t\ge0$ is an integer, and $m$ is
coprime to $p$. The zeros of the minimal polynomial
$$
T^n-1=(T^m-1)^{p^t}
$$
are then roots of unity of order dividing $m$. All of those occur as eigenvalues of $\phi$, but usually only in a bigger field (as we are interested in JCF, we were more or lessing willing to go to an algebraic closure anyway). All these eigenvalues are zeros of both the minimal and characteristic polynomial of multiplicity $p^t$, so we get $m$ Jordan  blocks of size $p^t$.
As an example consider the field $F_{16}$. The minimal polynomial of $\phi$ 
is then $T^4+1=(T+1)^4$. Thus $\lambda=1$ is the only eigenvalue of $\phi$. The kernels of the (composition) powers $(\phi-id)^k$, $k=1,2,3,4$ are the zeros of the polynomials $x^2+x$, $x^4+x$, $x^8+x^4+x^2+x$ and $x^{16}+x$ respectively. These can be identified as the subfield $F_2$, the subfield $F_4$, the kernel of the trace map $Tr:F_{16}\rightarrow F_2$, and the entire $F_{16}$ respectively.
