Rational Canonical Form over finite field Let $\phi: F_{p^n} \to F_{p^n}$ be a linear operator (where $F_{p^n}$ is considered as vector space over $F_p$) given by $x \mapsto x^p.$  In a previous exercise we prove that $\phi^n = I$ and that no lower power of $\phi$ is the identity.  
Then we are asked to give the rational canonical form of $\phi$ over $F_p.$  I am struggling with figuring out what the minimum polynomial is, I suspect that it is $x^n-1$ but I haven't been able to prove it.  Here is what I've tried so far:
We know the minimum polynomial $m(x)$ divides $x^n-1.$ Let $n= p^kr$ where $r$ is relatively prime to $p.$  Then 
$$x^n-1 = x^{p^kr} - 1 = (x^r-1)^{p^k}$$
so $m(x) | (x^r -1)^{p^k}.$  Now I know that
$$m(x) = (x^r-1)^l$$
where $l>p^{k-1}$ since if $l \leq p^{k-1}$ then we could multiply $m(x)$ be the appropriate factor to obtain 
$$m(x)p(x) = (x^r - 1)^{p^{k-1}} = x^{p^{k-1}} - 1.$$ 
and since $\phi$ satisfies $m(x)$ it satisfies the right side which is a contradiction to $n$ being the smallest power of $\phi$ which is the identity.  This is where I run out of ideas and cant seem to show $m(x) = x^n-1.$  Perhaps thats not the minimum polynomial but then it seems the possibilities for invariant factors becomes wildy complicated.
Could someone people provide some guidance?
Thanks!
 A: We have that $\,\phi^n=I\Longrightarrow \phi^n-I=0\Longrightarrow\,$  the polynomial $\,f(x)=x^n-1\,$ is the characteristic pol. of $\,\phi\,$ since it is monic and has degree equal to $\,n=\dim\Bbb F_{p^n}\,$ over $\,\Bbb F_p\,$ (and $\,\phi\,$ vanishes at it, of course).
Now, for any $\,k<n\,$ let us look at the polynomial
$$t(x):=\sum_{j=0}^ka_jx^{p^j}\in\Bbb F_p[x]$$
If $\,\phi\,$ is a zero of $\,t(x)\,$ , then for all $\,u\in\Bbb F_{p^n}\,$ . 
$$0=t(\phi)=\sum_{j=0}^ka_k\phi^{p^j}(u)=\phi\left(\sum_{j=0}^ka_ku^{p^j}\right)\iff \sum_{j=0}^ka_ku^{p^j}=0$$
the last equality following since the Frobeniux map is an automorphism, and then all the $\,p^n\,$elements of $\,\Bbb F_{p^n}\,$ are roots of a pol. of degree $\,p^k<p^n\,$ , and this is impossible over a field!
It follows the minimal polynomial of $\,\phi\,$ must be of degree $\,n\,$ and it thus must be $\,f(x)\,$ itself.
A: An easy induction argument proves that for $v\in F_{p^n},$  $\varphi^r(v) = v^{p^r}.$ 
Now let $f(x) = a_0 + a_1x + \cdots a_kx^k$ be a polynomial over $F_{p}$ of degree $k<n$ and 
suppose for a contradiction that $\varphi$ satisfies $f.$  Then for all $v\in F_{p^n}$ we have
$$0 = f(\varphi) = \sum_{j=0}^k a_j\phi^j(v) = \sum_{j=0}^k a_j v^{p^j}$$
But then all $p^n$ elements of $F_{p^n}$ satisfy a polynomial of degree $p^k < p^n$ which is impossible.  Thus $\varphi$ cannot satisfy any polynomial of degree less than $n.$  This prove that $x^n-1$ is the minimum polynomial of $\varphi$ and since it is also the characteristic polynomial we have that $x^n-1$ is the only invariant factor giving the following rational canonical form,
$$
\left[\begin{array}{ccccc}
0 & 0 & \ldots & 0 & 1 \\
1 & 0 & \ldots & 0 & 0\\
0 & 1 & \ldots & 0 & 0\\
\vdots &\vdots &\ddots & \vdots & \vdots \\
0 & 0 & \ldots & 1 & 0
\end{array}\right].
$$
