Jordan Can. Form of Frobenius map

[geor]

Hello all,

I am trying to solve this exercise here:

Let \phi denote the Frobenius map x |-> x^p on the finite field F_{p^n}. Determine the Jordan canonical form (over a field containing all the eigenvalues) for \phi considered as an F_p-linear transformation of the n-dimensional F_p-vector space F_{p^n}.

So, this is how I start:

Suppose that F_{p^n}=F_p(a1,a2,a3, ..., an) (those n elements will be powers of one element, but it doesn't matter). Now, since the Frobenius map is an isomorphism of F_{p^n} to itself, then \phi permutes a1, a2, ..., an.

Since a1, a2, ..., a3 form a basis of the n-dimensional F_p-vector space F_{p^n}, then the matrix of \phi in respect with that basis will be just a permutation matrix.

So the problem becomes equivalent with: "find the jordan canonical form of a permutation matrix".

Am I doing some obvious mistake here? Would the latter be something straightforward? I admit I can't see it...

Any help would be greatly appreciated.

 

[geor]

Okay, so this is a similar way that seems to work for me:

Suppose F_{p^n}=F_p(a), where a is a root of some irreducible polynomial over F_p of degree n.

Then,
a^(p^n-1), ..., a^{p^2}, a^p, a (= a^{p^n}) is a basis of the F_p-vector space F_p(a)

Then we notice that \phi(a^{p^i}) = a^{p^i+1}

So, in respect to the basis above, the matrix of \phi
becomes:

0 1 0 0 0 ... 0 0
0 0 1 0 0 ... 0 0
0 0 0 1 0 ... 0 0
.....
.....
0 0 0 0 0 ... 1 0
0 0 0 0 0 ... 0 1
1 0 0 0 0 ... 0 0

With some little effort, one can see that the characteristic polynomial
of this matrix is (t^n)-1.

That is, we have n distinct eigenvalues (all the nth roots of unity).

Thus, we will have n Jordan blocks, meaning that the Jordan canonical form
will be the diagonal matrix with the roots of unity in the diagonal

Could somebody please tell me if my arguments are correct or I miss something?
Or, if the above are correct, is there a simpler way to obtain this result?

Thanks a lot..