Proof: (11) We saw φn=1, so since [Fpn : Fp]=n we note φ is an n×n matrix and the invariant factors of φ all divide xn−1. Assume there are m>1 invariant factors, so h1(x) | h2(x) are invariant factors and write h2(x)=f(x)h1(x). Observe elements of the form ⨁mk=1ak where f(x) | ak when k=2, and ak=0 when k>2. These are all in the kernel of h1(x), and yet there are more than pa of them, implying there are more than pa solutions to the polynomial of degree pa represented by the linear transformation h1(x), contradiction. Therefore the sole invariant factor is one of degree n dividing xn−1, necessarily xn−1 itself and the Jordan canonical form is the matrix with 1s along the subdiagonal and a 1 in 1,n.
(12) Let n=pkm for p⧸|m. We see xn−1=(xm−1)pk in Fpn[x], so that by previous investigations the invariant factors are some powers of (x−α) where α is a power of an nth 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 mth primitive root in Fpn, iff m | pn−1) of φ is the matrix with Jordan blocks (x−ζm)pk.
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