When a linear automorphism of $\mathbb{F}_{q}^{n}$ has its minimal polynomial with greatest possible degree? Hello comrades I´m looking for the answer to a quesion, so I'll share it with you.
If $T:\mathbb{F}_{q}^{n}\rightarrow \mathbb{F}_{q}^{n}$ is a linear automorphism, and $r \in \mathbb{Z^{+}}$ is its order, i.e. $r$ is the smalest non negative number such that $T^r=Id$, then $T$ satisfies the polynomial $P(x)=x^r-1$, so the minimal polyomial $m_{T}(x)$ of $T$ divides $P(x)$. 
Question: Under what conditions does $m_{T}(x)=P(x)$? (the obvious answer is that they are equal if and only if they have the same degree, but I would like to listen other options).
Thanks for your help.
 A: I would approach this by applying the following tricks/ideas/observations.


*

*If $r>n$ then the answer is obviously that $T^r-1$ is not the minimal polynomial. The minimal polynomial is also a factor of the characteristic polynomial, and that has degree $n$.

*Next I would wish really hard that $\gcd(r,q)=1$ (IOW that the characteristic of your field is not a factor of $r$). In that case all the roots of $x^r-1$ are simple, and consequently it has no repeated factors in $\Bbb{F}_q[x]$. We can find the irreducible factors of $x^r-1$ as follows. We first construct the field $E=\Bbb{F}_q(\zeta)$ where $\zeta$ is a primitive root of unity of order $r$ (possibly $E$ is a proper extension of $\Bbb{F}_q$). So in the polynomial ring $E[x]$ we have
$$
x^r-1=\prod_{j=0}^{r-1}(x-\zeta^j),
$$
and we need to determine which factors we need to combine to get irreducible factors in $\Bbb{F}_q[x]$. The answer to that is given by Galois theory. We know that if $p(x)\mid x^r-1$ is a factor with coefficients in $\Bbb{F}_q$, then the set of zeros of $p(x)$ is stable under the Frobenius automorphism $F:z\mapsto z^q$. So the factor that has $\zeta^j$ as a zero, has as its zeros also the elements $F^\ell(\zeta^j)=\zeta^{q^\ell j}$. Here we can calculate $q^\ell j$ modulo $r$ because $\zeta^r=1$. Therefore the zeros that are lumped together form a so called cyclotomic coset
$$
C(r,j,q)=\{q^\ell j\bmod r\mid \ell=0,1,2,\ldots\}\subset \{0,1,\ldots,r-1\}.
$$
For example, if $r=13$ and $q=3$, then $C(13,2,3)=\{2,3\cdot2=6,3^2\cdot2=18\equiv5\}$ (and it stops there because $2\cdot 3^3=54\equiv 2$).
Anyway, the set $[r]:=\{0,1,\ldots,r-1\}$ is partitioned into cyclotomic cosets $C_t$ somehow, say $[r]=C_1\cup C_2\cup\cdots\cup C_s$, and the irreducible factors of $x^r-1$ in the ring $\Bbb{F}_q[x]$ are
$$
x^r-1=\prod_{i=1}^sp_i(x),
$$
where
$$
p_i(x):=\prod_{j\in C_i}(x-\alpha^j).
$$
Finally, we can conclude that $m_T(x)=x^r-1$ if and only if none of the polynomials $p^{(j)}(x):=\prod_{i\neq j}p_i(x)=(x^r-1)/p_j(x)$ annihilate $T$. This is the analogue to the result from elementary group theory: an element $g$ satisfying the equation $g^n=1$ is of order $n$ if and only if $g^{n/p_i}\neq1$ for all prime factors $p_i\mid n$.

*Finally if the characteristic $p$ is a factor of $r$, then we may have repeated factors. Write $r=r'p^a$ with $\gcd(r',p)=1$. In this case
$$
x^r-1=(x^{r'}-1)^{p^a}.
$$
In this case you need to find the factorization of $x^{r'}-1$ by the above procedure to find the irreducible factors of $x^r-1$ in $\Bbb{F}_q[x]$ (they all occur with multiplicity $p^a$). The rest is done as in part 2. Observe that you do need to check that the minimal polynomial is not a factor of any of $(x^r-1)/p_i(x)$ rather than $(x^{r'}-1)/p_i(x)$.


I don't know if this leads to anything that will be useful to you. If we would view the matrix of $T$ over an algebraic closure, then part 2 deals with the semisimple case of only diagonal Jordan blocks, but in part 3 we may see non-diagonal Jordan blocks.
