in the book that is reading, has the following excerpt:
"Each such $f$ has $\deg(f)$ distinct nonzero roots in E ,and each root $\beta$ of $f$ has as its characteristic polynomial over $\mathbb{F}_q$ the polynomial $f(x)^\tfrac{s}{\deg(f)} = x^s -c_{1}x^{s-1}...+(-1)^{s}c_s$."
The polynomial in question is monic and irreducible in $ \mathbb{F}_q $ and $E$ is an extension of this finite field. I believe that the characteristic polynomial related to a root $\beta$ of $f$ is the characteristic polynomial of the linear transformation matrix $ T: E \to E$ defined by $ T(x) = x\beta $. My question is: Why is this characteristic polynomial just $ f (x)^\tfrac {s}{\deg (f)} $?
Ps: $\deg(f)$ divides $s$.
Note that the minimal polynomial of the matrix $T$ is just the minimal polynomial of $\beta$ over $E$, which is $f$. So, if the roots of $f$ are $\beta_1,\dots,\beta_d$, the characteristic polynomial of $T$ must be $(x-\beta_1)^{e_1}\dots(x-\beta_d)^{e_d}$ for some exponents $e_1,\dots,e_d\geq 1$. Since the characteristic polynomial has coefficients in $\mathbb{F}_q$, the exponents $e_1,\dots,e_d$ must all be equal (for instance, you can see this by Galois theory: $Gal(E/\mathbb{F}_q)$ permutes $\beta_1,\dots,\beta_d$ transitively while fixing the characteristic polynomial). That means the characteristic polynomial is actually $f^e$ for some exponent $e$. Since the characteristic polynomial must have degree $[E:\mathbb{F}_q]$ (which I assume is what you mean by $s$), the exponent $e$ must be $[E:\mathbb{F}_q]/\deg(f)$.