Proof of normal basis theorem for finite fields

596 Views Asked by At

I'm trying to study this proof of normal basis theorem for finite fields to avoid to go through the formal study of modules.

At the very end of the proof they claim that since the minimal polynomial of $\pi$ the Frobenius map is $X^n - 1$ and it coincides with the minimal polynomial of certain element $v \in F$ in the field $F$, then the set of elements $\alpha,\pi\alpha,\ldots, \pi^{n-1}\alpha$ are linearly independent.

Why is the case?

1

There are 1 best solutions below

2
On BEST ANSWER

Suppose they are not linearly independent, then $f(\pi)(\alpha)=0$ for some polynomial $f$ of degree $<n-1$. We know that $\{1,\alpha,\dots,\alpha^{n-1}\}$ is a basis for $F$ over $P$. As $\pi:F\to F$ is an automorphism, $f(\pi)(\alpha)=0$ implies $f(\pi)=0$, but $\pi$ has degree $n$, which is impossible.