# Characteristic polynomial for the frobenious map.

Consider the Frobenius automorphism $\sigma:\mathbb{F}_{p^{n}} \to \mathbb{F}_{p^n}$ defined by $\sigma(x)=x^{p},$ where $\mathbb{F}_{p^n}$ is a finite field with $p^n$ elements. Then it is clearly $\sigma$ is $\mathbb{F}_p$ linear map. My question is what is the characteristic polynomial of $\sigma$ ?

Since $\mathbb{F}_{p^n}$ is the splitting field of $X^{p^n}-X$ over $\mathbb{F}_p,$ we have $\sigma ^n-1=0,$ thus the minimal polynomial of $\sigma$ will divide $X^n-1$ in $\mathbb{F}_p[X].$ But I can't find characteristic polynomial of $\sigma.$ Help me. Thanks.

• Oops. I had also answered this earlier. One more reason to CWify. Doubtful whether I should initiate closing this as a duplicate now :-( – Jyrki Lahtonen Mar 4 '18 at 19:48
• I missed that ..i got my answer..thanks Jyrki Lahtonen – user371231 Mar 4 '18 at 19:51

• The space is $n$-dimensional, so the characteristic polynomial has degree $n$.
• The minimal polynomial $m(T)$ is a factor of the characteristic polynomial (as well as of $T^n-1$).
• If $m(T)$ has degree $m$, say, $m(T)=\sum_{i=0}^ma_iT^i\in\Bbb{F}_p[T]$, then all the $p^n$ elements of the bigger field are zeros of the polynomial $M(x)=\sum_{i=0}^ma_ix^{p^i}$ of degree $p^m$. Therefore $m\ge n$.