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.

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

Combine the following pieces:

  • 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 $m\ge n$.
  • $\begingroup$ I have used the same argument here, and more recently here. Switching to CW, because I strongly believe that one should not seek to get paid twice for the same work even when the compensation is in the form of internet ego points. $\endgroup$ – Jyrki Lahtonen Mar 4 '18 at 19:41

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