0
$\begingroup$

I've got a field extension $F_{q^n} / F_q$ of degree n and the Frobenius morphism $f$: $ x \to x^q$ associated to $F_{q^n} / F_q$. Let m be an integer dividing n.

In terms of $f$, what's an expression of the Frobenius morphism associated to $F_{q^n} / F_{q^m}$?

$\endgroup$
  • 1
    $\begingroup$ You know that the Frobenius of $F_{q^n}/F_{q^m}$ is given by $x \mapsto x^{q^n}$ by the same fact you used. Now you can compute easily, how this is related to $f$. The divisibility is only needed to ensure that $F_{q^m}$ is a subfield of $F_{q^n}$. $\endgroup$ – kesa Mar 1 at 9:50
  • 1
    $\begingroup$ $f^m = f\circ ... \circ f$ $\endgroup$ – Max Mar 1 at 9:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.