Frobenius morphism associated to field extension

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 $$F_{q^n} / F_{q^m}$$?

• 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 $F_{q^n}$. – kesa Mar 1 at 9:50
• $f^m = f\circ ... \circ f$ – Max Mar 1 at 9:55