Let $K$ be a finite field of Characteristic $p$ and let $F$ be the Frobenius automprhism on $K$. Show that $F^k$ has at most $p^k$ fixed points in $K$.

$F: K \to K$ is an Automorphism, so $$F^k(x)=F(x^k)=x^{k^p}=x^{p^k}.$$

If $x$ is a fixed point of $F^k$ then, $$F^k(x)=x^{p^k}=x \iff x^{p^k}-x=0.$$ The polynomial $f(x)=x^{p^k}-x$ is of degree $p^k$ and has therefore at most $p^k$ roots. Does this proof work?


Your intermediate steps are incorrect, that is, in general $F^{k}(x) \ne F(x^{k})$, and $x^{k^{p}} \ne x^{p^{k}}$.

Rather, prove by induction on $k$ that $F^{k}(x) = x^{p^{k}}$.

Hint: $F^{2}(x) = F(F(x)) = F(x^{p}) = F(x)^{p} = (x^{p})^{p} = x^{p^{2}}$.

  • $\begingroup$ That doesn't hold in fields ? $\endgroup$ – badatmath May 16 '18 at 10:43
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    $\begingroup$ $(x^a)^b=x^{a\cdot b}=(x^b)^a\neq x^{(a^b)}\neq x^{(b^a)},$ whereas $(x^a)^a=x^{a\cdot a}=x^{(a^2)}$ $\endgroup$ – emma May 16 '18 at 12:22

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