1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ That doesn't hold in fields ? $\endgroup$ – badatmath May 16 '18 at 10:43
  • 1
    $\begingroup$ $(x^a)^b=x^{a\cdot b}=(x^b)^a\neq x^{(a^b)}\neq x^{(b^a)},$ whereas $(x^a)^a=x^{a\cdot a}=x^{(a^2)}$ $\endgroup$ – emma May 16 '18 at 12:22

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.