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Given: $p$ is prime. and $F$ is a finite field of characteristic $p$.

I've already shown that the map defined by $\phi(a) = a^p \quad \forall a \in F$ is an automorphism. (Also called: Frobenius automorphism)

My question: How to show that every subfield of F is the set of fixed points under some power of $\phi$?

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  • $\begingroup$ Any finite field arise as splitting field of polynomial $x^{p^{n}}-x$. Choose your $n$ accordingly. $\endgroup$ – Sunny Jun 10 at 11:02
  • $\begingroup$ @SunnyRathore You mean.. $\operatorname{Fix}(F) = I(F,\operatorname{id}_K) = \{x\in K : x^{p^n} = x\}$ is a subfield of $K$ for every field $K$ of characteristic $p$. Since $\operatorname{Fix}(F)$ is the set of zeros of the polynomial $P(X) = X^{p^n} - X$, it follows that $\operatorname{Fix}(F)$ has at most $\deg P = p^n$ elements. $\endgroup$ – Ilan Aizelman WS Jun 10 at 11:11
  • $\begingroup$ Where $I$ is defined: For any two fields $K,L$ and field homomorphisms $\varphi,\psi \colon K \to L$, the set $$I(\varphi,\psi) = \{ x\in K : \varphi(x) = \psi(x)\}$$ is a subfield of $K$. This is a generalisation of the often-used fact that the set of fixed points of a field endomorphism is a subfield of its domain [and this special case is what is used here; nevertheless, the more general fact is not harder to prove]. @SunnyRathore $\endgroup$ – Ilan Aizelman WS Jun 10 at 11:13
  • $\begingroup$ I simply mean that if $K$ is subfield of $F$ of cardinality $p^{m}$, where size of $F$ is $p^{n}$ and $m $ divides $n$, then choose $\phi(a) = a^{p^{m}}$ $\endgroup$ – Sunny Jun 10 at 11:16
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$F$ contains $\Bbb{F}_p$ so it is a $\Bbb{F}_p$-vector space, of finite dimension $n$, so it has $p^n$ elements. Thus $F^\times$ is a group with $p^n-1$ elements and any $a \in F$ is a root of $x(x^{p^n-1}-1) = x^{p^n}-x \in \Bbb{F}_p[x]$. But this polynomial has at most $p^n$ roots in the field $F$, thus $F$ is exactly the splitting field of $x^{p^n}-x \in \Bbb{F}_p[x]$ the latter being the unique field with $p^n$ elements.

Which means $F = \{ a \in \overline{\Bbb{F}}_p, \phi^n(a) = a\}$.

Also note if $F^\times$ is not cyclic, let $e < p^n-1$ its exponent, any $a \in F$ would be a root of $x^{e+1}-x$ having at most $e+1$ roots contradicting that $F$ has $p^n$ elements. Whence $F^\times$ is cyclic.

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  • $\begingroup$ Amazing :) Thank you. $\endgroup$ – Ilan Aizelman WS Jun 10 at 12:43
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Let $k$ be a subfield. What is its characteristic ? What is therefore its cardinal ? The cardinal of $k^\times$ ?

What does Lagrange's theorem then tell you about its elements ? And then about the elements of $k=k^\times\cup\{0\}$ ? How does that relate to powers of $\phi$ ?

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  • $\begingroup$ Well, I'm not sure :( $\endgroup$ – Ilan Aizelman WS Jun 10 at 11:20
  • $\begingroup$ Which part is a problem ? $\endgroup$ – Max Jun 10 at 11:34
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    $\begingroup$ I wrote an answer because I think your hints aren't enough $\endgroup$ – reuns Jun 10 at 11:36

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