# Show that every subfield of F is the set of fixed points under some power of $\phi$, $\phi(a) = a^p$

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 $$\phi$$?

• Any finite field arise as splitting field of polynomial $x^{p^{n}}-x$. Choose your $n$ accordingly. – Sunny Jun 10 at 11:02
• @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. – Ilan Aizelman WS Jun 10 at 11:11
• 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 – Ilan Aizelman WS Jun 10 at 11:13
• 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 $\phi(a) = a^{p^{m}}$ – Sunny Jun 10 at 11:16

$$F$$ 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.

• Amazing :) Thank you. – Ilan Aizelman WS Jun 10 at 12:43

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 $$\phi$$ ?

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