Exercise 10.3.4 (Fixed field for Galois group of finite field)

Question

What is the fixed field of $\langle 	heta angle \subseteq Gal(\mathbb{F}_{p^n}:\mathbb{F}_p)$ ?

Hassaan Naeem · Accepted Answer

extbf{Solution:}
We have that $Gal(\mathbb{F}_{p^n}:\mathbb{F}_p)$ is generated by the Frobenius automorphism of $\mathbb{F}_{p^n}$ which is just $	heta: g \mapsto g^p$,
hence we have $\{ 	heta, 	heta^2, ... \}$, the respective outputs from the map are then $g^p, g^{p^2}, ... , g^{p^n}$.
We also have that by definition $Fix(\langle 	heta angle) = \{ x \in X: 	heta x = x \ \forall 	heta \in \langle 	heta angle\}$.
If we let $m = |\langle 	heta angle|$ be the order, 
then we need to solve $g^{p^m} - g = 0 \ \forall g \in \mathbb{F}_{p^n}$. But we know that $\mathbb{F}_{p^n}$ is the splitting field of $g^{p^m} - g$.
Therefore $Fix(\langle 	heta angle) = \mathbb{F}_{p^m}$.
This also makes sense the the fundamental theorem since $\mathbb{F}_{p^n}:\mathbb{F}_{p^m}:\mathbb{F}_{p}$.

Richard Ganaye · Answer

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\begin{proof} 
If $	heta(x) = x$, then $	heta^k(x) = x$ for all integers $k$. Therefore
$$\mathrm{Fix}(\langle 	heta angle) = \{x \in \F_{p^n} \mid 	heta(x) = x\} = \{x \in \F_{p^n} \mid x^p = x\}.$$

Every $x \in \F_p$ satisfies $	heta(x) = x^p = x$, so $\F_p \subseteq \mathrm{Fix}(\langle 	heta angle) $.

Conversely, the polynomial $t^p -t$ has at most $p$ roots in the field $\F_{p^n} $. This shows that
$$|\mathrm{Fix}(\langle 	heta angle) | = |\{x \in \F_{p^n} \mid x^p = x\}| \leq p = |\F_p|.$$
Since $\F_p \subseteq \mathrm{Fix}(\langle 	heta angle)$, $ |\F_p| \leq |\mathrm{Fix}(\langle 	heta angle) | $, thus $p = |\F_p| = |\mathrm{Fix}(\langle 	heta angle) |$. Therefore the inclusion $\F_p \subseteq \mathrm{Fix}(\langle 	heta angle)$  between finite sets is an equality:
$$\mathrm{Fix}(\langle 	heta angle) = \F_p.$$
\end{proof}

Note: We know from Proposition 10.3.3 that the order of $	heta$ is $n$. If we use Galois theory on finite fields, the Fundamental Theorem gives for $H =\langle  	heta angle$,
$$[\F_{p^n} : \mathrm{Fix}(\langle 	heta angle)] = |\langle 	heta angle | = n = [\F_{p^n} : \F_p],$$
where $\F_p \subseteq \mathrm{Fix}(\langle 	heta angle)$, because $\F_p$ is the prime field of $M$. This shows anew that
$$\mathrm{Fix}(\langle 	heta angle) = \F_p.$$

Exercise 10.3.4 (Fixed field for Galois group of finite field)

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Exercise 10.3.4 (Fixed field for Galois group of finite field)

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