Let $F$ be a field of order $p^{23}$;$p$ is a prime .Let $\phi: F\to F$ be the field automorphism of $F$ sending $a$ to $a^p$. Let $K := \{a\in F:\phi (a)=a\}$.
What is the value of $[K : \Bbb F_p]$?
Clearly $\{0,1\}\subset K$.But to find $[K : \Bbb F_p]$ we have to first find $K$.
How should I find $[K : \Bbb F_p]$? Please give hints.
On
The set of elements of $\mathbf F_{p^{23}}$ invariant by the Frobenius endomorphism $\operatorname{Fr}\colon x\mapsto x^p$ contains the prime subfield $\mathbf F_p$, and is the set of roots of the polynomial $X^p-X$. As a polynomial of degree $p$ in an integral domain has at most $p$ roots , and $\lvert \mathbf F_p\rvert=p$, it follows the invariant subfield is exactly $\mathbf F_p$.
The elements of $K$ are just the roots of $x^p-x=0$. This is a polynomial over a field with degree $p$ and so it has $p$ roots, hence $K=\Bbb F_p$ and the dimension is $1$.
As a fun side-note $[F:\Bbb F_p]=23$ is prime, so it has only two sub-fields, both trivial (i.e. itself and the prime field). So as long as you can see that $\phi$ does not fix all of $F$, you know it fixes a proper subfield, hence it must be that $K=\Bbb F_p$.