Show that a finite field of order $p^n$ has exactly one subfield of $p^m$ elements for each divisor $m$ of $n$. Show that a finite field of order $p^n$ has exactly one subfield of $p^m$ elements for each divisor $m$ of $n$.
Suppose that $o(F)=p^n$ .Let $F$ has $\Bbb Z_p$ as its prime subfield. Let $n=km$. I will have to construct a subfield of order $p^m$.Frankly speaking I have not done enough to show you all.I could not get further .
I am finding it difficult where to start the problem.Any hints will be helpful
 A: Hint: Such a field is the splitting field of $X^{p^n}-X=0$ and contains the splitting field of $X^{p^m}-X$ if $m$ divides $n$.
A: Consider $S_m(F)=\{a\in F:a^{p^m}=a\}$
Step 1: Check that $S_m(F)$ is a subfield of $F$.Note that $0\in S_m(F)$ ;$o(S_m(F))\leq p^m\implies o(S_m^*(F))\leq p^m-1$
 where $(S_m^*(F))$ denotes set of all non -zero elements of $(S_m(F))$
Step 2:Let $n=md$;  $p^n-1=(p^m-1)(1+p^m+...)\implies p^m-1|p^n-1$.Consider $F^*$.We have $|F^*|=p^n-1$ and it is cyclic .So it has a unique subgroup of order $p^m-1$ and all its elements satisfy $x^{p^m}=x$.So we have exactly $p^m$ elements in $(S_m(F))$.So we get a subfield of order $p^m$.
A: Consider $\mathbb{Z}_p \leq \mathbb{Z}_{p^n} \leq \overline{\mathbb{Z}}_p$. Where the $\overline{\mathbb{Z}}_p$ is the algebraic closure of $\mathbb{Z}_p$.
Note that elements of the middle $\mathbb{Z}_{p^n}$ are those $p^n$ zeros of $x^{p^n}-x \in \overline{\mathbb{Z}}_p$.
Suppose $\mathbb{Z}_{p^n}$ contains two or more subfields $A_1,...,A_L,L\geq2$ of order $p^m$, then contradiction arises. Because those elements inside $A_l$ must satisfy $x^{p^m}-x=0.$ (This is because the units of a finite field form a cyclic group.) However, inside the largest $\overline{\mathbb{Z}}_p$ we can have $p^m$ such elements. Hence the only thing we need to care about is the existence.
Choose a generator $\alpha \in \mathbb{Z}_{p^n}^\times$. Then forall $a^j$, where $j \in \{1,2,...,p^n-1\}$, it is a root of $x^{p^m}-x=0 \iff jp^m-1 \equiv 0 \mod p^n-1.$ In other words, $j=k\cdot \frac{p^n-1}{p^m-1}$ for some appropriate $k\in \mathbb{Z}.$ Note that $\frac{p^n-1}{p^m-1}$ is an integer if $m$ divides $n$, because we can always factor the denominator from the numerator. Note that $k$ can be $1,2,...,p^m-1$.
Hence we have $p^m$ elements ($p^m-1$ nonzero elements plus the $0$) inside $\mathbb{Z}_{p^n}$ satisfies $x^{p^m}-x=0,$ which exhausts all possible roots in $\overline{\mathbb{Z}}_p$. And they forms a subfield of $\mathbb{Z}_{p^n}$. We are done.
