Show that a finite field of order $p^n$ has exactly one subfield of $p^m$ elements for each divisor $m$ of $n$.

Question

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

Answer 1

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 $p^m$.

Answer 2

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 $n$.