Splitting field in relation to finite fields

Question

I'm trying to prove that the subfields of a Galois Field $GF$ of order $p^n$ are isomorphic to a Galois field of order $p^r$ where $r|n$, and that there exists a unique subfield for each such $r$. I see that generally people use the Frobenius automorphism to prove this, but in my text I am not given any real relation between finite fields and the automorphism. I am given that any field $F$ has $p^n$ elements if and only if it is a splitting field for $f(t)=t^{p^n}-t$ over the prime subfield $\mathbf{Z}_p$. I'm also given the fact that for any $n\in \mathbb{N}$ and prime $p$ there is a unique finite field with $p^n$ elements.

From Lagrange's Theorem it follows that the order of any subfield of $GF$ divides $p^n$. If I could show that the order of any subfield of $GF$ must be $p^r$ where $r|n$ I think the rest would follow. I'm pretty sure all the ingredients are there but I just can't quite get the proof together. If anyone could help I'd be much obliged.

Answer 1

Hints (let me know if you want more than this):

1) For the order, consider that if $L$ is a finite field of order $p^n$ and $K$ is a subfield, then $L$ is a vector space over $K$ (of finite dimension since $L$ is finite!). What are the implications for $K$'s order?

2) For the existence of a subfield of given order $p^r$ with $r\mid n$, consider the set of elements of $L$ that are roots of $x^{p^r}-x$. Can you show that this polynomial splits in $L$ and that its $p^r$ roots form a subfield? (Hint for this: what is the group structure of the group of nonzero elements of $L^\times$ under multiplication?)

3) For uniqueness of the subfield, note that any element of a field of order $p^r$ must be a root of $x^{p^r}-x$. So any subfield of that order must be contained in the set of elements you considered in (2)...

Answer 2

If $r | n$, then the polynomial $x^{p^r}-x$ divides $x^{p^n}-x$, So, the set of zeros of $x^{p^r}-x$ is a subset of the set of zeros of $x^{p^n}-x$. The former set is $GF(p^r)$, and the latter set is $GF(p^n)$. This proves the existence of a subfield $GF(p^r)$ if $r | n$.

For uniqueness the part, observe that if $GF(p^n)$ had two or more subfields of order $p^r$, then the the number of zeros of $x^{p^r}-x$ in $GF(p^n)$ is more than $p^r$, a contradiction because the degree of this polynomial is $p^r$.

Finally, we prove that $r|n$ is necessary. We shall use the fact that if $F$ is a subfield of $E$ and $E$ is a subfield of $K$, then $[K:F]=[K:E][E:F]$. Let $E$ be a subfield of $GF(p^n)$. By Lagrange, the group $(E,+)$ has order $p^r$ for some $r \le n$. We have $[GF(p^n): E][E:GF(p] = [GF(p^n):GF(p)]$. The right hand side is $n$, and the left hand side is the product $[GF(p^n): GF(p^r)[GF(p^r):GF(p)] = xr$ for some positive integer $x$. Hence, $n=xr$ for some integer $x$. Thus, $r$ divides $n$.