# Splitting field in relation to finite fields

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.

## 2 Answers

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 $n$.

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)...

• Thanks so much, I understand 2) and 3) perfectly and can use them to derive the necessary results. Could you elaborate more on 1) though please, as my text does not mention vector spaces for finite fields at all. Is this $K$ a subfield of $GF$ or of $L$ here? – K.Power Jun 5 '16 at 14:35
• a) I had a typo, $L$ is supposed to be $GF(p^n)$. $K$ is a subfield of this. b) With (1) I'm trying to get at the fact that $L$'s order has to be a power of $K$'s. I think of this as being a consequence of the fact that $L$ is a finite-dimensional vector space over $K$ and is therefore isomorphic as a vector space (though not as a field) to $K^d$ for some $d$. This reasoning uses only the following basic result of linear algebra: a vector space has a basis. You don't need to known anything fancy about vector spaces. – Ben Blum-Smith Jun 5 '16 at 16:35
• It should be possible to derive that $L$ is isomorphic as an additive group to $K^d$ straight from the field properties. Try googling "every vector space has a basis" and trying to copy the proof, replacing the ground field (usually $\mathbb{R}$ or $\mathbb{C}$) with $K$ and the vector space with $L$. Once you know $L$ has a basis over $K$, it tells you every element of $L$ can be written uniquely as a linear combination with coefficients in $K$ of that basis, and you can explicitly count $L$'s elements by thinking about how many sets of $K$-coefficients are possible. – Ben Blum-Smith Jun 5 '16 at 16:39
• The goal is to conclude that $L$'s order is a power of $K$'s, which is the reason why $r$ has to divide $n$. – Ben Blum-Smith Jun 5 '16 at 16:40
• Thank you so much for the detailed explanation. I'll dig up the old linear algebra textbook and jog the memory, I now understand where the proof should go so thank you. – K.Power Jun 5 '16 at 19:01