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I am looking to prove that $GF(p^n)$ contains a unique subfield isomorphic to $GF(p^m)$ if and only if $m$ is a divisor of $n$. In the Wikipedia article:

https://en.wikipedia.org/wiki/Finite_field#Existence_and_uniqueness

They say that this statement has been proven by E. H. Moore in 1893.

How to go about proving the theorem? I know first I need to generate a subset, prove that it's a subfield, and then assume they are two and show it's unique.

But how to generate it in the first place? Consider the set of polynomials of the form $x^{p^m}-x=0$? Is there a way to show that it's a subfield without checking all the axioms?

dytin
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2 Answers2

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Note that the zeros of $x^{p^n}-x$ are exactly the elements of a finite field with $p^n$ elements. Since splitting fields are uniquely determined up to isomorphism, we can write $GF(p^n)$.

First, if $m$ divides $n$, then $p^m-1$ divides $p^n-1$ and moreover $x^{p^n-1}-1$ divides $x^{p^n-1}-1$. Thus $x^{p^m}-x$ divides $x^{p^n}-x$. Hence, $GF(p^m)$ is a subfield of $GF(p^n)$.

Conversely, if $GF(p^m)$ is a subfield of $GF(p^n)$, then $GF(p^n)$ is a vector space over $GF(p^m)$. Thus $p^n = (p^m)^k$ for some $k\geq 1$. Hence, $m$ is a divisor of $n$.

Added comment: Suppose $m$ divides $n$ with $n=km$. Then $\frac{p^n-1}{p^m-1}= \frac{(p^m)^k-1}{p^m-1} = (p^m)^{k-1} + (p^m)^{k-2} +\ldots+p^m+1$.

Wuestenfux
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Use the Galois correspondence. $GF(p^n)$ is a Galois extension of $GF(p)$, which has cyclic Galois group.

(Intermediate) extension fields correspond to subgroups (of the Galois group). We have a subgroup of order $m\iff m\mid n\iff$ we have an extension field of degree $m$.