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If $E$ and $F$ are subfields of a finite field $K$ and $E\cong F$, prove that $E = F$.

A finite field is a simple extension of each of its subfields and $\mathbb{Z}_p$ is a subfield of every finite field. Hence $E\cong \mathbb{Z}_p(u)$ and $F\cong \mathbb{Z}_p(v)$ for some $u,v\in K$. Proving that $u = v$ given $\mathbb{Z}_p(u)\cong \mathbb{Z}_p(v)$ may be stronger than I need though, since $u$ and $v$ could be different generators for the same set.

Can anyone help me with this? Many thanks.

Edit: This is a Galois Theory free zone.

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    Fyi: using $\mathbb{Z}_p$ to denote the field with $p$ elements is sort of like pushing mongo. Use $\mathbb{F}_p$.2012-07-07

2 Answers 2

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The isomorphism in particular implies that the orders of $E$ and $F$ are equal. That's all we need. Let us assume that you know that a finite field with $q=p^k$ elements is a splitting field for the polynomial $x^q-x=0$. Thus each of $E$ and $F$ consists of the roots (in $K$) of $x^q-x=0$. That implies that $E=F$.

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    It's Galois Theory free, so we should not use splitting fields :-). By simple group theory, the order of each element in the multiplicative group of each subfield satisfies $x^{q-1} = 1$. Thus each element in each subfield satisfies $x^q-x = 0$. This equation has at most $q$ roots, and since each subfield has $q$ elements, we have found all the roots, and they all lie in each subfield.2012-07-07
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Do you know Galois Theory? If so, it's easy: The Galois group of any finite field over any other is cyclic, so it has only one subgroup of any given order, so only one field of any given index.

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    Sorry I should have mentioned, I have no Galois Theory at my disposal. Group Theory up through Sylow Theorems, and Field Theory through splitting fields and general results on the classification of finite fields. Also undergrad ring theory, but that's probably not important.2012-07-07