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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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  • $\begingroup$ Note that the statement also holds when $K$ is an arbitrary field. $\endgroup$ Commented Aug 22 at 19:29

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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  • $\begingroup$ It seems intuitively obvious, but how do we know all the roots of $x^q - x$ are distinct? $\endgroup$ Commented Jul 7, 2012 at 4:58
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    $\begingroup$ @Nollie: The derivative of $x^q-x$ is $$qx^{q-1}-1=0\cdot x^{q-1}-1=-1$$ which is relatively prime to $x^q-x$. Therefore the polynomial $x^q-x$ is separable, i.e. its roots are distinct. $\endgroup$ Commented Jul 7, 2012 at 5:02
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    $\begingroup$ If we have a field extension $L/K$ and the polynomial $f\in K[x]$ splits completely in $L$ (i.e. $L$ contains every root of $f$), then there is a unique subfield $F\subseteq L$ that is the splitting field of $f$ over $K$ inside $L$, namely, $$F=K(\alpha_1,\ldots,\alpha_t)$$ where the $\alpha_i$ are the roots of $f$. Because both $E$ and $F$ are splitting fields for $x^q-x\in \mathbb{F}_q[x]$ inside the finite field $K$, they must be equal. $\endgroup$ Commented Jul 7, 2012 at 5:07
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    $\begingroup$ The theorem that a field of $q$ elements is a splitting field of $x^q-x$ over the ground field makes the fact that there are no multiple roots irrelevant. Which is why it is not mentioned in my answer. $\endgroup$ Commented Jul 7, 2012 at 5:08
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    $\begingroup$ 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. $\endgroup$ Commented Jul 7, 2012 at 7:17
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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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  • $\begingroup$ 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. $\endgroup$ Commented Jul 7, 2012 at 4:45
  • $\begingroup$ What if $K$ is infinite field? $\endgroup$ Commented Aug 22 at 19:26
  • $\begingroup$ @alg, an infinite field can certainly have two subfields that are isomorphic but not equal. E.g., let $L$ be any field, let $K=L(x,y)$ be the field of rational functions in indeterminates $x,y$ with coefficients in $L$, let $E=L(x)$, $F=L(y)$. $\endgroup$ Commented Aug 22 at 22:53
  • $\begingroup$ @alg, if you want an example with finite index, let $K={\bf Q}(\root3\of2,\rho)$, where $\rho^3=1\ne\rho$. Then ${\bf Q}(\root3\of2)$ and ${\bf Q}(\rho\root3\of2)$ are isomorphic but not equal. $\endgroup$ Commented Aug 22 at 23:05
  • $\begingroup$ @GerryMyerson Right, I agree. I was thinking about two finite subfields of an infinite field with equal cardinality. $\endgroup$ Commented Aug 22 at 23:11

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