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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

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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 seems intuitively obvious, but how do we know all the roots of $x^q - x$ are distinct?2012-07-07
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    @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](http://en.wikipedia.org/wiki/Separable_polynomial), i.e. its roots are distinct.2012-07-07
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    By looking at the derivative. However, the question makes me think you have not seen the result about $x^q-x$.2012-07-07
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    That result is one of my theorems yes, and Zev's comment answers my question (thank you Zev). But is knowing both contain all the roots of that poly enough without knowing it's separable? Not that it's particularly important, anyways thanks for the help.2012-07-07
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    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.2012-07-07
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    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.2012-07-07
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    Oh that's right, a splitting field is the smallest field possible, ok cool thanks. Your conciseness is legendary André, unfortunately us mere mortals sometimes require a bit of redundancy in the answers we receive. (said in jest =])2012-07-07
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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