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I apologize if the title seems to be misleading as I couldn't conjure up a more relevant title. My question is that suppose we have that a prime $p$ and $q = p^k$ for some positive integer $k > 1$. Suppose we have $\mathbb{F}_q$. If $x \in \mathbb{F}_q$ has the property such that $x^p = x$, then $x^{p-1} = 1$ when $p$ does not divide $x$. Suppose $p$ does not divide $x$. So far, the order of $x$ divides $p-1$ by some theorem in algebra. My question is that since $x \in \mathbb{F}_q$, does that mean it lies in $\mathbb{F}_p$ since it has order $\leq p-1$? If so, could you refer me to a theorem that comments on that?

Thanks in advance

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    What exactly do you mean when you ask whether an element of $\mathbb F_q$ lies in $\mathbb F_p$? Do you mean whether it lies in a subfield of $\mathbb F_q$ with $p$ elements that's isomorphic to $\mathbb F_p$?2012-08-01
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    It's true that $\{ x \in \mathbb F_q\, | \, x^p = x\}$ is $\mathbb F_p$ (Fermat's theorem gives you an inclusion, and a counting argument gives you the opposite one [a polynomial equation of degree $k$ cannot have more than $k$ solutions in a commutative field]). But I do not really understand your proof.2012-08-01
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    @PseudoNeo Sorry that my question was unclear but showing that $\mathbb{F}_p = \{x \in \mathbb{F}_q : x^p = p\}$ is my question. Thanks again!2012-08-01
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    @PseudoNeo: Could you spell out "Fermat's theorem gives you an inclusion"?2012-08-01
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    @PseudoNeo, your comment has all the right ingredients, so you could flesh it out to an answer. MathNewbie, I don't quite understand what you mean by "$p$ does not divide $x$"? In a field of characteristic $p$ we have $p=0$, so effectively you are saying that an element is not divisible by zero. That holds very often in a field, does it not?2012-08-01
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    Typo in my first comment: it should be $\mathbb{F}_p = \{x \in \mathbb{F}_q : x^p = x\}$.2012-08-01
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    @Jyrki Oh I see. I was trying to do something with FLT since $x^{p-1} = 1$ in $\mathbb{F}_p$.2012-08-01
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    @PseudoNeo, could you comment on a bit more about the reverse inclusion. I'm not seeing how I can conclude that if $x \in \mathbb{F}_q$ such that $x^p = x$, then $x$ must lie in $\mathbb{F}_p$. Any insights would be great.2012-08-01
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    That's all right. I think that PseudoNeo's idea is that the polynomial equation $x^p=x$ can have at most $p$ solutions in a field, and you have found them all already!2012-08-01
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    Update: Never mind, I was actually able to get this direction as well.2012-08-01

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