# Order of an element in $\mathbb{F}_q$ where $p^k = q$

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?

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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 $\mathbb F_p$? – joriki Aug 1 '12 at 10:29
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. – PseudoNeo Aug 1 '12 at 10:30
@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! – MathNewbie Aug 1 '12 at 10:35
@PseudoNeo: Could you spell out "Fermat's theorem gives you an inclusion"? – joriki Aug 1 '12 at 10:38
@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? – Jyrki Lahtonen Aug 1 '12 at 10:41

So, let $F$ be a finite field. As you said, the cardinal $q$ of $F$ is a prime power $q = p^e$ and $F$ is isomorphic to $\mathbb F_q$. The prime subfield of $F$ (the subfield generated by 1) is then isomorphic to $\mathbb F_p$. With a slight abuse of notation, I will from now on identify $F$ with $\mathbb F_q$ and its prime subfield with $\mathbb F_p$, so I consider the inclusion $\mathbb F_p \subseteq \mathbb F_q$.

Theorem. $\mathbb F_p = \{ x \in \mathbb F_q \, | \, x^p = x\}.$

Proof.

• First, $\mathbb F_p \subseteq \{ x \in \mathbb F_q \, | \, x^p = x\}$. Indeed, Fermat's little theorem says that $\forall a \in \mathbb Z, a^p \equiv a \, (\mathrm{mod}\, p)$. In more abstract terms, that means that $\forall x \in \mathbb F_p, x^p = x$, which is the inclusion I mentioned. (BTW, this finite field business gives a neat proof of Fermat's little theorem: Lagrange's theorem for the multiplicative group $\mathbb F_p^\times$ gives $\forall x \in \mathbb F_p \setminus \{0\}, x^{p-1} = 1$, and deducing Fermat's little theorem is now childplay.)

• To prove the opposite inclusion, note that it is enough to prove that $\{ x \in \mathbb F_q \, | \, x^p = x\}$ cannot have more than $p$ elements (because what we just did already gives $p$ elements). But that is a particular case of a more general result: in a (commutative) field, a polynomial of degree $d$ cannot have more than $d$ roots. (By Euclidean division, $a$ is a root of $P$ iff $(X-a)$ divides $P$ so a polynomial having $d$ roots is a multiple of $(X-a_1)\cdots (X-a_d)$. If it is nonzero, that implies $\deg P \geq d$.)

So the theorem is proved. The arguments I gave are enough to prove that more generally, if $K \subseteq F$ is an extension of finite fields, $K = \{ x \in F \, | \, x^q = x\}$, where $q = |K|$.

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