Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $q = p^m$. Suppose that $E/\mathbb F_q$ is an extension field and $\alpha \in E$ is algebraic over $\mathbb F_q$. Show that $[\mathbb F_q(\alpha) : \mathbb F_q] = $ the smallest positive integer $n$ such that $\alpha^{q^n} = \alpha$.

Scratch Work: So far I see that $\mathbb F_q$ is a splitting field of $x^{p^m} - x \in \mathbb F_p$. It follows that $[E : \mathbb F_p] = [E : \mathbb F_q][\mathbb F_q : \mathbb F_p]$. Since $\alpha \in E$ is algebraic over $\mathbb F_q$ and $\mathbb F_q/\mathbb F_p$ is a finite extension and moreover algebraic, it follows that $\alpha$ is algebraic over $\mathbb F_p$. Also it can be shown that $\mathbb F_q = \mathbb F_p(\alpha)$ so $$[E: \mathbb F_p] = [E : \mathbb F_p(\alpha)][\mathbb F_p(\alpha) : \mathbb F_p] = n[E : \mathbb F_p(\alpha)].$$ Moreover $[\mathbb F_q(\alpha) : \mathbb F_q] = [\mathbb F_q(\alpha) : \mathbb F_p(\alpha)]$.

But I'm not seeing any of this information implying what I want to prove. Any hints, proofs, would be appreciated.

share|cite|improve this question
Do you know Galois theory? – Potato Mar 13 '13 at 8:02
@Potato No. I'm working through Dummit and Foote and the next chapter is Galois Theory. – Robert Cardona Mar 13 '13 at 8:07
Your scratch work is wrong. $\mathbb{F}_p(\alpha)$ could be $E$, $\mathbb{F}_{p^{n}}$, or any field in-between, depending upon which generator $\alpha$ and how $m$ and $n$ relate. – Hurkyl Mar 13 '13 at 8:47
Do you know that there is only one finite field for every given order? – awllower Mar 13 '13 at 11:32


  1. $K=\mathbb{F}_q(\alpha)$ is an extension field of $F=\mathbb{F}_q$ of degree $n$ for some integer $n$. Therefore $|K|=q^n$, so all the elements $x\in K$ are solutions of the equation $x^{q^n}=x$. Here $\alpha\in K$, so $\ldots$
  2. If $\alpha$ is a zero of $f_k(x)=x^{q^k}-x$, then $\alpha$ is in the splitting field $E_k$ of $f_k(x)$ [Edit: over $F$ /Edit]. Consequently $K\subseteq E_k$. Here $|E_k|=q^k$, so does this allow us to compare $n$ and $k$?
share|cite|improve this answer

Hint: (I use $p,q$ but those are not as in the question) If $\mathbb{F}_{q}/\mathbb{F}_{p}$ is a finite extension of dimension $n$ then every element $\alpha$ in $\mathbb{F}_{q}$ satisfies $\alpha^{p^{n}}-\alpha=0$.

You should be able to find a proof easily as this is shown in the proof for uniqueness of a finite field.

share|cite|improve this answer
I have seen the proof for this before. But I am unsure as to how this leads to showing that $[\mathbb F_q(\alpha) : \mathbb F_q]$ is the smallest positive integer $n$ such that $\alpha^{q^n} = \alpha$. I understand that every element of $\mathbb F_q$ is a root for $x^{p^n} - x = 0$ but don't see its relation to $x^{q^n} - x = 0$. – Robert Cardona Mar 13 '13 at 8:13

Firstly I suppose some things already known:

I.A polynomial of degree $n$ can have at most $n$ roots in a field.
II."Freshman mistake:" $\alpha^q+\beta^q=(\alpha+\beta)^q$ in a finite field of characteristic $p$, where $q=p^n$.
III.Finite dimensional extensions of finite fields are still finite fields.

Since your $\alpha$ is algebraic over $\mathbb F_q$, it satisfies an irreducible polynomial $f(x)\in \mathbb F_q[x]$. Now, by II. we know that $f(x)^q=f(x^q)$ in $\mathbb F_q(\alpha)$. So, for any $n$, $\alpha^{q^n}$ is also a root of $f(x)$. If, for some $n, m<k$, where $q^k=|F_q[\alpha]|$, $\alpha^{q^n}=\alpha^{q^m}$, then, raising both sides to $q^{k-m}$-th power, we find that $\alpha^{q^{k+n-m}}=\alpha$, so that $\alpha$ generates a finite field of order $\le q^{k+n-m}$, a contradiction. Hence the set $\{\alpha,\alpha^q\ldots,\alpha^{q^{k-1}}\}$ is contained in the set of zeros of $f(x)$, and is of cardinality $k$. But $f(x)$ is of degree $k$ by assumption, so, by I. we find that $k=\text{deg}(f(x))$ is the smallest integer for which $\alpha^{q^k}=\alpha$. W.Z.B.W
If this is too ambiguous, tell me, so that I can improve upon it.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.