# $E/\mathbb F_q$ extension field. Show $[\mathbb F_q(\alpha) : \mathbb F_q]$ is smallest $n$ satisfying property.

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.

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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

Hints:

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 $k$?
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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.

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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.

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