Krdytkyu

My book says all extensions of finite fields are cyclic, but I could not find a proof (maybe I haven't looked hard enough). If it's straightforward, can you tell me why it's true? Thanks :)

I'm not sure that the cyclicity of the unit group of a finite field is being used here. (Not that I have any problem using it: see e.g. Section 2 of these notes for a proof.)

Let $K/mathbb{F}_q$ be a field extension of degree $n$, so $# K = q^n$. Let $sigma: K rightarrow K$ be $x mapsto x^q$. Recall:

(Lagrange's Little Theorem): Let $G$ be a finite abelian group of order $n$ and
$g in G$. Then the order of $g$ divides $n$.

The point is that LLT is a special case of Lagrange's Theorem which can be proved by the same argument which proves Fermat's Little Theorem -- i.e., the special case in which $G = mathbb{F}_p^{times}$. So one need not talk about cosets and such...

Let $x in mathbb{F}_q$. I claim that $x^q = x$. This is clear if $x = 0$, and otherwise apply LLT to $x in mathbb{F}_q^{times}$ to get $x^{q-1} = 1$, which implies $x^q = x$.

Therefore $sigma$ is an automorphism of $K/mathbb{F}_q$. As for its order, suppose
$sigma^i$ is equal to the identity: that is, for all $x in K$, $x^{q^i} = x$. We have $P_i(t) = t^{q^i} - t in K[t]$ is a polynomial of degree $q^i$ over the field $K$, so by the Root-Factor Theorem (a consequence of the division algorithm for polynomials), $P_i(t)$ has at most $q^i$ roots. It follows that the order of $sigma$ is equal to $n = log_q(# K)$. Thus the cyclic group generated by $sigma$ is a degree $n$ subgroup of $operatorname{Aut}(K/mathbb{F}_q)$. But by basic Galois theory, for an extension $K/F$ of degree $n$, we have $# operatorname{Aut}(K/F) leq n$, with equality holding if and only if $K/F$ is Galois. Therefore $K/mathbb{F}_q$
is a cyclic Galois extension.

If we like, we can now establish that there is a unique cyclic extension of degree $n$ for any $n in mathbb{Z}^+$: we can take
the splitting field of $t^{q^n} - t$, and splitting fields exist and are unique up to (nonunique) isomorphism over the ground field.

It seems to me that I have not used the cyclicity of $mathbb{F}_q^{times}$ anywhere...

Any finite extension of a finite field $mathbb{F}_q$ is cyclic. For such an extension $K$ first recall that the Frobenius map $x mapsto x^q$ is an $mathbb{F}_q$-linear endomorphism. If $x^q = y^q$ then $(x - y)^q = 0$, hence $x = y$, so the Frobenius map is injective. Since it is an injective linear map from a finite-dimensional vector space to itself, it is surjective, so it is an automorphism. Its fixed field is the subfield of roots of $x^q - x$, which are precisely the elements of the base field $mathbb{F}_q$. It follows that $K$ is Galois with Galois group the cyclic group generated by $x mapsto x^q$.

(I guess when I say $mathbb{F}_q$ I am being mildly circular. Interpret the above proof as follows: any finite extension of $mathbb{F}_p$ is cyclic, and in fact the above proof shows that they are all of the form $mathbb{F}_{p^n}$ using the fact that any finite subgroup of the multiplicative group of a field is cyclic, so finite fields $mathbb{F}_q$ really do have Frobenius maps like I just claimed they do, and then apply the proof again to $mathbb{F}_q$.)

Below is a complete, noncircular simple proof of the result mentioned by Qiaochu that avoids invoking the high-powered structure theorem for finite abelian groups.

Theorem $ $ A finite subgroup $rm:G:$ of the multiplicative group of a field is cyclic.

Proof $ $ The proposition below yields, with $rm,m = maxord(G) = expt(G),,$ that $rm, x^m = 1,$ has $rm:#G:$ roots. Since a polynomial $rm:f:$ over a field satisfies $rm:#roots f le deg f:$ we infer that $rm: #G le m.:$ But maxorder $rm:m le #G:$ since $rm:g^{#G} = 1:$ for all $rm:g in G:$ (Lagrange). $:$ Thus $rm:m = #G = maxord(G),:$ therefore $rm:G:$ has an element of order $rm#G,:$ hence $rm:G:$ is cyclic.

$begin{eqnarray}rm{bf Proposition}qquad maxord(G) &=&,rm expt(G) text{ for a finite abelian group} G, i.e.\[.5em]
rm max { ord(g) : : g in G} &=&,rm min { n>0 : : g^n = 1 forall g in G}end{eqnarray}$

Proof $ $ By the lemma below, $rm: S, =, { ord(g) : :g in G }$ is a finite
set of naturals closed under$rm lcm$.

Hence every $rm s in S:$ is a
divisor of the max elt $rm: m $ [else $rm: lcm(s,m) > m,$],$ $ so $rm m = expt(G)$.

Lemma $ $ A finite abelian group $rm:G:$ has an lcm-closed order set, i.e. with $rm: o(X) = $ order of $rm, X$

$$rm X,Y in G Rightarrow exists Z in G: o(Z) = lcm(o(X),o(Y))$$

Proof $ $ By induction on $rm: o(X), o(Y). $ If it's $:1:$ then trivially $rm:Z = 1$. $ $ Otherwise

write $rm, o(X), =: AP,: o(Y) = BP', P'mid P = p^m > 1, $ prime $rm: p:$ coprime to $rm: A,B.$

Then $rm, o(X^P) = A, o(Y^{P'}) = B. $ By induction there's a $rm: Z:$ with $rm : o(Z) = lcm(A,B)$

thus $rm, o(X^A Z) = P lcm(A,B) = lcm(AP,BP') = lcm(o(X),o(Y)).$