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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 :)

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Please make the body of your messages self-contained. "My books says this is true" is not a particularly informative first sentence. Titles/subject lines are no more part of the content of the message than the publisher's imprint on the spine of the book is part of the novel. –  Arturo Magidin May 10 '11 at 16:05

3 Answers 3

up vote 7 down vote accepted

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...

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sorry for the offtopic nonsense but I think it's great you came back.. –  quanta May 10 '11 at 19:51
@quanta: thanks. Let's see if we can avoid making a big deal out of it... –  Pete L. Clark May 10 '11 at 19:54
@PeteL.Clark : $G$ need not be abelian for FLT to hold , right ? –  Theorem Feb 8 '13 at 14:54
@Theorem: I'm not sure what "FLT" stands for here. Definitely Lagrange's Theorem holds for all finite groups. But the proof which generalizes the standard proof of Fermat's Little Theorem (hmm, FLT?) works only for finite abelian groups. So far as I know, in order to prove Lagrange's Theorem in the nonabelian case one needs to discuss cosets. –  Pete L. Clark Feb 8 '13 at 17:21

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 $\mathbb{F}_q$.)

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How about algebraic but infinite extensions of a finite field ? –  Fardad Pouran Nov 28 at 22:11
@Fardad: they are "topologically cyclic" in the sense that the subgroup of the Galois group generated by the Frobenius map is dense. –  Qiaochu Yuan Nov 28 at 22:20

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.

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

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'|\,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 $\rm \: o(Z) = lcm(A,B)$

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

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I think the OP was asking if the (finite) extension $K/\mathbb{F}_q$ is a cyclic extension, namely that $\mbox{Gal}(K/\mathbb{F}_q)$ is a cyclic group. Does this immediately follow from the fact that $K^\times$ is a cyclic group? –  Alon Amit May 10 '11 at 16:15
@Alon I've edited it to clarify that it was meant to support Qiaochu's answer. –  Bill Dubuque May 10 '11 at 18:04
Sadly, it looks like this nice answer is destined to be one of the gems at the tail end of the voting range. To work against that, I'm voting it up. –  t.b. May 10 '11 at 19:02

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