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I recently posted the following question, to which this question is a follow-up. Regardless, my question here will be self-contained.

Let $F$ be a finite field, and let $u,v$ be algebraic over $F$. Consider the fields $F(u,v),F(u)$ and $F(v)$. Must it be the case that there exist $a,b \in F$ for which $F(u,v) = F(au+bv)$?

If this question is more difficult than I suspect, I would still be interested in a specific solution for $F = \mathbb{F}_{2}$, so that $a,b \in \{0,1\}$.


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You are really asking: "Must it be the the case that there exist $a,b\in F$ such that $F(u,v)=F(au+bv)$?", right? Your phrasing is somewhat ambiguous. – Arturo Magidin May 13 '12 at 23:10
Yes. Edited accordingly. – Isaac Solomon May 13 '12 at 23:41
Similar questions have come up on MathOverflow, e.g.,… and…. Brawley, J. V., Carlitz, L. Irreducibles and the composed product for polynomials over a finite field, MR0893074 (89g:11118) looks like it might have an answer to the particular question raised here. – Gerry Myerson May 14 '12 at 0:43

1 Answer 1

up vote 2 down vote accepted

A partial result in this direction. I need to make an extra assumption.

My claim (thanks to KCd for the reformulation): If $F=F(u)\cap F(v)$, then $F(u,v)=F(u+v)$.

Let $|F|=q=p^n$. Then $|F(u)|=q^a$ and $|F(v)|=q^b$. The claim is interesting only, when $u,v\notin F$, so we assume that $a>1$ and $b>1$. We also have $q^{\gcd(a,b)}=|F(u)\cap F(v)|=q$, so $\gcd(a,b)=1$. Here we used the fact that inside any finite field the size of a subfield determines the subfield uniquely, and also the fact that a field of $q^t$ elements contains a subfield of size $q^k$ for any divisor $k\mid t$.

Let $\tau:x\mapsto x^q$ be the Frobenius automorphism. We have that $|F(u+v)|=q^m$, where $m$ is the smallest positive integer such $\tau^m(u+v)=u+v$. This equation implies that $$ u^{q^m}-u=v-v^{q^m}\in F(u)\cap F(v)=F.\tag{1} $$ I shall view $F(u,v)$ as a module over the polynomial ring $F[\tau]$ with $\tau$ acting as the Frobenius. Then $F$ consists precisely of the fixed points of $\tau$, so equation $(1)$ tells that both $u$ and $v$ are annihilated by $(\tau-1)(\tau^m-1)$. The extra assumption $\gcd(a,b)=1$ tells us that at least one of $a,b$ is coprime to $p$. Without loss of generality we can assume that $p\nmid a$. As $|F(u)|=q^a$ we know that $u$ is also annihilated by $\tau^a-1$. As $\gcd(a,p)=1$, the polynomial $\tau^a-1$ has no repeated factors. Therefore $u$ is annihilated by $$\gcd(\tau^a-1,(\tau^m-1)(\tau-1))=\gcd(\tau^a-1,\tau^m-1)=\tau^{\gcd(a,m)}-1.$$ But $u$ is not annihilated by any polynomial of the form $\tau^\ell-1$, $0<\ell<a$, so we can conclude that $a=\gcd(a,m)$ and $a\mid m$. Therefore $u\in F_{q^m}=F(u+v)$. Consequently also $v=(u+v)-u\in F(u+v)$. Therefore $F(u,v)\subseteq F(u+v)$. The reverse inclusion is trivial and the claim follows.

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Since $[F(u):F] = a$ and $[F(v):F] = b$, and the assumption that $F(u) \cap F(v) = F$ implies $a$ and $b$ are relatively prime, as you noted yourself, the conditions that $u \not\in F(v)$ and $v \not\in F(u)$ are not needed as explicit hypotheses. After you proved $u \in F(u+v)$ it follows that $F(u) \subset F(v)$, so $a|b$ and thus $a = 1$, so $u \in F$. Then obviously $F(u+v) = F(v) = F(u,v)$. – KCd May 14 '12 at 9:19
So a clean statement would be: if $F(u) \cap F(v) = F$ then $F(u,v) = F(u+v)$. That's obvious if $u$ or $v$ is in $F$, and your proof shows it is also true if neither $u$ nor $v$ is in $F$ too. – KCd May 14 '12 at 9:25
Your "partial" result has quite reasonable hypotheses considering what is known about the analogous situation for Galois extensions in characteristic 0, as seen at…. – KCd May 14 '12 at 9:56
Thanks, @KCd. I worked on this overnight. If I find the time, I will look up the paper by Brawley & Carlitz suggested by Gerry Myerson. Can't say I would be surprised to learn that Carlitz and his coauthors have settled this question. – Jyrki Lahtonen May 14 '12 at 10:09
@JyrkiLahtonen : I really appreciate you giving such a thorough answer! I don't have the time at the moment to look this over fully, but when I do (hopefully over the weekend) I might come back with some questions. Thanks again! – Isaac Solomon May 15 '12 at 6:03

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