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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 $a,b \in \{0,1\}$.

Thanks!

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    Similar questions have come up on MathOverflow, e.g., http://mathoverflow.net/questions/26832/degree-of-sum-of-algebraic-numbers and http://mathoverflow.net/questions/30144/algebraic-numbers-of-degree-3-and-6-whose-sum-has-degree-12/30145#30145. 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.2012-05-14

1 Answers 1

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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, 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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    @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!2012-05-15