Finite Field Extensions and the Sum of the Elements in Proper Subextensions (Follow-Up Question)

Solution 1:

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.