Finite Field Extensions and the Sum of the Elements in Proper Subextensions

Question

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 $F(u,v) = F(u)$ or $F(u,v) = F(u+v)$?

Answer 1

No: if $ \mathbb F_4=\mathbb F_2(a)$ and $\mathbb F_8=\mathbb F_2(b)$ then
$$\mathbb F_2(a, b-a)=\mathbb F_{64} \neq F_2(a)=\mathbb F_4 \quad \text {and}\quad \mathbb F_2(a, b-a)=\mathbb F_{64}\neq F_2(a+(b-a))=\mathbb F_8$$

Answer 2

No, neither statement is true. If the first equality were true, then we'd have to have $v \in F(u)$, so in particular, the degree of $v$ over $F$ would be $\leq$ the degree of $u$ over $F$. So if the degree of $v$ over $F$ is > degree of $u$ over $F$, then the first equality fails.

For the second one, choose $u$ to be any element algebraic over $F$, but $u \notin F$ and choose $v = -u$. Then $F(u,v) = F(u)$ while $F(u+v) = F(0) = F$ and $F(u) \neq F$ since $u \notin F$. Hope this helps.