Problem
Let $F$ be a field, and $K$ a field extension of $F$. Prove the following: Suppose $a,b \in K$ are algebraic over $F$ with degrees $m$ and $n$, where $m$ and $n$ are relatively prime. Then $F(a) \cap F(b) = F$.
Attempt
The equality $F(a) \cap F(b) = F$ implies that the only elements shared by $F(a)$ and $F(b)$ are contained within $F$ itself. $F(a)$ is the smallest field containing $F$ and $a$, and likewise with $F(b)$. Since the degrees of $a$ and $b$ over $F$ are relatively prime, one cannot be expressed as a multiple of the other. It follows that $b \not\in F(a)$ and $a \not\in F(b)$. Thus, $F(a) \cap F(b) = F$.
Questions
Does the fact that the degrees of $a$ and $b$ over $F$ are relatively prime imply that each is not contained in the field extension of the other? If not, what else am I missing?