Q:

Coprime degree of minimum polynomials

Let $F,K$ be fields. Suppose $a,b\in K$ are algebraic over $F$ with degrees $m,n$, where $(m,n)=1$. I want to show that $F(a,b)$ is of degree $mn$ over $F$, and $F(a)\cap F(b)=F$.
Consider $$[F(a,b):F]=m[F(a,b):F(a)]=[F(a,b):F(a)][F(a):F]=[F(a,b):F(b)][F(b):F]=n[F(a,b):F(b)]$$ so $[F(a,b):F]\geq mn$. On the other hand $\{a^ib^j\}_{0\leq i<m,0\leq j<n}$ is a basis, so $[F(a,b):F]=mn$.
Now, how could we show that $F(a)\cap F(b)=F$?

A:

Hint  If $c \in F(a) \cap F(b)$ then $[F(c):F]$ must divide $[F(a):F]$ and $[F(b):F]$.

Also, note that in the first part, you can simplify the argument this way:
Since $m,n |[F(a,b):F]$, their least common multiple divides $[F(a,b):F]$. But their least common multiple is $mn$.

