= Solution
Put $E=F(a,b)$. The <tower law for field extensions> shows that both $m$ and $n$ divide $[E:F]$. Since $m,n$ are <coprime>, their product divides $[E:F]$, so $[E:F]\geq mn$.
The <minimal polynomial> of $b$ over $F$ has degree $n$ and is still an annihilating <polynomial> over $F(a)$. The <minimal polynomial> over $F(a)$ divides it, so
$$
[E:F(a)]\leq n,\qquad [E:F]=[E:F(a)]m\leq mn.
$$
The upper and lower bounds agree:
$$
\boxed{[F(a,b):F]=mn}.
$$
Coprimality is the step forcing the lower bound to equal the product; without it the two generated <fields> can overlap substantially.
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