If deg $p(x)=m$ and $\deg q(x)=n$ and gcd of $(m,n) = 1$, prove that $[F(u,v):F] = mn$. Let $F$ be a subfield of $K$.
Assume that $u,v\in K$ are algebraic over $F$, with minimal polynomials $p(x)$ and $q(x)$.
If $\operatorname{deg}p(x)=m$ and $\operatorname{deg}q(x)=n$ and $\operatorname{gcd}(m,n) = 1$, prove that $[F(u,v):F] = mn$.
What this problem is basically saying is that
$[F(u,v):F]=[F(u,v):F(u)]\cdot[F(u):F]=mn$
which would imply that the minimal polynomial of $[F(u,v):F(u)]$ is in fact the same minimal polynomial as $[F(v):F]$ because otherwise, the degrees wouldn't be multiplicative. The question I have is, why would they be the same degree? Now the fact that $m$ and $n$ are coprime must be important because there are easy counterexamples if that were not the case, but I can't see how to use that to show the minimal polynomial is the same.
 A: From $p\in F[x]$, we get $p\in F(v)[x]$, hence since $p(u)=0$, it follows that the minimal polynomial for $u$ over $F(v)$ has degree at most $m$.

Then we get
$$
[F(u,v):F]=[F(u,v):F(v)][F(v):F]=[F(v)(u):F(v)][F(v):F]\le mn
$$
so it remains to to show $[F(u,v):F]\ge mn$.

From
$$
[F(u,v):F]=[F(u,v):F(u)][F(u):F]
$$
it follows that $[F(u,v):F]$ is a multiple of $m$, and from
$$
[F(u,v):F]=[F(u,v):F(v)][F(v):F]
$$
it follows that $[F(u,v):F]$ is a multiple of $n$.

But then since $m,n$ are relatively prime, and $[F(u,v):F]$ is a multiple of both $m$ and $n$, it follows that $[F(u,v):F]$ is a multiple of $mn$, hence $[F(u,v):F]\ge mn$.

Therefore $[F(u,v):F]=mn$.

As regards your question about minimal polynomials, from $[F(u,v):F]=mn$, we get
$$
\left\lbrace
\begin{align*}
&
[F(v)(u):F(v)]
=
[F(u,v):F(v)]
=
\frac{[F(u,v):F]}{F(v):F]}
=
\frac{mn}{n}
=
m
\\[4pt]
&
[F(u)(v):F(u)]
=
[F(u,v):F(u)]
=
\frac{[F(u,v):F]}{F(u):F]}
=
\frac{mn}{m}
=
n
\\[4pt]
\end{align*}
\right.
$$
hence

*

*The polynomial $p$, besides being the minimal polynomial for $u$ over $F$, is also the minimal polynomial for $u$ over $F(v)$.$\\[4pt]$

*The polynomial $q$, besides being the minimal polynomial for $v$ over $F$, is also the minimal polynomial for $v$ over $F(u)$.`

