Let $F$ be the field generated over the field $K$ by $u$ and $v$ of degress relatively prime $m$ and $n$, respectively, over $K$. Prove that $[F:K] = mn$
The way I did is like this: construct the diagram $K(u,v), K(u), K(v), K$. Then $[K(u):K] = m, [K(v):K] = n$. Since $u$ and $v$ have relatively prime degres, one cannot divide the other, and since they're irreducible, $[K(u,v):K(v)] = m$, so $[K(u,v):K] = [K(u,v):K(v)][K(v):K] = mn$
Am I right? I'm having trouble in the part one cannot divide the other, I don't know if that is really true
No, the relative primeness isn't really an indicator that the polynomials do not divide each other. If you have two distinct monic irreducible polynomials, they are guaranteed not to divide each other no matter what the degree is. On the other hand, if they're not irreducible, they can have coprime degrees and one can divide the other: like $x$ and $x^2$.
This is all completely decidable from the degrees of the extensions alone.
Because of the tower $K(u,v)\supseteq K(v)\supseteq K$, you know that $[K(v):K]=n$, and that $[K(u,v):K(v)]\leq m$. Then because we know how the degrees of towers of field extension are related, we can say $d=[K(u,v):K]\leq mn$.
But since $K(u)\subseteq K(u,v)$ we know $m|d$ and because $K(v)\subseteq K(u,v)$ we know $n|d$. Because $m,n$ are relatively prime, we can say $mn|d$. But the only natural number less or equal to $mn$ that is divisible by $mn$ is $mn$ itself. So $d=mn$.