1) Let $K$ be a field, $q=f/g\in K(X)-K$ with $f,g\in K[X]$ coprime. Show that $q$ is transcendental over $K$ and that $[K(X):K(q)]=\max(\deg f, \deg g)$. Calculate the minimal polynomial of $X$ over $K(q)$.
My approach: $0=\sum a_i(f/g)^i=\sum a_i g^{n-i}f^i/g^n\iff a:=\sum a_i g^{n-i}f^i=0$ but $\varphi(a)=a_nf^n=r\neq0$ where $\varphi:K[X]\rightarrow(K/(g))[X]$ canonical homomorphism and $a_nf^n=qg+r$ ($r\neq0$ because $f,g$ coprime and $a_n$ unit). So $f/g$ is transcendental over $K$.
Now for the dimension part I've tried to look at $\psi:K(q)[X]\rightarrow K(X), h\mapsto h(X)$. Now if I could calculate the minimal polynomial of $X$ and can show that the map is surjective I would be finished. But I can't quite see how the map is surjective and I can't think of the minimal polynomial!!! $h(X)=\sum_j(\sum_i a_{ij}(f/g)^i)X^i$. Is this correct till now? Is this the right way? Some tips and insights would be nice!
2) Let $a,b\in\mathbb C$, $m,n\in \mathbb N$ with $\gcd(m,n)=1$ and $a^m=2,b^n=3$ then $\mathbb Q(a,b)=\mathbb Q(ab)$ and calculate the minimal polynomial of $ab$ over $\mathbb Q$.
Now $(ab)^{mn}=(a^m)^n(b^n)^m=2^n3^m$ so $ab$ is root of $x^{mn}-2^n3^m$, so $[\mathbb Q(ab):\mathbb Q]\leq mn$. Now $(ab)^n=3a^n$. So $a^n\in\mathbb Q(ab)$ and so is $(a^n)^x/2^y=(a^n)^x/(a^m)^y=a^{nx-ym}$, and as $m,n$ are coprime $a\in\mathbb Q(ab)$. So is $b$. So $\mathbb Q(a,b)=\mathbb Q(ab)$. $x^{mn}-2^n3^m$ is also the minimal polynomial because $[\mathbb Q(a):\mathbb Q]=m, [\mathbb Q(b):\mathbb Q]=n$,so $n,m|[\mathbb Q(ab):\mathbb Q]$ so $[\mathbb Q(ab):\mathbb Q]=nm$. Is there an argument to show $[\mathbb Q(ab):\mathbb Q]= mn$ immediately (i.e. show its irreducibility)?