Suppose that $F$ is a field and that $u \in F(x):= \{PQ^{-1}:P,Q \in F[x], Q\neq 0 \}$, so that $F \subseteq F(u) \subseteq F(x)$. Is there a general method for determining $[F(x):F(u)]$?
For my homework problem I have been given the specific case of $u := \frac{x^3}{x+1} = x^2-x+1-\frac{1}{x+1}$ and we haven't even gone over anything remotely similar to a problem like this in class. The hint was to choose $v \in F(x)$ so that $F(u,v) = F(x)$.
If we had such a $v$ then $[F(x):F(u)] = [(F(u))(v):F(u)]$ which is equal to the minimum degree of a polynomial with coefficients in $F(u)$ which has $v$ as a root. But I don't know how to find such a polynomial, nor is it obvious that one exists.
If we take $v:=x$ then certainly $F(u,x)=F(x)$ but I didn't make any progress from this. I was thinking of writing a Mathematica script to try random polynomials but I don't think this is the intended method of solution.
I was also thinking that it might be significant that the roots of the numerator and denominator of $u$ are 0 and 1, namely they are elements of $F$.
Any help would be appreciated.