Intermediate field between $F$ and $F(x)$ 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.
 A: Hans gives you a method to proving $[F(x):F(u)]=3$ in your particular case. 
Here is the general answer. Write $u=P(x)/Q(x)\in F(x)$ with $\gcd(P, Q)=1$. Then 
$$ [F(x): F(u)]=\max\{ \deg P(x), \deg Q(x) \}.$$ 
Proof. As $F(u)=F(1/u)$, one can suppos $\deg P\ge \deg Q$. By changing $u$ with $u-c$ for a suitable $c\in F$, we can suppose $d=\deg P>\deg Q$. Consider the polynomial 
$$ H(T)=P(T)-uQ(T)\in F(u)[T].$$ 
It has degree $d$, and vanishes at $x\in F(x)$. It remains to show that $H(T)$ is irreducible in $F(u)[T]$. 
Note that $H(T)\in F[u][T]$. By Gauss Lemma, it is enough to show that $H$ is an irreducible element of the UFD ring $F[u][T]$. But $F[u][T]=F[u,T]=F[T][u]$. Viewed as a polynomial in coefficients in $F[T]$, $H$ has degree $1$, and its coefficients $P(T), Q(T)$ are coprime. So $H$ is an irreducible lement of $F[u,T]$ and we are done. 
Recall that if $A$ is an UFD, the irreducible elements of $A[T]$ are exactly the irreducible elements of $A$, and the polynomials $R(T)\in A[T]$ (of positive degree) whose coefficients are coprime and such that $R(T)$ is irreducible in $\mathrm{Frac}(A)[T]$. 
A: Observe $x$ satisfies the polynomial $f(y) = y^3 - \left(\frac{x^3}{x + 1}\right)(y + 1) \in F(u)[y]$.  This demonstrates $[F(x):F(u)] \leq 3$.  You should be able to show $[F(x):F(u)] \neq 1$ by arguing $x \not \in F(u)$.  I'll leave eliminating $[F(x) : F(u)] \neq 2$ to you.
