What is $[F(x):F(x^n)]$ where $F(x)$ if the field of fractions of polynomials over a field $F$? My intuition says it should be $n$, but I'm not precisely sure why. Perhaps I'm misunderstanding the definition, but what I'm trying to do is construct a basis $\{e_1, ..., e_n\}$ of elements in $F(x)$ which span $F(x)$ over the field $F(x^n)$. This means that each element of $F(x)$ can be written as a linear combination of $e_i$s with coefficients in $F(x^n)$. My problem is, I have no idea what such a basis should look like, since I have to account for polynomials in the numerator and denominator.
Is there something obvious I'm missing? Is this even the right way to approach the problem?
 A: You have $x$ transcendental over $F$. Set $u = x^n$. Let's first show $u$ is transcendental over $F$.
If $u$ were algebraic over $F$ then there is a nonconstant polynomial $f(T)$ in $F[T]$ such that $f(u) = 0$, so $f(x^n) = 0$. That makes $x$ algebraic over $F$ (with degree at most $n\deg f$), which contradicts $x$ not being transcendental over $F$.  So $u$ is transcendental over $F$.
Now we'll compute the degree of $x$ over $F(u)$.
The polynomial $T^n - u$ in $F(u)[T]$ has $x$ as a root: $x^n - u  = x^n - x^n = 0$, so $x$ has degree at most $n$ over $F(u)$. To show the degree is $n$, we'll show $T^n - u$ is irreducible over $F(u)$. That follows from the Eisenstein irreducibility criterion: apply it with $F[u]$ in place of $\mathbf Z$ and the irreducible element $u$ in place of a prime number: $T^n - u$ is Eisenstein at $u$ in $F[u][T]$ since its constant term is divisible by $u$ exactly once and the intermediate coefficients are $0$. Thus $T^n - u$ is irreducible in $F(u)[T]$.
If you've never seen the Eisenstein irreducibility criterion used outside of $\mathbf Z[T]$, check the proof of it remains valid with $\mathbf Z$ replaced by $F[u]$ and prime numbers in $\mathbf Z$ replaced by irreducible polynomials in $F[u]$.  For example, $T^5 - (u^3-u)T^2 + (u^3+u^2)$ is irreducible in $F(u)[T]$ because it is Eisenstein at the irreducible $u+1$: each non-leading coefficient is divisible by $u+1$ and the constant term $u^3 + u^2 = u^2(u+1)$ is divisible by $u+1$ exactly once.
In a similar way, $X^n + Y^n - 1$ is irreducible in $\mathbf Q[X,Y]$ for all $n \geq 1$ by viewing it as a polynomial in $X$ with constant term $Y^n - 1$, which is divisible by $Y-1$ exactly once: $X^n + Y^n - 1$ is monic in $X$ and Eisenstein at $Y-1$, so it is irreducible in $\mathbf Q[Y][X] = \mathbf Q[X,Y]$. For a generalization of this to other coefficient fields besides $\mathbf Q$, see my answer here.
A: Call $K=F(x^n)$. Then $F(x)=K(x)$, and $x$ is algebraic over $K$. The degree of $x$ over $K$ is $n$, because $x$ is the root of
$$X^n-x^n \in K[X]$$
Hence $[K(x):K]=n$.
