Let $F$ be field. Let $F[x]$ denote the polynomial ring over $F$, and $F(x)$ its field of fractions. Let $F(x^n)$ be the field of fractions of $F[x^n]$, the subring of polynomials in $x^n$. Consider the ring $F[x]\otimes F(x^n)$, where the tensor is over $F[x^n]$. I'm trying to show that this ring is isomorphic to $F(x)$ (equivalently, I'm trying to show that this ring is a field.)
For example, if $n=2$ we can prove this by showing that $p\otimes 1$ is invertible for all $p\in F[x]$ besides $0$. Write $p=e+o$ where $e$ has only even degree terms and $o$ has only odd degree terms. Then $p*(e-o) =(e^2-o^2)\in F[x^2] $ is nonzero. The inverse of $p\otimes 1$ is thus $(e-o)\otimes( e^2-o^2)^{-1}$. Thus $F[x]\otimes F(x^2)$ is a field.
This can be generalized to cover the case where $n$ is any power of $2$ by essentially repeating this process. Any ideas on how to do the case for general $n$?
Another possibly useful fact is that $(F[x]\otimes F(x^n))\otimes F((x^n))\cong F((x))$, where the tensors are over $F[x]$ and $F((x))$ refers to the ring of formal Laurent series over $F$, which we do know is a field.
Let $R=F[x^n]$ and $K=F(x^n)$. Note that as an $R$-algebra, $F[x]$ can naturally be identified with the quotient $R[t]/(t^n-x^n)$ (with $t$ mapping to $x$). So, $F[x]\otimes_R K$ can naturally be identified with $K[t]/(t^n-x^n)$. To conclude this is a field, you now just have to know that $t^n-x^n$ is an irreducible polynomial over $K$. One way to prove this is to observe that $t^n-x^n$ is irreducible over $R$ by Eisenstein's criterion (since $x^n$ is irreducible in $R$) and thus also over $K$ by Gauss's lemma.