Let $F$ be a field and let $R = F[x,x^2y,x^3y^2,...,x^ny^{n-1},...]$ be a subring of the polynomial ring $F[x,y]$. Prove that the fields of fractions of $R$ and $F[x,y]$ are isomorphic.
I know that the field of fractions of the polynomial ring $K[X_1,X_2,...,X_n]$ over a field $K$ is the field of rational functions $\{\frac{f(X_1,X_2,...,X_n)}{g(X_1,X_2,...,X_n)} : f,g \in K[X_1,X_2,...,X_n] , g \neq 0\}$. But I'm not exactly sure how to write down an isomorphism between the fields of fractions of $R$ and $F[x,y]$ ; we're essentially to cook up an isomorphism between a ratio of functions in $R$ and a ratio of functions in $F[x,y]$. But $R$ is a subring of $F[x,y]$, so it doesn't even seem to me like such a map would be surjective.
Any ideas would be appreciated.
Thanks!
Clearly the field of fractions of $R$ is contained in the field of fractions $F(x, y)$ of $F[x, y]$. So it remains to show that they are equal. Note that it is enough to show that $y$ is in the fraction field of $R$, as any subfield of $F(x, y)$ that contains $F$, $x$ and $y$ is necessarily equal (not just isomorphic) to $F(x, y)$, and $R$, and therefore its field of fractions, already contains $F$ and $x$.
Now since $x^2$ and $x^2y$ are both in $R$, we must have that $\frac{x^2y}{x^2} = y$ is in the fraction field of $R$ as well. This finishes the proof.