I am doing problem $8$ in section $9.1$ in Dummit and Foote:
Let $F$ be a field and let $R = F[x, x^2y, x^3y^2, ..., x^{n+1}y^n, ...]$ be a subring of the polynomial ring $F[x, y]$.
$a)$ Prove that the fields of fractions of $R$ and $F[x, y]$ are the same
I have found the solution to part $a$ in another question here. I would like to check that my understanding of the solution is correct.
I have figured out that $R$ is the set of polynomials in which every non-constant term has a higher $x$-degree than its $y$- degree.
If $K$ is the field of fractions of $R$, and $T$ is the field of fractions of $F[x, y]$, then it is clear that $K \subseteq T$. To show the reverse inclusion, the OP showed that $\dfrac x1$ and $\dfrac y1 \in K$, and concluded that $T \subseteq K$. It took me some time to figure out why $\dfrac x1, \dfrac y1 \in K \implies$ $T \subseteq K$, so I would like to double check that I understood it correctly. This is how I understand it:
For simplicity, let $F = \mathbb{Q}$. Given $\dfrac x1, \dfrac y1 \in K$, we would like to show that, for example, $\dfrac {2+x^2y+x^3y^9}{4+x^3y}$, an element of $T$, is also in $K$. From my characterization of $R$ we know that $4+x^3y \in R$ directly, so $\dfrac {1}{4+x^3y} \in K$. Now $2+x^2y \in R$ so $\dfrac {2+x^2y}{1} \in K$. Also $\dfrac {x^{10}y^9}{1} \in K$, and $\dfrac {1}{x^7} \in K$, so $\dfrac {x^{10}y^9}{1} \cdot \dfrac {1}{x^7} = \dfrac {x^3y^9}{1}\in K$, and adding/multiplying the appropriate things we get $\dfrac {2+x^2y+x^3y^9}{4+x^3y} \in K$.
Is this the right idea?