Proving that $S^{-1} A \cong \mathbb{R}[X, X^{-1}]$ (ring of fractions)

Question

Problem: Consider the ring $$A = \mathbb{R}[X,Y]/(XY)$$ and let $$S = \left\{ 1 + (XY), X + (XY), X^2 + (XY), \ldots \right\} $$ be the multiplicatively closed subset of $A$. Prove that $$S^{-1} A \cong \mathbb{R}[X, X^{-1}] $$ where $\mathbb{R}[X, X^{-1}]$ is the smallest subring of $\mathbb{R}(X)$ containing $\mathbb{R}, X$ and $X^{-1}$.

Attempt: I am really stuck here. I think the subring $\mathbb{R}[X, X^{-1}]$ looks like $$ \mathbb{R}[X, X^{-1}]= \left\{ \frac{f(X)}{X^n} \mid n \geq 0, f(X) \in \mathbb{R}[X] \right\}. $$ But I don't even know how to compute $S^{-1} A$, and hence how to define a map $\phi : S^{-1} A \rightarrow \mathbb{R}[X, X^{-1}]. $

Also, what I don't understand is, why it is necessary to divide out $\mathbb{R}[X,Y]$ by the ideal $(XY)$. For example, if we have $A^{'} = \mathbb{R}[X,Y]$ and $S^{'} = \left\{1, X, X^2, \ldots \right\}$, then $S^{' -1} A^{'} \cong \mathbb{R}[Y]$ ?

Answer 1

I think we can give an inverse homomorphism as follows: define a map $\psi\colon\mathbb{R} [X,X^{-1} ]\to S^{-1} A$ such that $\frac{f(X)}{X^k }\mapsto\frac{f(X)+(XY)}{X^k +(XY)} $, then a routine check shows that $\psi $ will be a ring homomorphism and what we need is that it is bijective.

Answer 2

You can define the map

$\psi: \mathbb{R}[X,X^{-1}]\to S^{-1}A$

that is the extension of the map such that $\psi(X)=(X+(XY)/ 1+(XY))$ and $\psi(X^{-1})=(1+(XY)/ X+(XY))$

You can observe that the map is surjective because $Y+(XY)$ is 0 in the localization of $A$ with respect $S$ because

$Y+(XY)=(XY)/X+(XY)=0$ because

$(Y+(XY))(X+(XY))=XY+(XY)=(XY)$