TITLE: Quotient of a polynomial ring localized
QUESTION [5 upvotes]: Question: 

Prove that $\mathbb{R}[x,y]/(xy)$ localised at $(x-a)$ is isomorphic to the ring $\mathbb{R}[x]$ localised at $(x-a)$.

Related question: 

What is the local ring at the point $(0,0)$, i.e. the ring $\mathbb{R}[x,y]/(xy)$ localized at $(x,y)$? 

I haven't been able to figure it out.
Edit: many thanks to anyone who answers but my main motivation in posting this is to  see that I am doing things properly so I would much appreciate an answer on my solution saying if I have understood something incorrectly or even a downvote/upvote.

REPLY [3 votes]: First recall that localisation commutes with quotients in the following sense. Suppose $R$ is a commutative ring and $S$ a multiplicatively closed subset. If $\mathfrak{a} \triangleleft R$, then $S^{-1}R/S^{-1}\mathfrak{a} \cong \bar{S}^{-1}(R/\mathfrak{a})$, where $\bar{S}$ is the image of $S$ in $R/\mathfrak{a}$.
 Here we are given that $R=\mathbb{R}[X,Y]$, $ \mathfrak{a}=(XY) $, and $\bar S= ~(x-a)^c.$ Thus $S=(XY,X-a)^c$. Now in the localisation $S^{-1}R$, the ideal $(XY)=(Y)$ and $(XY,X-a)=(Y,X-a)$ as $X$ has now become invertible. Thus, \begin{align}\biggl(\mathbb{R}[X,Y]/(XY)\biggr)_{(x-a)}&=\bar S^{-1}\left(R/\mathfrak a\right)\\&\simeq S^{-1}R/S^{-1}
\mathfrak{a}\\&\simeq\mathbb{R}[X,Y]_{(Y,X-a)}/(Y)_{(Y,X-a)}\\&\simeq\mathbb{R}[X]_{(X-a)}\end{align}
The last inequality again employing the commutativity of localisation with quotients.