An isomorphism about localisations Let $R=\mathbb{R}[x,y]/(xy)$ and $P\subset R$ the ideal generated by $x-1$. Show that $R_P\cong \mathbb{R}[x]_{(x-1)}$.
My attempt: first I tried to use the isomorphism between $\mathbb{R}[x,y]/(xy)$ and $\mathbb{R}[x]\oplus y\mathbb{R}[y]$, but this is a vector spaces isomorphism, so I can't consider in $\mathbb{R}[x]\oplus y\mathbb{R}[y]$ the ideal corresponding to the ideal $(x-1)$. 
Then I tried to use the universal property of localisation, but I need to define a map $R\rightarrow \mathbb{R}[x]_{(x-1)}$ which send elements not in $(x-1)$ in units, and here I have some difficulties... 
 A: Consider the composition series $$\mathbb R[x]\overset{i}{\longrightarrow}\mathbb R[x,y]\overset{\eta}{\longrightarrow}\mathbb R[x,y]/(xy)\overset{\text{loc}}{\longrightarrow}R_p$$ Where $i:\mathbb R[x]\to\mathbb R[x,y]$ is the unique inclusion mapping $x\mapsto x$, where $\eta:\mathbb R[x,y]\to\mathbb R[x,y]/(xy)$ is the canonical quotient morphism, and where $\text{loc}:\mathbb R[x,y]/(xy)\to R_p$ is the localization map. Let $f:\mathbb R[x]\to S$ be an arbitrary ring morphism such that $\forall q\in \mathbb R[x]\setminus (x-1),\ f(q)\in S^{\times}$.
By the universal properties of $\mathbb R[x]$ and $\mathbb R[x,y]$, there exists a unique map $f':\mathbb R[x,y]\to S$ such that $y\mapsto 0$ and $f'\circ i=f$.
Observe from the above that $f'(xy)=0$. By the universal property of $\mathbb R[x,y]/(xy)$, there exists a unique map $f'':\mathbb R[x,y]/(xy)\to S$ such that $f''\circ \eta=f'$.
Now note that $\eta(y)$ is in the ideal of $\mathbb R[x,y]/(xy)$ generated by $\eta(x-1)$. It follows that $\forall q\in \mathbb R[x,y]/(xy)\setminus (\eta(x-1)),\ f(q)\in S^{\times}$. By the universal property of $R_p$, there exists a unique map $f''':R_p\to S$ such that $f'''\circ \text{loc}=f''$.
So for each ring morphism $f:\mathbb R[x]\to S$ satisfying $\forall q\in \mathbb R[x]\setminus (x-1),\ f(q)\in S^{\times}$, there exists a unique lift $f''':R_p\to S$ such that $f'''\circ \left(\text{loc}\circ\eta\circ i\right)=f$. In other words, $R_p$ together with the map ${\text{loc}\circ\eta\circ i}:\mathbb R[x]\to R_p$ satisfy the universal property of $\mathbb R[x]_{(x-1)}$. 
We conclude that $R_p\cong \mathbb R[x]_{(x-1)}$ $\blacksquare$
