Let $R=\mathbb{R}[x,y]/(xy)$ and $P\subset R$ be the ideal generated by $x-1$. Show that
$$R_P\cong \mathbb{R}[x]_{(x-1)},$$
where $R_P$ is the localization of $R$ at $R\setminus P$ (since $R/P$ is a field, $P$ is maximal, hence prime) and $\mathbb{R}[x]_{(x-1)}$ is the localization of $\mathbb{R}[x]$ at $\{(x-1)^n:n\geq 1\}$.
Now, the proof starts as: $y\in P$, since $-y=xy-y=(x-1)y$, but $x\not\in P$. Thus $R_P\cong (R/y)/(P/y)$. I don't really get this. I guess, before this we have $R_P\cong R/P$ but I can't find an isomorphism. The proof continues.
$R/y$ is isomorphic to $\mathbb{R}[x]$ and $P/y$ is isomorphic to $\mathbb{R}[x](x-1)$.
Hence, $$R_P\cong \mathbb{R}[x]/\mathbb{R}[x](x-1)\cong \mathbb{R}[x]_{(x-1)}.$$ But I also don't understand this last isomorphism. Any help is appreciated. Thanks in advance!
A bad habit (especially when one works with quotient rings of polynomial rings) is to denote the indeterminates by small letters. So, let us write $R=K[X,Y]/(XY)$ as it should be. (Here $K$ is an arbitrary field.) Denote the residue classes of $X$ and $Y$ by $x$ and $y$, and thus $R=K[x,y]$. In this ring one considers the maximal ideal $P=(x-1)$. (In order to show that $P$ is maximal we consider $R/P=K[x,y]/(x-1)=\frac{K[X,Y]/(XY)}{(X-1,XY)/(XY)}$ which is isomorphic to $K[X,Y]/(X-1,XY)$. It is not hard to show that the later is isomorphic to $K$.)
Now come back to the main question: in fact, $R_P=K[x,y]_{(x-1)}$ is $(K[X,Y]/(XY))_{(X-1,XY)/(XY)}$, which at its turn is isomorphic to $K[X,Y]_{(X-1,XY)}/(XY)K[X,Y]_{(X-1,XY)}$. But $(XY)K[X,Y]_{(X-1,XY)}=(Y)K[X,Y]_{(X-1,XY)}$ since $X\notin(X-1,XY)$. Therefore we get $R_P\simeq K[X,Y]_{(X-1,XY)}/(Y)K[X,Y]_{(X-1,XY)}\simeq K[X]_{(X-1)}$.