My task is to understand the following example:
Let $R=\mathbb{R}[x,y]/(xy).$ Let $p=(x-1)$ be the ideal of $R$. Show that $(x-1)$ is a maximal ideal of $R$ and deduce that the localization at this maximal ideal $R_{p} \cong$ $\mathbb{R}[x]_{(x-1)}$. What does this geometrically mean?
My attempt:
$\mathbb{R}[x,y]/(xy)/(x-1) \cong \mathbb{R}[y]/(y) \cong \mathbb{R}$ which is a field. So the ideal $(x-1)$ is maximal hence prime. $R_{p}=R_{R\setminus p}=\left\{\frac{f(x,y)}{g(x,y)}\in\mathbb{R}[x,y]/(xy): g(x,y) \text{ is not divisible by the polynomial $x-1$}\right\}$. So we are looking at rational functions in $R=\mathbb{R}[x,y]/(xy)$ which are well defined in a neighborhood of the point $(1,0)$ and also at the point $(1,0)$. Is my argument valid?
$R$ is the ring of polynomial functions defined on the algebraic variety $V$ given by equation
$$xy = 0$$
The natural morphism $\mathbb{R}[x,y]\rightarrow R$ representing the restriction of functions. This variety is equal to the union of two irreducible varieties: one given by $x= 0$ (the vertical axis), and the other by $y=0$ (horizontal axis). So, $V$ is just a cross (pair of intersecting lines).
The ideal $(x-1)$ of $R$ is just the (maximal) ideal $(x-1,xy)$ of $\mathbb{R}[x,y]$ modulo $(xy)$. It corresponds to the intersection of the vertical line $x-1=0$ with the cross, which is the point $(1,0)$.
In a neighborhood of the point $(1,0)$, which is on the horizontal component of the cross, $V$ coincides with the line $y=0$ (the ring of functions on that line is just the polynomial ring in one variable $\mathbb{R}[x,y]/(y) \cong \mathbb{R}[x]$), and that is the geometric reason why the localization of $R$ (the cross) at the point $p$ is the same as the localization of $\mathbb{R}[x]$ (the horizontal line) at the point $x=1$ (because those varieties are the same near that point).
The proof of the isomorphism $R_p \cong \mathbb{R}[x]_{(x-1)}$ is as follows. Let $f$ be the localization morphism $f: R \rightarrow R_p$. The image of $x$ in this morphism is invertible because $x$ is not in $(x-1)$. So, from the relation $xy= 0$ (true in $R$) we get the same relation in $R_p$, where it implies $y = 0$ (because $x$ is invertible here).
This shows that $(y)\subset \ker f$, but the only non trivial ideals that contain $(y)$ are maximal ideals, and since $\ker f$ is not maximal (because the image $R_p$ is not a field), this means that $(y) = \ker f$.
By now, it should be clear that $R_p$ consists precisely of polynomials in $x$ (because $y = 0$) with the additional property that each polynomial not in $(x-1)$ is invertible. This ring is then equal to $\mathbb{R}[x]_{(x-1)}$. This is the algebraic reason.