Quotient of a polynomial ring localized

Question

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.

Answer 1

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.

Answer 2

Consider the ring homomorphism $\psi: \mathbb{R[x]}_{(x-a)}\rightarrow\left(\mathbb{R[x,y]}/(xy)\right)_{(x-a)}$ given by inclusion of rational functions $\frac{p(x)}{q(x)}$, where $(x-a)\nmid q(x)$ as this is an element of $\mathbb{R[x]}_{(x-a)}$. This is obviously a ring homomorphism and injective, so we just need to show it is a surjection

Edit to be more specific about inclusions of rational functions: $ \left(\mathbb{R[x,y]}/(xy)\right)_{(x-a)}$ consists of fractions so that the numerator and denominator are polynomials in $x$ and $y$. By "inculsion" I mean "$\frac{p(x)}{q(x)}\in \mathbb{R[x]}_{(x-a)}$ is also an element of $ \left(\mathbb{R[x,y]}/(xy)\right)_{(x-a)}$ so it's image under $\psi$ is just itself". To show it is injective assume that $\frac{p(x)}{q(x)}=0$ and let $n$ be the lowest power of $x$ in $p(x)$ with coefficent $c_n$. $\frac{1}{x^n}$ is a unit in $\left(\mathbb{R[x,y]}/(xy)\right)_{(x-a)}$ so we find $\frac{p(x)}{q(x)}=0 \leftrightarrow \frac{1}{x^n} \frac{p(x)}{q(x)}=0 \leftrightarrow up(x)x^{-n}=0$ for some $u$ in $\left(\mathbb{R[x,y]}/(xy)\right)_{(x-a)} \leftrightarrow c_n=0$, and we can repeat this to conclude all coeffeicents of $p(x)$ are zero, so $\frac{p(x)}{q(x)}$ is zero as required

Let $\frac{p(x,y)}{q(x,y)}\in\left(\mathbb{R[x,y]}/(xy)\right)_{(x-a)}$. Then we know 1) $(x-a)\nmid q(x)$ and 2) $xy\nmid q(x,y)$. The polynomial $x$ is invertible in this ring so we find $\frac{p(x,y)}{q(x,y)}=1.\frac{p(x,y)}{q(x,y)}=\frac{xp(x,y)}{xq(x,y)}=\frac{\tilde{p}(x)}{\tilde{q}(x)}$ as all '$y$'s will go to zero. This is an element in the image of $\psi$, so $\psi$ is surjective and we are done (I think). The second question is a total mess for me so I wont bother writing it..