Multiplicative closed set for localization of $R=\mathbb{R}[x,y]$ at the prime ideal $P=(x,y)$

Question

The following is taken from Abstract Algebra: A First Course by Stephen Lovett

Background

Consider the ring $R=\mathbb{R}[x,y]$ and the prime ideal $P=(x,y)$. Prove that the elements in the localization $R_p$ are rational expressions of the form $$\frac{r(x,y)}{1+xp(x,y)+yq(x,y)} \quad\text{for }p(x,y),q(x,y),r(x,y)\in \mathbb{R}[x,y]$$

Question:

For the question in the background section above, I want to ask what the multiplicative closed set is suppose to be? I understand that the ring of fractions is suppose to be

$\mathrm{Frac}(R)=\{\frac{r(x,y)}{1+xp(x,y)+yq(x,y)}\mid r(x,y),1+xp(x,y)+yq(x,y)\in R, \wedge 1+xp(x,y)+yq(x,y)\not\in P=(x,y) \}$.

My guess for the multiplicative closed set would be we can let a subset $S\subset\mathbb{R}[x,y]$ consisting of all two variable polynomials with non zero constant terms and $S^{-1}R=\{\sum_{i,j}a_{ij}x_i^{\alpha_i}x_j^{\alpha_j}\mid \alpha_i, \alpha_j\in \mathbb{N}\}$.

I have not done much concrete examples for the topic of localization. Every time I see the phrase "localization at a prime" or anytime the word localization comes up, multiplicative closed set don't get mentioned even when most of the time, where there are mentions of examples tat are not from polynomial rings. And when it does make mentions of them, it often times is in the context of algebraic geometry. I want to get good at computational examples in the context of ring of fractions, basically nothing to do with algebraic geometry, like algebraic curves.

Thank you in advance

Answer 1

When you localize at a prime ideal $P$, the multiplicatively closed subset is the complement of $P$, i.e. $R_P:= S^{-1}R$ with $S=R\backslash P$.

In your specific example, $P=(x,y)$ consists of all polynomials with zero constant term, and the complement are those polynomials with non-zero constant term (clearly multiplicatively closed since $\mathbb{R}$ has no zero divisors). The form of the rational functions in your first display scales the denominator to have constant term one.