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Let $I=[x,y]$ be the prime ideal generated by the polynomials $x,y$ with real coefficients and let $R_I$ be the localization of the ring $R=\mathbb{R}[x,y]$ in $I$. Can someone help my to visualize/to interpret the elements of $R_I$ as rational functions on $\mathbb{R}^2$ which are defined in 0 and to determine the domain of these rational functions?

In particular I don't understand the part with defined in $0$. Isn't it the case, that all elements of $R_I$ just sets of rational functions, where the numerator is some polynomial of $R=\mathbb{R}[x,y]$ and the denominator just a constant polynomial ? And isn't the domain just $\mathbb{R}$?

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up vote 4 down vote accepted

First, the localization $R_I$ is not a quotient (as your title suggests). In this case, $R$ injects into $R_I$. (Warning: the localization map is not always injective!)

$R_I$ is precisely what you said at the end of your first paragraph; $R_I$ consists of rational functions $f(x,y)/g(x,y)$ such that $g(0,0) \neq 0$. The latter condition is what it means for the rational function $f(x,y)/g(x,y)$ to be defined at $(0,0)$. You should check that the sum and product of two rational functions defined at $(0,0)$ is a rational function defined at $(0,0)$. The domain of such a rational function consists of those $(x,y)$ where $g(x,y) \neq 0$.

The way to think about elements of $R_I$ is that they are (algebraic) germs of functions at $(0,0)$. That is, each rational function in $R_I$ is defined in some neighborhood of $(0,0)$ (exercise: prove this); however, there is no single neighborhood of $(0,0)$ on which all of the functions in $R_I$ are defined.

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