Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathbb{R}$?

share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.