If $R=\mathbb{C}[x,y]$ is the polynomial ring in two variables $x$ and $y$ then we know that the localization of R at the multiplicative set $S=[1,x,x^2,x^3,...]$ is given by $R_x=\mathbb{C}[x,x^{1},y]$. Now, what will be the localization of $R$ at the prime ideal $(x)$. i.e. what will $R_{(x)}$ be?

1$\begingroup$ If you localize at a prime ideal $p$ in a ring $R$, then the multiplicative set that gets inverted is $S = R  p$. So for example, $x1$ is invertible in $R_{(x)}$. This is explained in any basic commutative algebra text. $\endgroup$– cfrancSep 24, 2010 at 13:28

$\begingroup$ It's funny that I have voted to close a question about localization as "too localized". $\endgroup$– Harry GindiSep 24, 2010 at 13:30

2$\begingroup$ I don't know if this will satisfy you, since it is just a restatement of the definition, but $R_{(x)}$ is the ring of rational functions on $\mathbb{C}^2$ which don't have a pole along the hyperplane $x=0$. $\endgroup$– David E SpeyerSep 24, 2010 at 14:14
1 Answer
I doubt that there is a nice description which will satisfy you. As a $\mathbb{C}$algebra, $R_{(x)}$ is not finitely generated. Anyway every localization of a factorial domain at a principle prime ideal is a discrete valuation domain. In particular, $R_{(x)}$ is such a domain with prime ideal $(x)$. Every nonzero element has a unique representation $x^n u$, where $u$ is a fraction, such that $x$ is coprime to both the numerator and the denominator of $u$. Probably an algebraic geometer would call this ring the local ring of the affine plane at the line $x=0$.