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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?

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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, $x-1$ is invertible in $R_{(x)}$. This is explained in any basic commutative algebra text. – cfranc Sep 24 '10 at 13:28
It's funny that I have voted to close a question about localization as "too localized". – Harry Gindi Sep 24 '10 at 13:30
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 $x=0$. – David Speyer Sep 24 '10 at 14:14
up vote 7 down vote accepted

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 $x=0$.

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