Localization of $\mathbb{C}[x,y]/(x^{3}-y^{3})$ Consider the ring $R=\mathbb{C}[x,y]/(x^{3}-y^{3})$ and let $S$ be the set of all non-zero divisors of $R$. How to find $S^{-1}A$?
I guess the idea is to find a ring which is isomorphic to (or perhaps that contains a subring isomorphic to $\mathbb{C}[x,y]/(x^{3}-y^{3})$) but not sure what is one. Can you please help?
 A: Some general remarks which might help:
The localization of a ring $A$ at the set of all non-zero divisors is called the total qoutient ring of $A$.  If $A$ is Noetherian and reduced (i.e. contains no non-zero divisors) then it is naturally isomorphic to the product of the localizations of $A$ at its (finite set of) minimal primes.

Applying these to your case:
Your ring has three minimal primes, and so the localization you want will be a product of three fields.  
A: For any $f\in\mathbb{C}[x,y]$, let $\bar{f}\in R$ be the image of $f$ in $R$ under the quotient map. Here are some things that may help you get started:


*

*The prime factorization of $x^3-y^3$ in $\mathbb{C}[x,y]$ is $(x-y)(x-\zeta_3 y)(x-\zeta_3^2 y)$.

*$\bar{f}=\bar{0}$ if and only if $f\in(x^3-y^3)$ .

*$\bar{f}\in R$ is a zero-divisor if and only if $f$ ______ (fill in the blank).

*For any ring $A$ and multiplicatively closed set $U$, the elements of $U^{-1}A$ are just things of the form $\frac{a}{u}$ where $a\in A$ is arbitrary and $u\in U$.
