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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?

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up vote 3 down vote accepted

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 $u\in U$.

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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.

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Doesn't a ring being reduced mean that it contains no nilpotents? I think the statement that "if $A$ is Noetherian then its total quotient ring is the product of the localizations of $A$ at its minimal primes" holds without the assumption of $A$ being reduced, since the total quotient ring is both Noetherian and dimension 0, so it is Artinian, and thus a product of local Artinian rings. Of course, in the case of $A$ is reduced, these local Artinian rings will be fields, which is the case here. – Jiangwei Xue Apr 30 '11 at 12:51
@Jiangwei: Dear Jianwei, If a Noetherian ring $A$ has nilpotents, it need not be the case that the total quotient ring of $A$ is Artinian, because there can be zero divisors whose support is not a union of irreducible components (or, equivalently, which are not contained in any minimal prime). Consider e.g. the example $\mathbb C[x,y]/(xy, y^2).$ In short, it is necessary to restrict to a reduced ring (or, at least, a ring which satisfies Serre's condition $S_0$) in order for the total quotient ring to be of dimension zero, and hence Artinian. Regards, – Matt E May 1 '11 at 1:19
Indeed I was wrong. For some reason I thought the set of zero divisors are the union of the minimal primes. As you have pointed out clearly, this is not the case in general. – Jiangwei Xue May 5 '11 at 6:13

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