Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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 $u\in U$.

share|improve this answer

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.

share|improve this answer
    
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

Your Answer

 
discard

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.