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.

Is the localization of a reduced ring (no nilpotents) still reduced?

share|improve this question
2  
Interestingly, while localization preserves reducedness (as Mariano has shown below), the somewhat analogous and in theory equally "inoffensive" operation of completion in general does not. One example is $\mathbb{C}[x]/(x^2+1)$ completed at $(x)$; then $x^2+1$ will have $n$th roots in the completion (by the binomial expansion). Nonetheless, if you complete most rings you encounter daily at ideals contained in the Jacobson radical, then reducedness will be preserved. (The condition you need is called excellence.) –  Akhil Mathew Oct 28 '10 at 5:43
1  
@Akhil: don't believe the hype -- although they are roughly analogous, completion is a significantly more powerful and less completely benign construction than localization. There are lots of tricky things that can happen if you look at completion in the general case. –  Pete L. Clark Oct 28 '10 at 6:21
2  
By the way, this is a question which (in most American universities at least) is at the beginning graduate level. As Mariano's answer shows, it has an answer which is straightforward and consists of little more than understanding the definitions. Thus I think it would have been better for the OP to solve it on his/her own. –  Pete L. Clark Oct 28 '10 at 6:25

2 Answers 2

up vote 9 down vote accepted

Let $A$ be a ring, $S\subset A$ a multiplicatively closed subset, and suppose that $0\neq a/b\in A_S$ is nilpotent. Then there exists $n$ such that $(a/b)^n=0$, i.e., such that there exists $t\in S$ with $ta^n=0$. But then $ta$ is nilpotent in $A$. If it is zero, then $a/b=0$ in $A_S$, which it isn't.

share|improve this answer

EDIT: This argument is incorrect but I feel others could learn from Mariano Suárez-Alvarez's comments so I've made the post CW.

If I understand correctly you are asking does reduce ring imply reduced localization.

Argue by contrapositive, assume that localization is not reduced, i.e. contains nilpotents. Since elements of the localization of our ring are of the form $\frac{r}{s}$ where s is a subset of our ring that does not contain 0. Then choose a nilpotent element in the localization $(\frac{r}{s})^{n}=0$ Since 0 is not in S it must be the case that $r^{n}=0$, i.e. r is nilpotent. Hence the original ring is nilpotent.

share|improve this answer
1  
This is not right: $r^n/s^n$ may be zero without $r^n$ being zero. An example, localize $k[x,y]/(xy)$ at $S=\{x^i:i\geq0\}$ and look at $y/x$. –  Mariano Suárez-Alvarez Oct 28 '10 at 5:36
    
@Mariano Suárez-Alvarez I must misunderstand something. I thought the zero element of the localization was 0/s. Can you point me to a reference to understand this better? –  WWright Oct 28 '10 at 5:39
    
@WWring: any textbook on commutative algebra which gives the construction of localization! That is the zero element: your problem is with the way equality is defined. –  Mariano Suárez-Alvarez Oct 28 '10 at 5:40
    
I see. I'm afraid I need to consult my books. –  WWright Oct 28 '10 at 5:48

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.