Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hello All,is This conclusion true?

If $(R,m)$ is a local ring and $ Min Ass R=Ass R$ then can we conclude that $Min Ass \hat{R}=Ass \hat{R}$? ($\hat{R}$ is $m$-adic completion of $R$)

$MinAss$ means minimal primes in $Ass(R)$. "$Min Ass R = Ass R$" means that $R$ has no embedded prime ideals. In fact, if every associated prime ideal of $R$ is minimal then every associated prime ideal of $\hat{R}$ is minimal?

share|improve this question
    
What does the equality "Min Ass R = Ass R" exactly mean ? –  Ralph Feb 23 '12 at 3:42
    
It would be nice to have a definition of MinAss here... –  darij grinberg Feb 23 '12 at 3:43
2  
MinAss means minimal primes in Ass(R). "Min Ass R = Ass R" means R has no embedded prime ideals. –  Mahdi Majidi-Zolbanin Feb 23 '12 at 4:04
    
MinAss means minimal primes in Ass(R). "Min Ass R = Ass R" means R has no embedded prime ideals –  Stella Feb 23 '12 at 8:49

1 Answer 1

up vote 4 down vote accepted

The answer is no in general. In the paper Fibres formelles d'un anneau local noethérien D. Ferrand and M. Raynaud give an example of a two-dimensional local domain whose $\mathfrak{m}$-adic completion has embedded prime ideals. In the same paper, they mention that the answer is yes in certain special cases, such as, when $R$ is a quotient of a Cohen-Macaulay ring, or when $R$ is universally Japanese.

share|improve this answer
    
oh dear, I don't understated language of this paper. Can you explain it more. –  Stella Feb 24 '12 at 13:14
1  
If by language you mean French, then you can look at the paper Local domains with bad sets of formal prime divisors By Brodmann and Rotthaus: sciencedirect.com.rpa.laguardia.edu:2048/science/article/pii/… –  Mahdi Majidi-Zolbanin Feb 24 '12 at 14:23
1  
The link I posted above doesn't seem to work. Brodmann and Rotthaus paper is published in Journal of Algebra, Volume 75, Issue 2, April 1982, Pages 386–394. –  Mahdi Majidi-Zolbanin Feb 24 '12 at 16:01

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.