2
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ What does the equality "Min Ass R = Ass R" exactly mean ? $\endgroup$
    – Ralph
    Feb 23, 2012 at 3:42
  • $\begingroup$ It would be nice to have a definition of MinAss here... $\endgroup$ Feb 23, 2012 at 3:43
  • 2
    $\begingroup$ MinAss means minimal primes in Ass(R). "Min Ass R = Ass R" means R has no embedded prime ideals. $\endgroup$ Feb 23, 2012 at 4:04
  • $\begingroup$ MinAss means minimal primes in Ass(R). "Min Ass R = Ass R" means R has no embedded prime ideals $\endgroup$
    – Stella
    Feb 23, 2012 at 8:49

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ oh dear, I don't understated language of this paper. Can you explain it more. $\endgroup$
    – Stella
    Feb 24, 2012 at 13:14
  • 1
    $\begingroup$ 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/… $\endgroup$ Feb 24, 2012 at 14:23
  • 1
    $\begingroup$ 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. $\endgroup$ Feb 24, 2012 at 16:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.