# minimal prime devisor(MinAss R)

Hello All,is This conclusion true? if $(R,m)$ be a local ring & $Min Ass R=Ass R$ then we Can 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 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.

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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
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 –  anna Feb 23 '12 at 8:49
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.