Question

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.

Answer 1

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.