Hello All,is This conclusion true?



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> 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 $R$)
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$MinAss$ 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?



Minimal prime divisors (MinAss R)
The answer is *no* in general. In the paper *[Fibres formelles d'un anneau local noethérien](http://archive.numdam.org/ARCHIVE/ASENS/ASENS_1970_4_3_3/ASENS_1970_4_3_3_295_0/ASENS_1970_4_3_3_295_0.pdf)* 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.


