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