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