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Let $\hat{R}$ is $m$-adic completion of a local ring $(R,m)$.What is the relation between $Min R$ and $Min \hat{R}$. we know that $\hat{R}$ is faithfully flat $R$-module. $Min R$=set of all minimal prime divisors of zero. I think

$p\in Min(R)$ iff $\hat{p}=p\hat{R}\in Min\hat{R}$.

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If $p$ is a minimal prime of $R$, then $p \hat{R}$ need not be prime. See… –  Dustin Cartwright May 12 '12 at 8:53

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What you are sayinng is not true in general. For example, the completion of a domain may not be a domain. An example is given in exercises of Bourbaki's Commutative Algebra.

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To be more precise, the example in Bourbaki exercises is the following: $K$ is an algebraically closed field of char $0$, $R=K[X,Y]$, $p=X(X^2+Y^2)+(X^2-Y^2)$. Then one can show the ideal $pR$ is prime. So $R/pR$ is a domain. If $\mathfrak{m}$ is the maximal ideal of $R/pR$ generated by $X$ and $Y$, then the exercise asks to show that the $\mathfrak{m}$-adic completion of $R/pR$ is not a domain. The given hint is to show that $p$ decomposes into a product of two formal power series in $K[[X,Y]]$. –  Mahdi Majidi-Zolbanin May 15 '12 at 3:12

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