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Suppose, $(R,m)$ is a Noetherian ring that is not a domain. Can $\hat{R}_m$ be a domain?

I think this cannot be the case for if $a,b\in R$ s.t. $a,b\neq 0$ and $ab=0$. Then, this $R$ is a Noetherian local ring, by the Krull intersection theorem, $\cap_i m^i=0$ and the LHS of this is the kernel of the canonical map $R\to \hat{R}_m$. So this map is injective. So $(a,a,a,...)$ and $(b,b,b,...)$ are non-zero elements of $\hat{R}_m$ whose product of zero.

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Your proof is completely correct. Just for fun and because it seems related, Theorem 7.9 from Eisenbud says that if $R$ is a local ring with maximal ideal $m$ that is a localization of a ring finitely generated over a field or $\mathbb{Z}$, then $\hat{R}$ has no nilpotent elements. – Jesko Hüttenhain Dec 2 '11 at 16:38

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