Completion of the homomorphic image of a Cohen-Macaulay ring

Suppose $R=S/I$ is the homomorphic image of a local Cohen-Macaulay ring $(S,m)$. I know the completion of $R$ as an $S$-module is just $R \otimes \hat S$, where $\hat S$ is the $m$-adic completion of $S$. Is this also the completion of $R$ as a ring? (I think this comes down to showing that the inverse limit of $S/(m^i +I)$ is the same as the (inverse limit of $S/m^i$) tensored with $S/I$.)

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1 Answer

If $S$ is any local ring with maximal ideal $m_S$, and $R = S/I$ is a quotient of $S$, then the maximal ideal $m_R$ of $R$ is $m_S/I$, and its $i$th power (for any $i \geq 0$) is equal to $m_S^i + I /I$.

The $m_R$-adic completion of $R$ is equal to

inverse limit of $R/m_R^i$ = inverse limit of $S/(I + m_S^n)$.

The completion of $R$ as an $S$-module is equal to

inverse limit of $R\otimes_S (S/m_S^i)$ = inverse limit of $S/(I +m_s^i)$

(which by the Artin--Rees theorem is the same as $R\otimes_S \hat{S}$).

Thus the two notions of completion are indeed the same (and there is no need to assume that $S$ is Cohen--Macaulay).

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