Lemma 15.27.2. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $A$ be a set. Assume $R$ is Noetherian. The completion $(\bigoplus \nolimits _{\alpha \in A} R)^\wedge $ is a flat $R$-module.
Proof. Denote $R^\wedge $ the completion of $R$ with respect to $I$. As $R \to R^\wedge $ is flat by Algebra, Lemma 10.97.2 it suffices to prove that $(\bigoplus \nolimits _{\alpha \in A} R)^\wedge $ is a flat $R^\wedge $-module (use Algebra, Lemma 10.39.4). Since
\[ (\bigoplus \nolimits _{\alpha \in A} R)^\wedge = (\bigoplus \nolimits _{\alpha \in A} R^\wedge )^\wedge \]
we may replace $R$ by $R^\wedge $ and assume that $R$ is complete with respect to $I$ (see Algebra, Lemma 10.97.4). In this case Lemma 15.27.1 tells us the map $(\bigoplus \nolimits _{\alpha \in A} R)^\wedge \to \prod _{\alpha \in A} R$ is universally injective. Thus, by Algebra, Lemma 10.82.7 it suffices to show that $\prod _{\alpha \in A} R$ is flat. By Algebra, Proposition 10.90.6 (and Algebra, Lemma 10.90.5) we see that $\prod _{\alpha \in A} R$ is flat. $\square$
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