Lemma 15.27.5. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume

$I$ is finitely generated,

$R/I$ is Noetherian,

$M/IM$ is flat over $R/I$,

$\text{Tor}_1^ R(M, R/I) = 0$.

Then the $I$-adic completion $R^\wedge $ is a Noetherian ring and $M^\wedge $ is flat over $R^\wedge $.

**Proof.**
By Algebra, Lemma 10.98.8 the modules $M/I^ nM$ are flat over $R/I^ n$ for all $n$. By Algebra, Lemma 10.95.3 we have (a) $R^\wedge $ and $M^\wedge $ are $I$-adically complete and (b) $R/I^ n = R^\wedge /I^ nR^\wedge $ for all $n$. By Algebra, Lemma 10.96.5 the ring $R^\wedge $ is Noetherian. Applying Lemma 15.27.4 we conclude that $M^\wedge = \mathop{\mathrm{lim}}\nolimits M/I^ nM$ is flat as an $R^\wedge $-module.
$\square$

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