Completion is flat

Remarks

By the Artin–Rees lemma, completion $M \mapsto \hat M$ is an exact functor on finitely generated $R$-modules. Both $M \mapsto \hat R \otimes_R M$ and $M \mapsto \hat M$ are right exact and agree on finite free modules, so they agree on finitely presented modules, which gives clause 1. Then for an ideal $J$ of $R$ the map $\hat R \otimes_R J \to \hat R$ is the injective map $\hat J \to \hat R$, which gives clause 2 by the ideal criterion for flatness. For clause 3, each maximal ideal $\mathfrak n$ contains $I$, so $\hat R/\mathfrak n \hat R \cong R/\mathfrak n \neq 0$. Clause 3 can fail without its hypothesis. The $p$-adic completion $\Z_p$ of $\Z$ is flat over $\Z$, but $\Z_p \otimes_\Z \Z/q\Z = 0$ for a prime $q \neq p$, because $q$ is a unit in $\Z_p$. So $\Z_p$ is not faithfully flat over $\Z$.

Going deeper

For a student

Flatness lets one move properties between a Noetherian local ring $R$ and its completion $\hat R$. For instance $\dim \hat R = \dim R$, and $R$ is regular, or Cohen–Macaulay, if and only if $\hat R$ is. Combined with the Cohen structure theorem, this reduces many questions about local rings to quotients of power series rings.