Lemma 15.43.8. Let $A \to B$ be a local homomorphism of Noetherian local rings. Then the induced map of completions $A^\wedge \to B^\wedge$ is flat if and only if $A \to B$ is flat.

Proof. Consider the commutative diagram

$\xymatrix{ A^\wedge \ar[r] & B^\wedge \\ A \ar[r] \ar[u] & B \ar[u] }$

The vertical arrows are faithfully flat. Assume that $A^\wedge \to B^\wedge$ is flat. Then $A \to B^\wedge$ is flat. Hence $B$ is flat over $A$ by Algebra, Lemma 10.39.9.

Assume that $A \to B$ is flat. Then $A \to B^\wedge$ is flat. Hence $B^\wedge /\mathfrak m_ A^ n B^\wedge$ is flat over $A/\mathfrak m_ A^ n$ for all $n \geq 1$. Note that $\mathfrak m_ A^ n A^\wedge$ is the $n$th power of the maximal ideal $\mathfrak m_ A^\wedge$ of $A^\wedge$ and $A/\mathfrak m_ A^ n = A^\wedge /(\mathfrak m_ A^\wedge )^ n$. Thus we see that $B^\wedge$ is flat over $A^\wedge$ by applying Algebra, Lemma 10.99.11 (with $R = A^\wedge$, $I = \mathfrak m_ A^\wedge$, $S = B^\wedge$, $M = S$). $\square$

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