Lemma 15.42.8. Let $A \to B$ be a flat local homomorphism of Noetherian local rings such that $\mathfrak m_ A B = \mathfrak m_ B$ and $\kappa (\mathfrak m_ A) = \kappa (\mathfrak m_ B)$. Then $A \to B$ induces an isomorphism $A^\wedge \to B^\wedge $ of completions.

**Proof.**
By Algebra, Lemma 10.96.7 we see that $B^\wedge $ is the $\mathfrak m_ A$-adic completion of $B$ and that $A^\wedge \to B^\wedge $ is finite. Since $A \to B$ is flat we have $\text{Tor}_1^ A(B, \kappa (\mathfrak m_ A)) = 0$. Hence we see that $B^\wedge $ is flat over $A^\wedge $ by Lemma 15.27.5. Thus $B^\wedge $ is a free $A^\wedge $-module by Algebra, Lemma 10.77.4. Since $A^\wedge \to B^\wedge $ induces an isomorphism $\kappa (\mathfrak m_ A) = A^\wedge /\mathfrak m_ A A^\wedge \to B^\wedge /\mathfrak m_ A B^\wedge = B^\wedge /\mathfrak m_ B B^\wedge = \kappa (\mathfrak m_ B)$ by our assumptions (and Algebra, Lemma 10.95.3), we see that $B^\wedge $ is free of rank $1$. Thus $A^\wedge \to B^\wedge $ is an isomorphism.
$\square$

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