**Proof.**
Let $F_\bullet \to A/\mathfrak m_ A$ be a resolution by finite free $A$-modules. Observe that $\text{Tor}^ A_ p(B, A/\mathfrak m_ A)$ is the $p$th homology of the complex $F_\bullet \otimes _ A B$. Let $F_\bullet ^\wedge = F_\bullet \otimes _ A A^\wedge $ be the completion. Then $F_\bullet ^\wedge $ is a resolution of $A^\wedge /\mathfrak m_{A^\wedge }$ by finite free $A^\wedge $-modules (as $A \to A^\wedge $ is flat and completion on finite modules is exact, see Algebra, Lemmas 10.97.1 and 10.97.2). It follows that

\[ F_\bullet ^\wedge \otimes _{A^\wedge } B^\wedge = F_\bullet \otimes _ A B \otimes _ B B^\wedge \]

By flatness of $B \to B^\wedge $ we conclude that

\[ \text{Tor}^{A^\wedge }_ p(B^\wedge , A^\wedge /\mathfrak m_{A^\wedge }) = \text{Tor}^ A_ p(B, A/\mathfrak m_ A) \otimes _ B B^\wedge \]

In this way we see that the condition in (1) on the local ring map $A \to B$ is equivalent to the same condition for the local ring map $A^\wedge \to B^\wedge $. Thus we may assume $A$ and $B$ are complete local Noetherian rings (since the other conditions are formulated in terms of the completions in any case).

Assume $A$ and $B$ are complete local Noetherian rings. Choose a diagram

\[ \xymatrix{ S \ar[r] & B \\ R \ar[u] \ar[r] & A \ar[u] } \]

as in More on Algebra, Lemma 15.39.3. Let $I = \mathop{\mathrm{Ker}}(R \to A)$ and $J = \mathop{\mathrm{Ker}}(S \to B)$. The proposition now follows from Lemma 23.7.6.
$\square$

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