Lemma 91.8.3. Let $A \to B$ be a ring map such that $\text{Tor}^ A_ i(B, B) = 0$ for $i > 0$ and such that $L_{B/B \otimes _ A B} = 0$. Then $L_{B/A} = 0$ in $D(B)$.

Proof. By Lemma 91.6.2 we see that $L_{B/A} \otimes _ B^\mathbf {L} (B \otimes _ A B) = L_{B \otimes _ A B/B}$. Now we use the distinguished triangle (91.7.0.1)

$L_{B \otimes _ A B/B} \otimes ^\mathbf {L}_{(B \otimes _ A B)} B \to L_{B/B} \to L_{B/B \otimes _ A B} \to L_{B \otimes _ A B/B} \otimes ^\mathbf {L}_{(B \otimes _ A B)} B[1]$

associated to the ring maps $B \to B \otimes _ A B \to B$ and the vanishing of $L_{B/B}$ (Lemma 91.4.7) and $L_{B/B \otimes _ A B}$ (assumed) to see that

$0 = L_{B \otimes _ A B/B} \otimes ^\mathbf {L}_{(B \otimes _ A B)} B = L_{B/A} \otimes _ B^\mathbf {L} (B \otimes _ A B) \otimes ^\mathbf {L}_{(B \otimes _ A B)} B = L_{B/A}$

as desired. $\square$

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