Remark 91.12.6. Let $A \to B$ be a ring map. Let $P_\bullet$ be a resolution of $B$ over $A$ (Remark 91.5.5). Set $J_ n = \mathop{\mathrm{Ker}}(P_ n \to B)$. Note that

$\text{Tor}_2^{P_ n}(B, B) = \text{Tor}_1^{P_ n}(J_ n, B) = \mathop{\mathrm{Ker}}(J_ n \otimes _{P_ n} J_ n \to J_ n^2).$

Hence $H_2(L_{B/A})$ is canonically equal to

$\mathop{\mathrm{Coker}}(\text{Tor}_2^{P_1}(B, B) \to \text{Tor}_2^{P_0}(B, B))$

by Remark 91.11.5. To make this more explicit we choose $P_2$, $P_1$, $P_0$ as in Example 91.5.9. We claim that

$\text{Tor}_2^{P_1}(B, B) = \wedge ^2(\bigoplus \nolimits _{t \in T} B)\ \oplus \ \bigoplus \nolimits _{t \in T} J_0\ \oplus \ \text{Tor}_2^{P_0}(B, B)$

Namely, the basis elements $x_ t \wedge x_{t'}$ of the first summand corresponds to the element $x_ t \otimes x_{t'} - x_{t'} \otimes x_ t$ of $J_1 \otimes _{P_1} J_1$. For $f \in J_0$ the element $x_ t \otimes f$ of the second summand corresponds to the element $x_ t \otimes s_0(f) - s_0(f) \otimes x_ t$ of $J_1 \otimes _{P_1} J_1$. Finally, the map $\text{Tor}_2^{P_0}(B, B) \to \text{Tor}_2^{P_1}(B, B)$ is given by $s_0$. The map $d_0 - d_1 : \text{Tor}_2^{P_1}(B, B) \to \text{Tor}_2^{P_0}(B, B)$ is zero on the last summand, maps $x_ t \otimes f$ to $f \otimes f_ t - f_ t \otimes f$, and maps $x_ t \wedge x_{t'}$ to $f_ t \otimes f_{t'} - f_{t'} \otimes f_ t$. All in all we conclude that there is an exact sequence

$\wedge ^2_ B(J_0/J_0^2) \to \text{Tor}_2^{P_0}(B, B) \to H^{-2}(L_{B/A}) \to 0$

In this way we obtain a direct proof of a consequence of Quillen's spectral sequence discussed in Remark 91.12.5.

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