Remark 92.12.5. In the situation of Theorem 92.12.4 let I = \mathop{\mathrm{Ker}}(A \to B). Then H^{-1}(L_{B/A}) = H_1(\mathcal{C}_{B/A}, \Omega ) = I/I^2, see Lemma 92.11.2. Hence H_ k(\mathcal{C}_{B/A}, \text{Sym}^ k(\Omega )) = \wedge ^ k_ B(I/I^2) by Remark 92.12.2. Thus the E_1-page looks like
\begin{matrix} B
\\ 0
\\ 0
& I/I^2
\\ 0
& H^{-2}(L_{B/A})
\\ 0
& H^{-3}(L_{B/A})
& \wedge ^2(I/I^2)
\\ 0
& H^{-4}(L_{B/A})
& H_3(\mathcal{C}_{B/A}, \text{Sym}^2(\Omega ))
\\ 0
& H^{-5}(L_{B/A})
& H_4(\mathcal{C}_{B/A}, \text{Sym}^2(\Omega ))
& \wedge ^3(I/I^2)
\end{matrix}
with horizontal differential. Thus we obtain edge maps \text{Tor}_ i^ A(B, B) \to H^{-i}(L_{B/A}), i > 0 and \wedge ^ i_ B(I/I^2) \to \text{Tor}_ i^ A(B, B). Finally, we have \text{Tor}_1^ A(B, B) = I/I^2 and there is a five term exact sequence
\text{Tor}_3^ A(B, B) \to H^{-3}(L_{B/A}) \to \wedge ^2_ B(I/I^2) \to \text{Tor}_2^ A(B, B) \to H^{-2}(L_{B/A}) \to 0
of low degree terms.
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