The Stacks project

Proof. Recall that $B$ is the quotient of $A[x_2, \ldots , x_ t]/(a_1x_2 - a_2, \ldots , a_1x_ t - a_ t)$ by its $a_1$-torsion, see Algebra, Lemma 10.70.6. Let

viewed as a chain complex sitting in degrees $(t - 1), \ldots , 0$. The complex $B_\bullet [1/a_1]$ is isomorphic to the Koszul complex on $x_2 - a_2/a_1, \ldots , x_ t - a_ t/a_1$ which is a regular sequence in $A[1/a_1][x_2, \ldots , x_ t]$. Since regular sequences are Koszul regular, we conclude that the augmentation

is a quasi-isomorphism after inverting $a_1$. Since the homology modules of the cone $C_\bullet$ on $\epsilon$ are finite $A[x_2, \ldots , x_ n]$-modules and since $C_\bullet$ is bounded, we conclude that there exists a $c \geq 0$ such that $a_1^ c$ annihilates all of these. By Derived Categories, Lemma 13.12.5 this implies that, after possibly replacing $c$ by a larger integer, that $a_1^ c$ is zero on $C_\bullet$ in $D(A)$. The proof is finished once the reader contemplates the distinguished triangle

Namely, the first term is represented by $B_\bullet \otimes _ A M$ which is sitting in homological degrees $(t - 1), \ldots , 0$ in view of the fact that the terms in the Koszul complex $B_\bullet$ are free (and hence flat) $A$-modules. Whence $\text{Tor}_ i^ A(B, M) = H_ i(C_\bullet \otimes _ A^\mathbf {L} M)$ for $i > t - 1$ and this is annihilated by $a_1^ c$. Since $a_1^ cB = I^ cB$ and since the tor module is a module over $B$ we conclude. $\square$