Lemma 15.89.9. Assume \varphi : R \to S is a flat ring map and I = (f_1, \ldots , f_ t) \subset R is an ideal such that R/I \to S/IS is an isomorphism. Let M be an R-module. Then the complex (15.89.8.1) is exact.
Proof. First proof. Denote \check{\mathcal{C}}_ R \to \check{\mathcal{C}}_ S the quasi-isomorphism of extended alternating Čech complexes of Lemma 15.89.4. Since these complexes are bounded with flat terms, we see that M \otimes _ R \check{\mathcal{C}}_ R \to M \otimes _ R \check{\mathcal{C}}_ S is a quasi-isomorphism too (Lemmas 15.59.7 and 15.59.12). Now the complex (15.89.8.1) is a truncation of the cone of the map M \otimes _ R \check{\mathcal{C}}_ R \to M \otimes _ R \check{\mathcal{C}}_ S and we win.
Second computational proof. Let m \in M. If \alpha (m) = 0, then m \in M[I^\infty ], see Lemma 15.88.3. Pick n such that I^ n m = 0 and consider the map \varphi : R/I^ n \to M. If m \otimes 1 = 0, then \varphi \otimes 1_ S = 0, hence \varphi = 0 (see Lemma 15.89.3) hence m = 0. In this way we see that \alpha is injective.
Let (m', m'_1, \ldots , m'_ t) \in \mathop{\mathrm{Ker}}(\beta ). Write m'_ i = m_ i/f_ i^ n for some n > 0 and m_ i \in M. We may, after possibly enlarging n assume that f_ i^ n m' = m_ i \otimes 1 in M \otimes _ R S and f_ j^ nm_ i - f_ i^ nm_ j = 0 in M. In particular we see that (m_1, \ldots , m_ t) defines an element \xi of H_1(M, (f_1^ n, \ldots , f_ t^ n)). Since H_1(M, (f_1^ n, \ldots , f_ t^ n)) is annihilated by I^{tn + 1} (see Lemma 15.89.7) and since R \to S is flat we see that
by Lemma 15.89.2 The existence of m' implies that \xi maps to zero in the last group, i.e., the element \xi is zero. Thus there exists an m \in M such that m_ i = f_ i^ n m. Then (m', m'_1, \ldots , m'_ t) - \alpha (m) = (m'', 0, \ldots , 0) for some m'' \in (M \otimes _ R S)[(IS)^\infty ]. By Lemma 15.89.3 we conclude that m'' \in M[I^\infty ] and we win. \square
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