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The Stacks project

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

H_1(M, (f_1^ n, \ldots , f_ t^ n)) = H_1(M, (f_1^ n, \ldots , f_ t^ n)) \otimes _ R S = H_1(M \otimes _ R S, (f_1^ n, \ldots , f_ t^ n))

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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