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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