Lemma 15.88.8. Assume $\varphi : R \to S$ is a flat ring map and $I \subset R$ is a finitely generated ideal such that $R/I \to S/IS$ is an isomorphism. Let $M$, $N$ be $R$-modules. Assume $M$ is $I$-power torsion. Given an short exact sequence

$0 \to N \otimes _ R S \to \tilde E \to M \otimes _ R S \to 0$

there exists a commutative diagram

$\xymatrix{ 0 \ar[r] & N \ar[r] \ar[d] & E \ar[r] \ar[d] & M \ar[r] \ar[d] & 0 \\ 0 \ar[r] & N \otimes _ R S \ar[r] & \tilde E \ar[r] & M \otimes _ R S \ar[r] & 0 }$

with exact rows.

Proof. As $M$ is $I$-power torsion we see that $M \otimes _ R S = M$, see Lemma 15.88.2. We will use this identification without further mention. As $R \to S$ is flat, the base change functor is exact and we obtain a functorial map of $\mathop{\mathrm{Ext}}\nolimits$-groups

$\mathop{\mathrm{Ext}}\nolimits _ R(M, N) \longrightarrow \mathop{\mathrm{Ext}}\nolimits _ S(M \otimes _ R S, N \otimes _ R S),$

see Homology, Lemma 12.7.3. The claim of the lemma is that this map is surjective when $M$ is $I$-power torsion. In fact we will show that it is an isomorphism. By Lemma 15.87.2 we can find a surjection $M' \to M$ with $M'$ a direct sum of modules of the form $R/I^ n$. Using the long exact sequence of Homology, Lemma 12.6.4 we see that it suffices to prove the lemma for $M'$. Using compatibility of $\mathop{\mathrm{Ext}}\nolimits$ with direct sums (details omitted) we reduce to the case where $M = R/I^ n$ for some $n$.

Let $f_1, \ldots , f_ t$ be generators for $I^ n$. By Lemma 15.88.6 we have a commutative diagram

$\xymatrix{ 0 \ar[r] & \mathop{\mathrm{Ext}}\nolimits _ R(R/I^ n, N) \ar[r] \ar[d] & H_1(N, f_\bullet ) \ar[r] \ar[d] & \mathop{\mathrm{Hom}}\nolimits _ R(K, N) \ar[d] \\ 0 \ar[r] & \mathop{\mathrm{Ext}}\nolimits _ S(S/I^ nS, N \otimes S) \ar[r] & H_1(N \otimes S, f_\bullet ) \ar[r] & \mathop{\mathrm{Hom}}\nolimits _ S(K \otimes S, N \otimes S) }$

with exact rows where $K$ is as in Lemma 15.88.5. Hence it suffices to prove that the two right vertical arrows are isomorphisms. Since $K$ is annihilated by $I^ n$ we see that $\mathop{\mathrm{Hom}}\nolimits _ R(K, N) = \mathop{\mathrm{Hom}}\nolimits _ S(K \otimes _ R S, N \otimes _ R S)$ by Lemma 15.88.3. As $R \to S$ is flat we have $H_1(N, f_\bullet ) \otimes _ R S = H_1(N \otimes _ R S, f_\bullet )$. As $H_1(N, f_\bullet )$ is annihilated by $I^ n$, see Lemma 15.88.7 we have $H_1(N, f_\bullet ) \otimes _ R S = H_1(N, f_\bullet )$ by Lemma 15.88.2. $\square$

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