A killed first Tor obstruction yields flatness after local Noetherian base change

We record the first comparison. For ring maps $A\to B\to C$ and an $A$-module $N$ with $N{\otimes }_{A}B$ flat over $B$, the natural map ${\mathrm{Tor}}_1^A(B,N){\otimes }_{B}C\to {\mathrm{Tor}}_1^A(C,N)$ is surjective. Choose a free resolution $F_2\to F_1\to F_0\to N\to 0$, and put $K_B=\ker (F_1{\otimes }_{A}B\to F_0{\otimes }_{A}B)$. The sequence $0\to K_B\to F_1{\otimes }_{A}B\to F_0{\otimes }_{A}B\to N{\otimes }_{A}B\to 0$ remains exact through $F_1{\otimes }_{A}C$ after tensoring with $C$: flatness of $N{\otimes }_{A}B$ kills the first Tor obstruction, and the kernel of its free cover is flat by the long exact Tor sequence. Consequently $K_B{\otimes }_{B}C$ surjects onto $K_C=\ker (F_1{\otimes }_{A}C\to F_0{\otimes }_{A}C)$. First Tor is the quotient of these kernels by the image of $F_2$, so the claimed comparison is surjective.

We record the second comparison. For $A\to B$, an $A$-module $N$, and an ideal $J\subseteq B$, the natural map ${\mathrm{Tor}}_1^A(B\mathrm{/}J,N)\to {\mathrm{Tor}}_1^B(B\mathrm{/}J,N{\otimes }_{A}B)$ is surjective. Use the same free $A$-resolution $F_2\to F_1\to F_0\to N$. After tensoring it with $B$, $F_1{\otimes }_{A}B\to F_0{\otimes }_{A}B\to N{\otimes }_{A}B\to 0$ is still exact. Add a free $B$-module in degree $2$ to kill any extra kernel of its degree-one map, obtaining a free $B$-resolution of $N{\otimes }_{A}B$ through degree $2$. Upon reducing both complexes modulo $J$, their degree-one cycles are the same, while the $B$-resolution has at least the boundaries from the $A$-resolution. Thus its degree-one homology is a quotient of the latter, proving surjectivity.

Put $T={\mathrm{Tor}}_1^R(R\mathrm{/}I,M)$ and $T^{\mathrm{\prime }}={\mathrm{Tor}}_1^{R^{\mathrm{\prime }}}(R^{\mathrm{\prime }}\mathrm{/}{I}^{\mathrm{\prime }},M^{\mathrm{\prime }})$. Apply step 1.1 to $R\to R\mathrm{/}I\to R^{\mathrm{\prime }}\mathrm{/}{I}^{\mathrm{\prime }}$; the required flatness of $M{\otimes }_{R}R\mathrm{/}I=M\mathrm{/}IM$ over $R\mathrm{/}I$ is a hypothesis. It makes $T{\otimes }_{R\mathrm{/}I}{R}^{\mathrm{\prime }}\mathrm{/}{I}^{\mathrm{\prime }}\to {\mathrm{Tor}}_1^R(R^{\mathrm{\prime }}\mathrm{/}{I}^{\mathrm{\prime }},M)$ surjective. Step 1.2, with $A=R$, $B=R^{\mathrm{\prime }}$, $J=I^{\mathrm{\prime }}$, then makes the map from this last Tor group onto ${\mathrm{Tor}}_1^{R^{\mathrm{\prime }}}(R^{\mathrm{\prime }}\mathrm{/}{I}^{\mathrm{\prime }},M{\otimes }_R{R}^{\mathrm{\prime }})$ surjective. Since $M^{\mathrm{\prime }}=M{\otimes }_S{S}^{\mathrm{\prime }}$ is a localization of $M{\otimes }_R{R}^{\mathrm{\prime }}$ as a module over $S{\otimes }_R{R}^{\mathrm{\prime }}$, exact localization of a free $R^{\mathrm{\prime }}$-resolution identifies $T^{\mathrm{\prime }}$ with the corresponding localization of this final Tor group. Hence the natural map from $T$ to $T^{\mathrm{\prime }}$ has image generating $T^{\mathrm{\prime }}$ as an $S^{\mathrm{\prime }}$-module. The assumed zero map therefore forces $T^{\mathrm{\prime }}=0$.

The quotient $M^{\mathrm{\prime }}\mathrm{/}{I}^{\mathrm{\prime }}{M}^{\mathrm{\prime }}$ is a localization of $(M\mathrm{/}IM){\otimes }_{R\mathrm{/}I}(R^{\mathrm{\prime }}\mathrm{/}{I}^{\mathrm{\prime }})$, so it is flat over $R^{\mathrm{\prime }}\mathrm{/}{I}^{\mathrm{\prime }}$ by scalar extension and localization [F2]. The ring $S^{\mathrm{\prime }}$ is Noetherian local and $M^{\mathrm{\prime }}$ is finite over it. By [F2], $T^{\mathrm{\prime }}=0$ means $I^{\mathrm{\prime }}{\otimes }_{R^{\mathrm{\prime }}}{M}^{\mathrm{\prime }}\to M^{\mathrm{\prime }}$ is injective. Thus [F1] applies to $R^{\mathrm{\prime }}\to S^{\mathrm{\prime }}$ and $I^{\mathrm{\prime }}$ and gives that $M^{\mathrm{\prime }}$ is flat over $R^{\mathrm{\prime }}$.

If $M^{\mathrm{\prime }}=0$, the conclusion is immediate and the same Tor argument still applies. AC is inherited by [F1] and by the use of free resolutions in [F2]; every generator selection in steps 1.1–2.1 is finite at the degree being used. [F1, F2, step 3.1] $\mathrm{\square }$