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I was trying to prove the following statements:

Let $R \rightarrow R'$ be a commutative ring homomorphism and $n \geq 1$ be an integer. Let $F$ be an $R$-module such that $\mathrm{Tor}_i^R(R',F)=0$ for all $1 \leq i \leq n$. Then: 1) for any $R'$-module $C$ and all $0 \leq i \leq n$ there is a natural isomorphism $\mathrm{Ext}_R^i(F,C) \cong \mathrm{Ext}_{R'}^i(R' \otimes F,C)$; 2) for any $R'$-module $C$ and all $0 \leq i \leq n$ there is a natural isomorphism $\mathrm{Tor}^R_i(F,C) \cong \mathrm{Tor}^{R'}_i(R' \otimes F,C)$.

My attempt was the following:

Let $P_{n+1} \rightarrow P_n \rightarrow \dots \rightarrow P_1 \rightarrow P_0$ be the initial fragment of a projective resolution of the $R$-module $F$. Then $R' \otimes P_{n+1} \rightarrow R' \otimes P_n \rightarrow \dots \rightarrow R' \otimes P_1 \rightarrow R' \otimes P_0$ is the initial fragment of a projective resolution of the $R'$-module $R' \otimes F$. Indeed, the fact that the $R'$-modules appearing in the sequence are projective comes simply from the fact that extensions of scalars preserve projectivity, as $\mathrm{Hom}_{R'}(R' \otimes P_i, -) \cong \mathrm{Hom}_R(P_i, -)$. The fact that the augmented complex $P_\bullet \rightarrow F \rightarrow 0$ stays exact up to spot $n$ after tensoring with $R'$ is instead ensured by the assumption. In fact, call $K_{i+1} := \mathrm{Ker}(P_i \rightarrow P_{i-1})$ for all $i \geq 1$, $K_1:=\mathrm{Ker}(P_0\rightarrow F)$ and $K_0:=F$ the syzygies of $F$; then for every $0 \leq i \leq n$ the short exact sequence $0 \rightarrow K_{i+1} \rightarrow P_i \rightarrow K_i \rightarrow 0$ stays exact after tensoring with $R'$ because for all $0 \leq i \leq n$ one has $\mathrm{Tor}_1^R(R',K_i) \cong \mathrm{Tor}_{i+1}^R(R',F)=0$, using the assumption. Moreover, the exact sequence $P_{n+1} \rightarrow K_n \rightarrow 0$ stays exact since tensor product is right exact. So gluing together the short exact sequences above for $0 \leq i \leq n$ we build the desired acyclic complex. Now, at least for $1 \leq i \leq n$, we can compute \begin{align*} \mathrm{Ext}_{R'}^i(R' \otimes F,C) & = H^i(\mathrm{Hom}_{R'}(R' \otimes P_\bullet,C)) \cong \\ & \cong H^i(\mathrm{Hom}_R(P_\bullet,C)) = \mathrm{Ext}_R^i(F,C) \end{align*} and similarly for $\mathrm{Tor}$.

But I realized that I would need also $\mathrm{Tor}_R^{n+1}(R',F)=0$ as assumption. So now I am stuck figuring out what's actually wrong: am I missing something or is $\mathrm{Tor}_R^{n+1}(R',F)=0$ really needed?

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