Lemma 10.75.1. Given a flat ring map $R \to R'$ and $R$-modules $M$, $N$ the natural $R$-module map $\text{Tor}_ i^ R(M, N)\otimes _ R R' \to \text{Tor}_ i^{R'}(M \otimes _ R R', N \otimes _ R R')$ is an isomorphism for all $i$.

## 10.75 Functorialities for Tor

In this section we briefly discuss the functoriality of $\text{Tor}$ with respect to change of ring, etc. Here is a list of items to work out.

Given a ring map $R \to R'$, an $R$-module $M$ and an $R'$-module $N'$ the $R$-modules $\text{Tor}_ i^ R(M, N')$ have a natural $R'$-module structure.

Given a ring map $R \to R'$ and $R$-modules $M$, $N$ there is a natural $R$-module map $\text{Tor}_ i^ R(M, N) \to \text{Tor}_ i^{R'}(M \otimes _ R R', N \otimes _ R R')$.

Given a ring map $R \to R'$ an $R$-module $M$ and an $R'$-module $N'$ there exists a natural $R'$-module map $\text{Tor}_ i^ R(M, N') \to \text{Tor}_ i^{R'}(M \otimes _ R R', N')$.

**Proof.**
Omitted. This is true because a free resolution $F_\bullet $ of $M$ over $R$ stays exact when tensoring with $R'$ over $R$ and hence $(F_\bullet \otimes _ R N)\otimes _ R R'$ computes the Tor groups over $R'$.
$\square$

The following lemma does not seem to fit anywhere else.

Lemma 10.75.2. Let $R$ be a ring. Let $M = \mathop{\mathrm{colim}}\nolimits M_ i$ be a filtered colimit of $R$-modules. Let $N$ be an $R$-module. Then $\text{Tor}_ n^ R(M, N) = \mathop{\mathrm{colim}}\nolimits \text{Tor}_ n^ R(M_ i, N)$ for all $n$.

**Proof.**
Choose a free resolution $F_\bullet $ of $N$. Then $F_\bullet \otimes _ R M = \mathop{\mathrm{colim}}\nolimits F_\bullet \otimes _ R M_ i$ as complexes by Lemma 10.11.9. Thus the result by Lemma 10.8.8.
$\square$

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