The Stacks project

10.76 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.

  1. 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.

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

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

Lemma 10.76.1. 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 $i$.

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 10.76.2. 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 10.12.9. Thus the result by Lemma 10.8.8. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00M7. Beware of the difference between the letter 'O' and the digit '0'.