Lemma 10.98.12. Let $R \to R' \to R''$ be ring maps. Let $M$ be an $R$-module. Suppose that $M \otimes _ R R'$ is flat over $R'$. Then the natural map $\text{Tor}_1^ R(M, R') \otimes _{R'} R'' \to \text{Tor}_1^ R(M, R'')$ is onto.

**Proof.**
Let $F_\bullet $ be a free resolution of $M$ over $R$. The complex $F_2 \otimes _ R R' \to F_1\otimes _ R R' \to F_0 \otimes _ R R'$ computes $\text{Tor}_1^ R(M, R')$. The complex $F_2 \otimes _ R R'' \to F_1\otimes _ R R'' \to F_0 \otimes _ R R''$ computes $\text{Tor}_1^ R(M, R'')$. Note that $F_ i \otimes _ R R' \otimes _{R'} R'' = F_ i \otimes _ R R''$. Let $K' = \mathop{\mathrm{Ker}}(F_1\otimes _ R R' \to F_0 \otimes _ R R')$ and similarly $K'' = \mathop{\mathrm{Ker}}(F_1\otimes _ R R'' \to F_0 \otimes _ R R'')$. Thus we have an exact sequence

By the assumption that $M \otimes _ R R'$ is flat over $R'$, the sequence

is still exact. This means that $K' \otimes _{R'} R'' \to K''$ is surjective. Since $\text{Tor}_1^ R(M, R')$ is a quotient of $K'$ and $\text{Tor}_1^ R(M, R'')$ is a quotient of $K''$ we win. $\square$

## 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Your email address will not be published.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.

## Comments (0)