The Stacks project

Lemma 15.59.6. Let $R$ be a ring. Let $0 \to K_1^\bullet \to K_2^\bullet \to K_3^\bullet \to 0$ be a short exact sequence of complexes. If $K_3^ n$ is flat for all $n \in \mathbf{Z}$ and two out of three of $K_ i^\bullet $ are K-flat, so is the third.

Proof. Let $L^\bullet $ be a complex of $R$-modules. Then

\[ 0 \to \text{Tot}(L^\bullet \otimes _ R K_1^\bullet ) \to \text{Tot}(L^\bullet \otimes _ R K_2^\bullet ) \to \text{Tot}(L^\bullet \otimes _ R K_3^\bullet ) \to 0 \]

is a short exact sequence of complexes. Namely, for each $n, m$ the sequence of modules $0 \to L^ n \otimes _ R K_1^ m \to L^ n \otimes _ R K_2^ m \to L^ n \otimes _ R K_3^ m \to 0$ is exact by Algebra, Lemma 10.39.12 and the sequence of complexes is a direct sum of these. Thus the lemma follows from this and the fact that in a short exact sequence of complexes if two out of three are acyclic, so is the third. $\square$

Comments (0)

There are also:

  • 4 comment(s) on Section 15.59: Derived tensor product

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Your email address will not be published.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 0BYH. Beware of the difference between the letter 'O' and the digit '0'.