The Stacks project

Lemma 15.95.6. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet $ be a complex of $f$-torsion free $A$-modules. There is a canonical map of complexes

\[ \eta _ fM^\bullet \otimes _ A A/fA \longrightarrow H^\bullet (M^\bullet /f) \]

which is a quasi-isomorphism where the right hand side is the complex (15.95.5.2).

Proof. Let $x \in (\eta _ fM)^ i$. Then $x = f^ ix' \in f^ iM$ and $d^ i(x) = f^{i + 1}y \in f^{i + 1}M^{i + 1}$. Thus $d^ i$ maps $x' \otimes f^ i$ to zero in $M^{i + 1} \otimes f^ iA/f^{i + 1}A$. All tensor products are over $A$ in this proof. Hence we may map $x$ to the class of $x' \otimes f^ i$ in $H^ i(M^\bullet \otimes f^ iA/f^{i + 1}A)$. It is clear that this rule defines a map

\[ (\eta _ fM)^ i \otimes A/fA \longrightarrow H^ i(M^\bullet \otimes f^ iA/f^{i + 1}A) \]

of $A/fA$-modules. Observe that in the situation above, we may view $x' \otimes f^ i$ as an element of $M^ i \otimes f^ iA/f^{i + 2}A$ with differential $d^ i(x' \otimes f^ i) = y \otimes f^{i + 1}$. By the construction of $\beta $ above we find that $\beta (x' \otimes f^ i) = y \otimes f^{i + 1}$ and we conclude that our maps are compatible with differentials, i.e., we have a map of complexes.

To finish the proof, we observe that the construction given in the previous paragraph agrees with the maps $(\eta _ fM)^ i \otimes A/fA \to Z^ i/B^ i$ discussed in Remark 15.95.5. Since we have seen that the kernel of these maps is an acyclic subcomplex of $\eta _ fM^\bullet \otimes A/fA$, the lemma is proved. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 15.95: An operator introduced by Berthelot and Ogus

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Your email address will not be published.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 0F7T. Beware of the difference between the letter 'O' and the digit '0'.