The Stacks project

Remark 60.6.7. Let $A \to B$ be a ring map and let $(J, \delta )$ be a divided power structure on $B$. Set $\Omega _{B/A, \delta }^ i = \wedge ^ i_ B \Omega _{B/A, \delta }$ where $\Omega _{B/A, \delta }$ is the target of the universal divided power $A$-derivation $\text{d} = \text{d}_{B/A} : B \to \Omega _{B/A, \delta }$. Note that $\Omega _{B/A, \delta }$ is the quotient of $\Omega _{B/A}$ by the $B$-submodule generated by the elements $\text{d}\delta _ n(x) - \delta _{n - 1}(x)\text{d}x$ for $x \in J$. We claim Algebra, Lemma 10.132.1 applies. To see this it suffices to verify the elements $\text{d}\delta _ n(x) - \delta _{n - 1}(x)\text{d}x$ of $\Omega _ B$ are mapped to zero in $\Omega ^2_{B/A, \delta }$. We observe that

\[ \text{d}(\delta _{n - 1}(x)) \wedge \text{d}x = \delta _{n - 2}(x) \text{d}x \wedge \text{d}x = 0 \]

in $\Omega ^2_{B/A, \delta }$ as desired. Hence we obtain a divided power de Rham complex

\[ \Omega ^0_{B/A, \delta } \to \Omega ^1_{B/A, \delta } \to \Omega ^2_{B/A, \delta } \to \ldots \]

which will play an important role in the sequel.


Comments (0)

There are also:

  • 8 comment(s) on Section 60.6: Module of differentials

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Your email address will not be published.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 07HZ. Beware of the difference between the letter 'O' and the digit '0'.