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.

There are also:

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

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).