Remark 60.6.9. Consider a commutative diagram of rings

$\xymatrix{ B \ar[r]_\varphi & B' \\ A \ar[u] \ar[r] & A' \ar[u] }$

Let $\Omega _{B/A} \to \Omega$ and $\Omega _{B'/A'} \to \Omega '$ be quotients satisfying the assumptions of Algebra, Lemma 10.132.1. Assume there is a map $\varphi : \Omega \to \Omega '$ which fits into a commutative diagram

$\xymatrix{ \Omega _{B/A} \ar[r] \ar[d] & \Omega _{B'/A'} \ar[d] \\ \Omega \ar[r]^{\varphi } & \Omega ' }$

where the top horizontal arrow is the canonical map $\Omega _{B/A} \to \Omega _{B'/A'}$ induced by $\varphi : B \to B'$. In this situation, given any pair $(M, \nabla )$ where $M$ is a $B$-module and $\nabla : M \to M \otimes _ B \Omega$ is a connection we obtain a base change $(M \otimes _ B B', \nabla ')$ where

$\nabla ' : M \otimes _ B B' \longrightarrow (M \otimes _ B B') \otimes _{B'} \Omega ' = M \otimes _ B \Omega '$

is defined by the rule

$\nabla '(m \otimes b') = \sum m_ i \otimes b'\text{d}\varphi (b_ i) + m \otimes \text{d}b'$

if $\nabla (m) = \sum m_ i \otimes \text{d}b_ i$. If $\nabla$ is integrable, then so is $\nabla '$, and in this case there is a canonical map of de Rham complexes (Remark 60.6.8)

60.6.9.1
\begin{equation} \label{crystalline-equation-base-change-map-complexes} M \otimes _ B \Omega ^\bullet \longrightarrow (M \otimes _ B B') \otimes _{B'} (\Omega ')^\bullet = M \otimes _ B (\Omega ')^\bullet \end{equation}

which maps $m \otimes \eta$ to $m \otimes \varphi (\eta )$.

Comment #4914 by Rubén Muñoz--Bertrand on

At the beginning aren't we supposed to read $\Omega_{B'/A'}\to\Omega'$ instead of $\Omega_{B'/A}\to\Omega'$?

There are also:

• 7 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).