Remark 60.6.9. Consider a commutative diagram of rings

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

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

is defined by the rule

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)

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

## Comments (2)

Comment #4914 by Rubén Muñoz--Bertrand on

Comment #5184 by Johan on

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