Proving $C\otimes_A\Omega^1_{A/R} \cong \Omega^1_{C/B}$

Question

I am completely stuck on this so any help would be great.

Let $R$ be a commutative ring and let $A$ and $B$ be $R$-algebras. Let $C:=A\otimes_RB$. Show that $C\otimes_A\Omega^1_{A/R} \cong \Omega^1_{C/B}$ as $C$-modules.

Answer 1

One uses universal properties. The $R$-derivation $A \xrightarrow{can} C \xrightarrow{d} \Omega^1_{C/B}~|_A$ extends to some $A$-linear map $\Omega^1_{A/R} \to \Omega^1_{C/B}~|_A$ and hence to some $C$-linear map $C \otimes_A \Omega^1_{A/R} \to \Omega^1_{C/B}$. Conversely, one checks that the $R$-linear map $C \to C \otimes_A \Omega^1_{A/R}$ defined by $a \otimes b \mapsto b \otimes d(a)$ is actually a $B$-derivation and therefore extends to some $C$-linear map $\Omega^1_{C/B} \to C \otimes_A \Omega^1_{A/R}$. The two constructed maps are inverse to each other. For a more conceptual proof which doesn't use any calculations, and even works in arbitrary tensor categories, see Corollary 4.5.5 in my thesis.