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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.

baltazar
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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.

Martin Brandenburg
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  • Just one detail: for the "conversely" part you defined the map $a \otimes b \mapsto b \otimes d(a)$, but the first on the right should be an element in $C$, a tensor? Maybe $a \otimes b \mapsto (1 \otimes b) \otimes d(a)$ ? – baltazar Nov 29 '14 at 21:06
  • Yes. You could also identify $C \otimes_A \Omega^1_{A/R}$ with $B \otimes_R \Omega^1_{A/R}$ right away. – Martin Brandenburg Nov 29 '14 at 22:09
  • How do you identify them? Sorry for these too specific questions, it's now apparent to me that I didn't have enough practice with tensor products. – baltazar Nov 29 '14 at 23:23
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    There is a general "cancellation" law $(B \otimes_R A) \otimes_A M \cong B \otimes_R M|_R$ for $A$-modules $M$. This is easy to check. – Martin Brandenburg Nov 29 '14 at 23:44