# Kaehler differentials of tensor products over another ring

Is this result well-known (or even true)? Let \begin{align*} f:A & \to B\\ g:A & \to C \end{align*} be homomorphisms of finitely generate $k$-algebras and let

$R=B\otimes_{k}C$ and $S=B\otimes_{A}C$

Then there is a surjective homomorphism $$\xi:R\to S$$ and $$\Omega_{S/k}=S\otimes_{B}\Omega_{B/k}\oplus_{\Omega_{A/k}}S\otimes_{C}\Omega_{C/k}$$ Where the $\oplus$ with a subscript represents a push-out.

• why do you think that it is true? maybe you can kindly give a non-trivial example? – user74900 Mar 26 '18 at 17:17
• That is true, but it follows from two other more primitive results. First, $\Omega_{B/A}$, resp. $\Omega_{C/A}$, is the cokernel of the natural map $B\otimes_A \Omega_{A/k} \to \Omega_{B/k}$, resp. $C\otimes_A \Omega_{A/k} \to \Omega_{C/k}$, cf. Proposition II.8.3A, p. 173, Algebraic geometry, R. Hartshorne. Second, $\Omega_{S/B}$ is isomorphic to $S\otimes_C \Omega_{C/A}$, resp. $\Omega_{S/A}$ is isomorphic to $S\otimes_B \Omega_{B/A}$, cf. Proposition 8.2A. Finally, using Proposition 8.3A once more, the result follows. – Jason Starr Mar 26 '18 at 17:29
• Thank you! My proof simply uses the conormal exact sequence corresponding to the map $\xi$. – Justin Smith Mar 26 '18 at 19:18