First fundamental Exact sequence

Question

I am reading matsumura, in that if $k \xrightarrow{f} A \xrightarrow{g}B$ are $k$ algebra morphisms. Then $$\Omega_{A/k}\otimes_A B \xrightarrow{\alpha} \Omega_{B/k} \xrightarrow{\beta} \Omega_{B/A} \to 0$$ is an exact sequence of $B$ modules. Where $\alpha(d_{A/k} a \otimes b) = b d_{B/k} g(a)$

How is this map well defined and a $B$ module morphism?

Answer 1

In general, one has an $A$-linear map $\Omega_{A/k}\to\Omega_{B/k}$ (see what's written in Section 00RM after Diagram 00RQ). Now use the fact that extension of scalars is left adjoint to restriction of scalars to obtain a $B$-linear map $\Omega_{A/k}\otimes_A B\xrightarrow{\alpha}\Omega_{B/k}$.