Kähler differentials commute with scalar base change

Restriction and extension of derivations. Let $M$ be a $B^{\mathrm{\prime }}$-module. Restriction along $b\mapsto b\otimes 1$ sends a derivation in ${\mathrm{Der}}_{A^{\mathrm{\prime }}}(B^{\mathrm{\prime }},M)$ to an element of ${\mathrm{Der}}_A(B,M)$, because the composite is additive, $A$-constant and satisfies Leibniz. Conversely, given $D\in {\mathrm{Der}}_A(B,M)$, the map ${\beta }_D\text{\hspace{0.17em}⁣}:B\times A^{\mathrm{\prime }}\to M$, ${\beta }_D(b,a^{\mathrm{\prime }}):=a^{\mathrm{\prime }}D(b)$, is $A$-bilinear: it is additive in each variable and ${\beta }_D(\alpha b,a^{\mathrm{\prime }})=a^{\mathrm{\prime }}\alpha D(b)={\beta }_D(b,\alpha a^{\mathrm{\prime }})$ for $\alpha \in A$. By [F2] it factors through a group homomorphism $\tilde{D}\text{\hspace{0.17em}⁣}:B^{\mathrm{\prime }}\to M$ with $\tilde{D}(b\otimes a^{\mathrm{\prime }})=a^{\mathrm{\prime }}D(b)$; this is $A^{\mathrm{\prime }}$-linear because $\tilde{D}((b\otimes a^{\mathrm{\prime }})a^{\mathrm{\prime }\mathrm{\prime }})=a^{\mathrm{\prime }}{a}^{\mathrm{\prime }\mathrm{\prime }}D(b)=a^{\mathrm{\prime }\mathrm{\prime }}\tilde{D}(b\otimes a^{\mathrm{\prime }})$, and it is a derivation, since $\tilde{D}((b\otimes a^{\mathrm{\prime }})(b^{\mathrm{\prime }}\otimes c^{\mathrm{\prime }}))=a^{\mathrm{\prime }}{c}^{\mathrm{\prime }}D(b{b}^{\mathrm{\prime }})=a^{\mathrm{\prime }}{c}^{\mathrm{\prime }}(bD(b^{\mathrm{\prime }})+b^{\mathrm{\prime }}D(b))=(b\otimes a^{\mathrm{\prime }})\tilde{D}(b^{\mathrm{\prime }}\otimes c^{\mathrm{\prime }})+(b^{\mathrm{\prime }}\otimes c^{\mathrm{\prime }})\tilde{D}(b\otimes a^{\mathrm{\prime }})$. Also $\tilde{D}(1\otimes a^{\mathrm{\prime }})=a^{\mathrm{\prime }}D(1)=0$, so $\tilde{D}$ is $A^{\mathrm{\prime }}$-constant. The two assignments are inverse: restriction of $\tilde{D}$ gives $b\mapsto D(b)$, and an extension of a restricted derivation agrees with $\tilde{D}$ on the pure tensors $b\otimes a^{\mathrm{\prime }}$, which generate $B^{\mathrm{\prime }}$ over $A^{\mathrm{\prime }}$ by [F3]. So restriction is a natural bijection ${\mathrm{Der}}_{A^{\mathrm{\prime }}}(B^{\mathrm{\prime }},M)\cong {\mathrm{Der}}_A(B,M)$ for every $B^{\mathrm{\prime }}$-module $M$.

The canonical map. The composite $B\to B^{\mathrm{\prime }}\stackrel{{\mathrm{d}}^{\mathrm{\prime }}}{\to }{\mathrm{\Omega }}_{B^{\mathrm{\prime }}\mathrm{/}{A}^{\mathrm{\prime }}}$ is an $A$-derivation of $B$ into the $B^{\mathrm{\prime }}$-module ${\mathrm{\Omega }}_{B^{\mathrm{\prime }}\mathrm{/}{A}^{\mathrm{\prime }}}$, so by [F1] it corresponds to a $B$-linear $\rho \text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{B\mathrm{/}A}\to {\mathrm{\Omega }}_{B^{\mathrm{\prime }}\mathrm{/}{A}^{\mathrm{\prime }}}$ with $\rho (\mathrm{d}b)={\mathrm{d}}^{\mathrm{\prime }}(b\otimes 1)$. The map ${\mathrm{\Omega }}_{B\mathrm{/}A}\times B^{\mathrm{\prime }}\to {\mathrm{\Omega }}_{B^{\mathrm{\prime }}\mathrm{/}{A}^{\mathrm{\prime }}}$, $(\omega ,b^{\mathrm{\prime }})\mapsto b^{\mathrm{\prime }}\rho (\omega )$, is $B$-balanced, so by [F2] it factors through a group homomorphism $\alpha \text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{B\mathrm{/}A}{\otimes }_B{B}^{\mathrm{\prime }}\to {\mathrm{\Omega }}_{B^{\mathrm{\prime }}\mathrm{/}{A}^{\mathrm{\prime }}}$ with $\alpha (\mathrm{d}b\otimes a^{\mathrm{\prime }})=a^{\mathrm{\prime }}{\mathrm{d}}^{\mathrm{\prime }}(b\otimes 1)$; it is $B^{\mathrm{\prime }}$-linear by construction.

The inverse map. Let $N:={\mathrm{\Omega }}_{B\mathrm{/}A}{\otimes }_B{B}^{\mathrm{\prime }}$, a $B^{\mathrm{\prime }}$-module, and let $D\text{\hspace{0.17em}⁣}:B\to N$ be $D(b):=\mathrm{d}b\otimes 1$; this is an $A$-derivation, since $b\mapsto \mathrm{d}b$ is one and $-\otimes 1$ is additive. By step 1.1 there is a unique $A^{\mathrm{\prime }}$-derivation $\tilde{D}\text{\hspace{0.17em}⁣}:B^{\mathrm{\prime }}\to N$ with $\tilde{D}(b\otimes 1)=D(b)$ and $\tilde{D}(b\otimes a^{\mathrm{\prime }})=a^{\mathrm{\prime }}(\mathrm{d}b\otimes 1)$. By [F1] applied to $A^{\mathrm{\prime }}\to B^{\mathrm{\prime }}$ it corresponds to a $B^{\mathrm{\prime }}$-linear map $\beta \text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{B^{\mathrm{\prime }}\mathrm{/}{A}^{\mathrm{\prime }}}\to N$ with $\beta ({\mathrm{d}}^{\mathrm{\prime }}(b\otimes a^{\mathrm{\prime }}))=a^{\mathrm{\prime }}(\mathrm{d}b\otimes 1)$.

The maps are inverse. On the one hand $\beta (\alpha (\mathrm{d}b\otimes a^{\mathrm{\prime }}))=\beta (a^{\mathrm{\prime }}{\mathrm{d}}^{\mathrm{\prime }}(b\otimes 1))=a^{\mathrm{\prime }}(\mathrm{d}b\otimes 1)=\mathrm{d}b\otimes a^{\mathrm{\prime }}$, and the elements $\mathrm{d}b\otimes a^{\mathrm{\prime }}$ generate ${\mathrm{\Omega }}_{B\mathrm{/}A}{\otimes }_B{B}^{\mathrm{\prime }}$ over $B^{\mathrm{\prime }}$ because the elements $\mathrm{d}b$ generate ${\mathrm{\Omega }}_{B\mathrm{/}A}$ over $B$; hence $\beta \circ \alpha =\mathrm{i}\mathrm{d}$. On the other hand $\alpha (\beta ({\mathrm{d}}^{\mathrm{\prime }}(b\otimes a^{\mathrm{\prime }})))=\alpha (a^{\mathrm{\prime }}(\mathrm{d}b\otimes 1))=a^{\mathrm{\prime }}{\mathrm{d}}^{\mathrm{\prime }}(b\otimes 1)={\mathrm{d}}^{\mathrm{\prime }}(b\otimes a^{\mathrm{\prime }})$, where the last equality uses $b\otimes a^{\mathrm{\prime }}=(b\otimes 1)(1\otimes a^{\mathrm{\prime }})$, the Leibniz rule and ${\mathrm{d}}^{\mathrm{\prime }}(1\otimes a^{\mathrm{\prime }})=0$; since the elements ${\mathrm{d}}^{\mathrm{\prime }}(b\otimes a^{\mathrm{\prime }})$ generate ${\mathrm{\Omega }}_{B^{\mathrm{\prime }}\mathrm{/}{A}^{\mathrm{\prime }}}$ over $B^{\mathrm{\prime }}$ by [F3] and [F4], we get $\alpha \circ \beta =\mathrm{i}\mathrm{d}$. Hence $\alpha$ is an isomorphism. The construction used only the given base-change maps, so no statement is made about an arbitrary ring map $B\to C$.