Let $B$ and $A'$ be commutative $A$ algebras, and set $B'=B\otimes_AB'$. I am trying to show that as $B$ modules, the modules of Kähler differentials satisfy:
$$\Omega_{B/A}\otimes_AA'\cong \Omega_{B'/A'}$$
Now, via the base change property of tensor products, we clearly have that:
$$\Omega_{B/A}\otimes_AA'\cong \Omega_{B/A}\otimes_BB'=\Omega_{B/A}\otimes_B(B\otimes A')$$
so it suffices to provide an isomorphism $\Omega_{B/A}\otimes_BB'\rightarrow \Omega_{B'/A'}$.
Note that $B'$ is an $A'$ algebra via the inclusion $\iota_{A}:a'\mapsto 1\otimes a'$, and that there $B'\otimes A'$ is a $B$ algebra via the natural inclusion $\iota_B:b\mapsto b\otimes 1$. It follows by the universal property of Kähler differentials that there exists a $B$-module homomorphism $\xi:\Omega_{B/A}\rightarrow \Omega_{B'/A'}$ satisfiying:
\begin{align}
\xi\circ d_{B/A}=d_{B'/A'}\circ \iota_B
\end{align}
By the universal property of the tensor product, we have an obvious map $B$-linear map:
\begin{align}
f:\Omega_{B/A}\otimes_BB'&\longrightarrow \Omega_{B'/A'}\\
x\otimes b'&\longmapsto b'\cdot \xi(x)
\end{align}
but showing this an isomorphism is tricky. Since $\Omega_{B'/A'}$ is generated by $d_{B'/A'}(B')$, and $A'$ acts kinda stupidly on $B'$, it's easy to see that $\Omega_{B'/A'}$ is generated by elements of the form $d_{B'/A'}(b\otimes 1)=\xi(d_{B/A}(b))$, so $f$ is clearly surjective.
Showing $f$ has proven trickier, and I'm not quite sure how to do it. I thought that since $\Omega_{B'/A'}$ is generated by $d_{B'/A'}(b\otimes 1)$, I could define an inverse by:
\begin{align}
\sum_ib_i'\cdot d_{B'/A'}(b_i\otimes 1)\longmapsto \sum_id_{B/A}(b_i)\otimes b_i'
\end{align}
but showing that this well defined has proven just as tricky.
Is there anyway to do this succinctly? When trying to show the isomorphism directly via $\Omega_{B/A}\otimes_AA'$ I ran into the same injectivity problems.
Best Answer
Kahler differentials have a universal property, and this is the key to doing what you ask succinctly. The universal property is that for a ring morphism $\varphi: R\to S$ and an $S$-module $M$, the map $\operatorname{Hom}_S(\Omega_{S/R},M)\to \operatorname{Der}_R(S,M)$ by $\alpha\mapsto \alpha\circ d$ is an isomorphism of functors (where $d:S\to\Omega_{S/R}$ is the map $s\mapsto ds$).
This is done at Stacks 00RV; here is an outline of the proof.
In your notation, write $d':B'=B\otimes_A A' \to \Omega_{B/A}\otimes_A A'$ for the map $d'(\sum(b_i \otimes a_i))= \sum d(b_i)\otimes a_i$, which exists as $B\times A'\to \Omega_{B/A} \otimes_A A'$ by $(b,a')\mapsto d(b)\otimes_A a'$ is $A$-bilinear. It's not difficult to check this is an $A'$-derivation too. For an $A'$-derivation $D:B'\to M'$ with $M'$ a $B'$-module, the composition $B\to B'\to M'$ is an $A$-derivation, so we get an $A$-linear map $\varphi_D:\Omega_{B/A}\to M'$, which tensors up to a map $\varphi_D':\Omega_{B/A}\otimes_A A' \to M'$ satisfying $D=\varphi_D'\circ d'$. Therefore these two pairs satisfy the same universal property, so they're isomorphic up to unique isomorphism.