Kernel of the induced map $\bar\phi:\Omega_{S/R}\to\Omega_{S'/R'}$

Question

$\DeclareMathOperator{\Lin}{Lin}$ $\DeclareMathOperator{\Ima}{Im}$ I have the same problem as the OP in this thread, however I additionaly looking for some simpler proof of one fact. Lemma 10.130.6 from this Stacks Project site makes one claim too quickly for me.

Consider following diagram:

$$ \require{AMScd} \begin{CD} \Omega_{S/R}@>{\bar\phi}>>\Omega_{S'/R'} \\ @AA{d_{S/R}}A @AA{d_{S'/R'}}A\\ S @>{\phi}>> S'\\ @AA{\alpha}A @AA{\beta}A\\ R @>{\psi}>> R' \end{CD} $$

where the lower square is given and the upper one is induced. Stacks Project constructs $\bar\phi$ directely, but one can easly do so using universal property of $(\Omega_{S/R},d_{S/R}).$ Then the mentioned lemma comes in:

Lemma (130.6): Suppose that $\phi : S \longrightarrow S'$ is surjective and we write $ I $ for its kernel. Then the map $ \Omega_{S/R} \longrightarrow \Omega_{S'/R'}$ is surjective and it's kernel is generated as an $S$-module by elements of the form $ ds$ where $s \in S$ such that $ \phi(s) = \beta (r')$ for some $ r' \in R'$.

Surjectivity follows from the functorial nature of the assignemnt $\phi\mapsto\bar\phi.$ The hard part is the explicit form of the $\ker\bar\phi.$ Everything boils down to the following claim made in the proof of the above lemma:

Claim: "The following elements generate the kernel as an $S$-module for sure: $ida,i\in I,a\in S,$ and $da,$ with $a\in S$ such that $\phi(a)=\beta(r')$ for some $r′\in R′.$"

So it says that $$\tag{1}\ker\bar\phi = \Lin_S\{ids,da:i\in I, s\in S, \phi(a)\in\Ima\beta \}$$ and argument is such that we have to chase the diagram which is above the Lemma (130.6) in this already mentioned site.

First, I am not comfortable with chasing such diagrams made of so many free modules. But secondly, I feel that it should be possible to prove the claim using only the diagram written above and universal property of Kahler differentials. My feeling is based on the fact, that there are other models for Kahler differentials and for them the Lemma should also hold.

Question Is there a proof of the equality $(1)$ which uses only the diagram written above and universal property of Kahler differentials?

Answer 1

Probably not. The universal property of Kahler differentials says that given an $R$-linear derivation $\psi$ from an $R$-algebra $S$ to an $R$-module $M$, this factors first as the universal derivation $d:S\to \Omega_{S/R}$ followed by a map $\Omega_{S/R}\to M$. The only derivations in the diagram that interact with $\Omega_{S/R}$ or $\Omega_{S'/R'}$ are $d_{S/R},d{S'/R'},$ and $ d_{S'/R'}\circ\phi$. But applying the universal property to each of these just gives us back the maps that are in the diagram.

I would implore you to give the diagram chase another chance. All it says (via 10.130.5) is that $fdg\in \Omega_{S/R}$ gets mapped to $\phi(f)d\phi(g)$ in $\Omega_{S'/R'}$, from which it follows that either $\phi(f)$ or $\phi(g)$ must be zero for $fdg$ to be in the kernel.