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I am trying to find out how to complete the proof of the following statement.

Let $R, S, R', S'$ be commutative rings and $(\psi, \varphi) : (R, S) \to (R', S')$ be a morphism from the algebra $\alpha : R \to S$ to $\beta : R' \to S'$, i.e. $\psi : R \to R'$, $\varphi : S \to S'$ are ring maps with $\beta \circ \psi = \varphi \circ \alpha$.

Assume that $\varphi$ is surjective. Prove that the kernel of the induced map on differentials (which is obviously surjective)

$\Omega_{S/R} \to \Omega_{S'/R'}, \qquad s\mathrm{d}s' \mapsto \varphi(s)\mathrm{d}\varphi(s')$

is generated by all $a$ such that $\varphi(a)$ is in the image of $\beta$.

This result is taken from http://stacks.math.columbia.edu/tag/00RR and relates to the proof of Proposition 16.3 in Eisenbud, when he states that

"This is the same as the description by generators and relations of $\Omega_{T/R}$, except that $\mathrm{d}f$, $f \in I$ are replaced by $0$..."

The context is a bit different, but in both cases it is unclear why the kernel is really generated by those elements (it is obvious that they belong in there!). When I get down to explicitly writing the details, some cancelling may occur and I don't know how to manage it.

I would be thankful if anyone can clarify this.

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This is one of thousands concrete statements which are best (or in fact only) explained in the functorial approach to mathematics: Instead of looking at an object $X$ "internally", look at its represented functor $\hom(X,-)$, which is usually described by means of a universal property of $X$. Put differently: Forget about elements, think about morphisms! All this is justified by the Yoneda Lemma.

The module of differentials $\Omega^1_{S/R} \in \mathsf{Mod}(S)$ is defined by $$\hom_S(\Omega^1_{S/R},M) = \mathrm{Der}_R(S,M)$$ for $S$-modules $M$. Now let $\phi : S \to S'$ be a surjective homomorphism "over" $\psi : R \to R'$ (diagram). Then we have for $S'$-modules $M$: $$\hom_{S'}(\Omega^1_{S'/R'},M) = \mathrm{Der}_{R'}(S',M)$$ I am a bit lazy and don't write down the functors which restrict scalars, but they will be all over the place. Let $I := \ker(\phi)$. Then we have an isomorphism of $R$-modules $S' \cong S/I$. Hence, $$\mathrm{Der}_{R'}(S',M) \subseteq \hom_{R'}(S',M) \subseteq \hom_{R}(S',M) \cong \{\delta \in \hom_R(S,M) : \delta|_I = 0\}.$$ The image consists of all $R$-linear maps $\delta : S \to M$ vanishing on $I$ such that the $R$-linear map $\delta' : S' \to M$ defined by $\delta' \circ \phi = \delta$ satisfies the Leibniz rule and vanishes on $R'$. This means that $\delta$ satisfies the Leibniz rule and that $\delta$ vanishes on $T := \{s \in S : \phi(s) \in R' \cdot 1\}$. This contains $I$. Thus, if $d : S \to \Omega^1_{S/R}$ is the universal differential, $$\mathrm{Der}_{R'}(S',M) \cong \{\delta \in \mathrm{Der}_R(S,M) : \delta|_T = 0\}$$ $$\cong \{f \in \hom_S(\Omega^1_{S/R},M) : f(d(T))=0\} \cong \hom_S(\Omega^1_{S/R}/\langle d_S(T) \rangle ,M)$$ Since $\Omega^1_{S/R}/\langle d_S(T) \rangle$ is killed by $I$, it carries the structure of an $S'$-module. It follows $$\hom_{S'}(\Omega^1_{S'/R'},-) \cong \hom_{S'}(\Omega^1_{S/R}/\langle d_S(T) \rangle,-)$$ and hence $\Omega^1_{S'/R'} \cong \Omega^1_{S/R}/\langle d_S(T) \rangle$ by Yoneda. $\square$

Try to prove this "directly" using elements in one of the explicit constructions of $\Omega^1$. Does it work at all? If yes, how many pages?

In order to illustrate the technique, let me also show a quick proof of the next lemma: If $A \to B \to C$ are homomorphisms of commutative rings, then there is an exact sequence $$C \otimes_B \Omega^1_{B/A} \to \Omega^1_{C/A} \to \Omega^1_{C/B} \to 0.$$ Proof: By definition of a cokernel and the Yoneda Lemma, the claim is equivalent to an exact sequence $$0 \to \hom_C(\Omega^1_{C/B},M) \to \hom_C(\Omega^1_{C/A},M) \to \hom_C(C \otimes_B \Omega^1_{B/A},M)$$ naturally in $M \in \mathsf{Mod}(C)$. By definition of $\Omega^1$ and the adjunction between scalar extension and restriction, this simplifies to $$0 \to \mathrm{Der}_B(C,M) \to \mathrm{Der}_A(C,M) \to \mathrm{Der}_A(B,M).$$ The map on the left is the obvious inclusion, the map on the right is given by pullback with $B \to C$. Exactness is clear. $\square$

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  • $\begingroup$ Thanks, this was wonderful. Now that I think of it, it was a bit silly to try to prove something so general using the specific construction. $\endgroup$ – Chindea Filip Nov 29 '13 at 14:48

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