I am reading the Eisenbud, commutative algebra, proof of Theorem 16.24 and some question arises.
First, let me note some associated definition and theorem.
Definiton. If $S$ is a ring and $M$ is an $S$-module, then a map (of abelian groups ) $d : S\to M$ is a derivation if it satisfies the Leibniz rule $d(fg) = fdg + gdf$ for $f,g \in S$. If $S$ is an $R$-algebra, then we say that $d$ is $R$-linear if it is a map of $R$-modules.
Proposition 16.11. Let $\varphi : S \to S'$ be a map of $R$-algebras, and let $\delta : S \to S'$ be a map of abelian groups. If $\delta(S)^2 =0$, then $\varphi + \delta$ is a homomorphism of $R$-algebras iff $\delta$ is an $R$-linear derivation in the sense that $$\delta(b_1b_2)= \varphi(b_1)\delta(b_2) + \varphi(b_2)\delta(b_1).$$
Now let me state the Theorem 16.24 and its proof.
Theorem 16.24. Let $I$ be the kernel of the multiplication map $\mu : S\otimes_R S \to S$. If $e : S \to I/I^2$ is the map defined by $b \mapsto [1\otimes b - b\otimes 1]$, then there is an isomorphism $\varphi : \Omega_{S/R} \to I/I^2$ of $S$-modules such that $\varphi d =e $; that is, the pair $(d, \Omega_{S/R})$ is in a natural sense isomorphic to $(e, I/I^2)$.
Proof. We first show that $e$ is a ($R$-linear) derivation so that by the univaersal property there is a unique map $\varphi : \Omega_{S/R} \to I/I^2$ satisfying $e= \varphi d$. Indeed, $e$ is the difference of the two maps $e_1 : b\mapsto [1\otimes b]$ and $e_2 : b \mapsto [b\otimes 1]$, which are algebra maps splitting the sequence $$ I/I^2 \hookrightarrow(S\otimes_R S)/I^2 \to S \to 0.$$
Quite generally, if $T\to S$ is an algebra map whose kernel $J$ has square $0$, then $J$ is naturally an $S$-module ( So we may deduce that $I/I^2$ is an $S$-module ? ). By the Proposition 16.11 above, any two splittings $e_1 , e_2 : S \to T$ differ by derivation.
Q.1. First, I can't understand why the $e = e_1 - e_2$ above is $R$-linear derivation from the bold statement. My first attempt is that letting $\varphi := e_2 $ and $\delta : =e$ in the Proposition 16.11, if $e_1 = e_2 + e = \varphi + \delta : S \to I/I^2$ is a homomorphism of $R$-algebra and $e(S)^2 = 0$, then $e$ is $R$-linear derivation (up to $\varphi= e_2$) in the sense that
$$ e(b_1b_2) = e_2 (b_1)e(b_2)+e_2(b_2)e(b_1):= [b_1 \otimes_R 1]e(b_2) + [b_2 \otimes_R 1] e(b_1) $$
My question is, from this, can we deduce that $e$ is indeed $R$-linear derivation? ; i.e., does it satisfy the Leibniz rule and is it really a map of $R$-modules? What is the (natural) $S$-module structure on $I/I^2$? I think that I am somewhat confused about notational issue comming from the definitions ( maybe there is a possibility of difference) of honest $R$-linearlity and $R$-linearlity upto $\varphi= e_2$. Is there a point that I missed or made mistake? Or is there any other route to show that $e$ is indeed a $R$-linear derivation?
Next, we continue to his argument for proof of Theorem 16.24 as follows :
Because of the universal property of $d$ and $\Omega_{S/R}$, there is a unique map $\varphi : \Omega_{S/R} \to I/I^2$ satisfying $e= \varphi d$ ; that is, with $\varphi(db) = 1\otimes b - b\otimes 1$. It remains to prove that $\varphi$ is an isomorphism. which we shall do by identifying its inverse. Let $T:=S \ltimes \Omega_{S/R}$ be the "trivial extension of $S$ by $\Omega_{S/R}$" ; i.e., the ring which, as an abelian group, is the direct sum of $S$ and $\Omega_{S/R}$, and whose multiplication is defined, for $b, b' \in S$, and $u, u' \in \Omega_{S/R}$ by $$ (b,u)(b',u') = (bb', bu' + b'u).$$
We claim that there is a ring homomorphism $$ \psi : S\otimes_R S \to T ; \psi : a \otimes b \mapsto (ab, adb) $$ for $a,b\in S$.
Since $\psi(1\otimes b - b\otimes 1) = (0,db)$, we see that the restriction of $\psi$ to $I$ induces the desired inverse on $I/I^2$.
Q.2. I don't understand the bold statement. Why the induced homomorphism on $I/I^2$ from the restriction of $\psi$ to $I$ has codomain $\Omega_{S/R}$ ? And why the induced morphism $I/I^2 \to \Omega_{S/R}$ is indeed $S$-module homomorphism?
Thanks for reading. Can anyone helps?