Module of differentials of a finitely generated $R$-algebra I'm reading Commutative Algebra with a View Toward Algebraic Geometry written by Eisenbud. In chapter 16(Module of Differentials) there is a proposition:

If $S$ is a finitely generated $R$-algebra, say $S=R[x_1,...,x_r]/I$,
and if $I=(f_1,...,f_s)$, then $S
 \otimes_R\Omega_{R[x_1,...,x_r]/R}=\oplus_iSdx_i$.

My attempt: Let $T=R[x_1,...,x_r]$, we have the following map with $I$ as its kernel: $T\to S\to 0$. $\{d(x_i+I)\}_{I\in\{1,...,r\}}$ generate $\Omega_{S/R}$ therefore, $\Omega_{S/R}=\oplus_iSd(x_i+I)$. Using the conormal sequence, there is a surjective morphism $D\pi$ from $S \otimes_R\Omega_{R[x_1,...,x_r]/R}$ to $\Omega_{S/R}$. $D\pi$ is not necessarily an isomorphism...
 A: If $A:=R[x_1,..,x_r]$ is a polynomial ring over $R$ in $r$ variables it follows there is an isomorphism of $A$-modules $\Omega_{A/R} \cong \oplus_i Adx_i$. The sum is a direct sum, hence $\Omega_{A/R}$ is a free $A$-module of rank $r$ on the elements $dx_i$. It follows $S\otimes_R \Omega_{A/R} \cong \oplus_i S\otimes_R Adx_i$ and $S\otimes_R A \neq S$ are non-isomorphic rings, hence it seems your formula is incorrect. If the elements $f_1,..,f_l$ generate your ideal $I$, and you want a formula for the $S$-module $\Omega_{S/R}$ the following holds:
Formula 1. $\Omega_{S/R} \cong \oplus_i Sdx_i/C,$
where the module $C$ is the left $S$-module generated by the differentials $df_i$ for all $i=1,..,l$. You may find details on this construction in a commutative algebra book where they discuss the module of derivations and differentials. There are "fundamental exact sequences" involving the module of differentials, and these sequences can be used to prove such results.
Mastumuras book "Commutative Ring Theory" studies derivations and differentials and gives explicit proofs of properties of this construction. Formula 1 is proved.
