Lemma 10.131.9. In diagram (10.131.4.1), suppose that $S \to S'$ is surjective with kernel $I \subset S$, and assume that $R' = R$. Then there is a canonical exact sequence of $S'$-modules

$I/I^2 \longrightarrow \Omega _{S/R} \otimes _ S S' \longrightarrow \Omega _{S'/R} \longrightarrow 0$

The leftmost map is characterized by the rule that $f \in I$ maps to $\text{d}f \otimes 1$.

Proof. The middle term is $\Omega _{S/R} \otimes _ S S/I$. For $f \in I$ denote $\overline{f}$ the image of $f$ in $I/I^2$. To show that the map $\overline{f} \mapsto \text{d}f \otimes 1$ is well defined we just have to check that $\text{d} f_1f_2 \otimes 1 = 0$ if $f_1, f_2 \in I$. And this is clear from the Leibniz rule $\text{d} f_1f_2 \otimes 1 = (f_1 \text{d}f_2 + f_2 \text{d} f_1 )\otimes 1 = \text{d}f_2 \otimes f_1 + \text{d}f_1 \otimes f_2 = 0$. A similar computation show this map is $S' = S/I$-linear.

The map $\Omega _{S/R} \otimes _ S S' \to \Omega _{S'/R}$ is the canonical $S'$-linear map associated to the $S$-linear map $\Omega _{S/R} \to \Omega _{S'/R}$. It is surjective because $\Omega _{S/R} \to \Omega _{S'/R}$ is surjective by Lemma 10.131.6.

The composite of the two maps is zero because $\text{d}f$ maps to zero in $\Omega _{S'/R}$ for $f \in I$. Note that exactness just says that the kernel of $\Omega _{S/R} \to \Omega _{S'/R}$ is generated as an $S$-submodule by the submodule $I\Omega _{S/R}$ together with the elements $\text{d}f$, with $f \in I$. We know by Lemma 10.131.6 that this kernel is generated by the elements $\text{d}(a)$ where $\varphi (a) = \beta (r)$ for some $r \in R$. But then $a = \alpha (r) + a - \alpha (r)$, so $\text{d}(a) = \text{d}(a - \alpha (r))$. And $a - \alpha (r) \in I$ since $\varphi (a - \alpha (r)) = \varphi (a) - \varphi (\alpha (r)) = \beta (r) - \beta (r) = 0$. We conclude the elements $\text{d}f$ with $f \in I$ already generate the kernel as an $S$-module, as desired. $\square$

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