The Stacks project

Lemma 10.131.9. In diagram (, 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. $\square$

Comments (0)

There are also:

  • 10 comment(s) on Section 10.131: Differentials

Post a comment




All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00RU. Beware of the difference between the letter 'O' and the digit '0'.