Lemma 10.131.6. In diagram (10.131.4.1), suppose that $S \to S'$ is surjective with kernel $I \subset S$. Then $\Omega _{S/R} \to \Omega _{S'/R'}$ is surjective with kernel generated as an $S$-module by the elements $\text{d}a$, where $a \in S$ is such that $\varphi (a) \in \beta (R')$. (This includes in particular the elements $\text{d}(i)$, $i \in I$.)

Proof. Consider the map of presentations above. Clearly the right vertical map of free modules is surjective. Thus the map is surjective. A diagram chase shows that the following elements generate the kernel as an $S$-module for sure: $i\text{d}a, i\in I, a \in S$, and $\text{d}a$, with $a \in S$ such that $\varphi (a) = \beta (r')$ for some $r' \in R'$. Note that $\varphi (i) = \varphi (ia) = 0 = \beta (0)$, and that $\text{d}(ia) = i\text{d}a + a \text{d}i$. Hence $i\text{d}a = \text{d}(ia) - a \text{d}i$ is an $S$-linear combination of elements of the second kind. $\square$

There are also:

• 12 comment(s) on Section 10.131: Differentials

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).