Kernel of map of Kahler differentials This is lemma 10.130.6, stacks project


*

*I understand those objects describe lie in the middle.

*I have trouble understanding how this "diagram" chase is done. Especially when we are dealing with relations. It would be nice if an elaboration is given.

*Is there an easier to see this  (i.e. by categorical arguments?)

 A: Start with the commutative square of ring homomorphisms
$$\require{AMScd}
\begin{CD}
R @>{\alpha}>> S\\
@V{\psi}VV @VV{\varphi}V\\
R' @>>{\beta}> S'
\end{CD}$$
where $\varphi$ is surjective  with kernel $I$.
We are given free presentations of the modules of Kähler differentials, and so obtain an exact commutative diagram
$$\require{AMScd}
\begin{CD}
@. @. F_1 @>>> F'_1\\
@. @. @VVV @VVV\\
0 @>>> K_0 @>>> F_0 @>>> F'_0 @>>> 0\\
@. @VVV @VVV @VVV\\
0 @>>> \mathrm{Ker} @>>> \Omega_{S/R} @>>> \Omega_{S'/R'} @>>> 0\\
@. @. @VVV @VVV\\
@. @. 0 @. 0
\end{CD}$$
As in the Snake Lemma, we have a surjection from $\mathrm{Coker}(F_1\to F'_1)$ to the cokernel of $K_0\to\mathrm{Ker}$.
Now
$$ F_1 = \bigoplus_{(a,b)\in S^2} S[(a,b)] \oplus \bigoplus_{(f,g)\in S^2} S[(f,g)] \oplus \bigoplus_{r\in R}S[r], $$
and similarly for $F'_1$. Since $S$ surjects onto $S'=S/I$, the cokernel of $F_1\to F'_1$ is the same as the cokernel of the map
$$ \bigoplus_{r\in R}S[r] \to \bigoplus_{r'\in R'}S'[r']. $$
Also, this map factors through the free module $\bigoplus_{r'\in R'}S[r']$. So we can lift to obtain a surjection
$$ K_0 \oplus\bigoplus_{r'\in R'}S[r'] \twoheadrightarrow \mathrm{Ker}, $$
where the map $S[r']\to\mathrm{Ker}$ sends $[r']$ to $da$ for some choice of $a\in S$ with $\varphi(a)=\beta(r')$
Finally, an element of $F_0$ can be written as $\sum s_{ij}[a_i+x_j]$ where $x_j\in I$ and the $\varphi(a_i)$ are distinct. This lies in $K_0$ if and only if $\sum_j\varphi(s_{ij})=0$ for all $i$, equivalently $y_i:=\sum_js_{ij}\in I$ for all $i$. We can therefore rewrite it as
$$ \sum_{ij}\big(s_{ij}[a_i+x_j]-s_{ij}[a_i]\big) + \sum_iy_i[a_i]. $$
Thus $K_0$ is generated by $[a+x]-[a]$ and $x[a]$ for all $a\in S$ and $x\in I$.
This shows that $\mathrm{Ker}$ is generated by $d(a+x)-da=dx$ and $xda$ for $a\in S$ and $x\in I$, together with $da$ for $a\in S$ with $\varphi(a)\in\mathrm{Im}(\beta)$.
Since $xda=d(ax)-adx$ and $\varphi(ax)=0=\varphi(x)$, we see that $\mathrm{Ker}$ is generated by terms of the form $da$ with $\varphi(a)\in\mathrm{Im}(\beta)$.
