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I am trying to understand the proof of Lemma 130.6 in CommAlg of the stacks project. I will try to recall everything now:

Suppose we have a diagram of commutative rings

$$\require{AMScd} \begin{CD} S @>{\varphi}>> S'\\ @AA{\alpha}A @AA{\beta}A\\ R @>{\psi}>> R' \end{CD}$$

This induces a commutative diagram

$$\require{AMScd} \begin{CD} \Omega_{S/R} @>>> \Omega_{S'/R'}\\ @AA{d}A @AA{d}A\\ S @>{\varphi}>> S' \end{CD}$$

Lemma (130.6): Suppose that $\varphi : S \longrightarrow S'$ is surjective and we write $ I $ for its kernel. Then the map $ \Omega_{S/R} \longrightarrow \Omega_{S'/R'}$ is surjective and it's kernel is generated as an $S$-module by elements of the form $ ds$ where $s \in S$ such that $ \varphi(s) = \beta (r')$ for some $ r' \in R'$.

The surjectivity statement is clear to me. What I don't understand is the argument about the generators of the Kernel. In the proof the following is claimed:

Claim: A diagram chase shows that the kernel is certainly generated by elements of the form $ ids$ where $ i \in I$ and $ s \in S$ such that $ \varphi(s) = \beta (r')$ for some $ r' \in R$.

So my question is: Why is the above claim true?

Edit: Here is one idea I had that gets around the argument in the stacks project. Suppose that $ f dg \in Ker( \Omega_{S/R} \longrightarrow \Omega_{S'/R'}) $. If $ \varphi(f) = 0$ then we can write $ fdg = gdf-d(fg)$ and we win. Else if $ \varphi(f) \neq 0$ then it follows that $ \varphi(g) = \beta(r')$ for some $r' \in R'$. Is it convincing?

darij grinberg
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  • Have you tried doing the diagram chase? Do you get stuck in any one particular step of it? – KReiser Aug 09 '16 at 04:04
  • Yes, I've tried doing the diagram chase and the part Im stuck on is still why the kernel is generated by those elements. It's not at all clear to me. –  Aug 09 '16 at 04:56
  • I think it would be good to note that he gives you a big diagram of presentations to chase. http://stacks.math.columbia.edu/tag/00RM it looks like he throws in two types of elements to begin with. – Hoot Aug 09 '16 at 06:36
  • Did you even bother to read the 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: $ida$,$i\in I$,$a\in S$, and $da$, 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 $d(ia)=ida+adi$. Hence $ida=d(ia)−adi$ is an $S$-linear combination of elements of the second kind." Verbatim from the link. – KReiser Aug 09 '16 at 20:23
  • @KReiser In the proof that you paste there is this "unclear claim". Unfortunetly I also have the same problem recently as OP, but I am looking for the proof which avoids the diagram chase. If you are still interested in this topic plese have a look at: https://math.stackexchange.com/questions/2541319/kernel-of-the-induced-map-bar-phi-omega-s-r-to-omega-s-r – Fallen Apart Nov 28 '17 at 15:55

1 Answers1

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To do it by diagram chase, use the diagram they give in the proof of lemma 10.130.6 of http://stacks.math.columbia.edu/tag/00RM (I am too lazy to reproduce it) (be careful, I use the notation there not yours)

Call the left vertical map $k$ et $j$ the right one, and $b$ the upper horizontal map and $a$ the lower one. To simplify the notations, we suppose $R\subset S$ and $R'\subset S'$. This does not change the proof.

Case 1 We suppose the map $R\to R'$ surjective.

By hypothesis, the vertical maps $k$ and $j$ are surjective, which will allow basic diagram chasing.

If an element $\alpha$ goes to zero through the induced map $\psi$ \begin{align} \Omega_{S/R} &\longrightarrow \Omega_{S'/R'} \\ fdg &\longmapsto \phi(f) d'\phi(g) \end{align} let us call $\beta$ one of its antecedent in $\bigoplus S[a]$. Then the image of $j(\beta)$ in $\Omega_{S'/R'}$ being zero, there exists $$\gamma\in\bigoplus S'[(a', b')] \oplus \bigoplus S'[(f', g')] \oplus \bigoplus S'[r']$$ with $b(\gamma)=j(\beta)$. Let $$\delta\in\bigoplus S[(a, b)] \oplus \bigoplus S[(f, g)] \oplus \bigoplus S[r]$$ such that $k(\delta)=\gamma$. Then $$j(\beta)=b(k(\delta))=j(a(\delta))$$ Therefore, $j(\beta-a(\delta))=0$, with $\beta-a(\delta)\in\bigoplus S[a]$. Since the domain and codomain of $j$ are free module over respectively $S$ and $S'$ the only way for its image by $j$ to be zero is that each of its components of the type $s[g]$ ($s, g\in S$) is sent to zero, which means that $\phi(s)[\phi(g)]$ is zero in $S'[\phi(g)]$. Once again because this is a free module, we have necessarily $\phi(s)=0$ or $\phi(g)=0$, and so $s\in I$ or $g\in I$.

You conclude by saying that $\beta$ is a sum an element that goes to zero by quotienting in $\Omega_{S/R}$ and elements of the form $s[g]$ with $s$ or $g$ in $I$ therefore $\alpha$ is a sum of elements of the type $s\ dg$, with $s\in I$ or $g\in I$. Then you use the little trick given in the text to reach the following conclusion: the kernel of $\psi$ is $S\ dI$ (which is a smaller set of generators than $S\ d\phi^{-1}(R')$ because we are in case 1)

Case 2 The general case

To be able to do diagram chasing, we have to change the domain of the map $k$. We call $R''=\phi^{-1}(R')$. Then the new left hand bottom corner of the diagram is $$\bigoplus S[(a, b)] \oplus \bigoplus S[(f, g)] \oplus \bigoplus_{r\in R''} S[r]$$ With this new domain, the map $k$ is surjective, and all the reasoning of case 1 applies, except at the end where $\beta$ is the sum of an element that goes to zero in $\Omega_{S/R}$ (that belongs to the original domain of $k$), an element that belongs to $\oplus_{r\in R''-R}S[r]$, and elements of the form $s[g]$ with $s\in I$ or $g\in I$. Therefore $\alpha$ is a sum of elements of the type $s\ dg$, with $s\in I$ or $g\in I$ like before, and a sum of elements of the type $s\ dr$ where $r\in R''-R$.

Therefore, the kernel of $\psi$ is indeed $S\ d\phi^{-1}(R')$ in the general case.

brunoh
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