2

Assume that all rings are commutative with identity. Let $k$ be a fixed ring with $k$-algebras $\varphi: k \rightarrow R$ and $\tau: k \rightarrow A$. Let $\tau' : R \rightarrow B$ define an $R$-algebra and $\psi: A \rightarrow B$ a morphism of rings so that the square commutes, that is $$ \tau' \varphi = \psi \tau. $$ Suppose further that $\psi: A \rightarrow B$ is surjective. Let $d: A \rightarrow \Omega_{A/k}$ denote the canonical derivation of $A$ over $k$ and let $d': B \rightarrow \Omega_{B/R}$ be the canonical derivation of $B$ over $R$. It is clear to me that the induced map, \begin{align} \Omega_{A/k} &\longrightarrow \Omega_{B/R} \\ fdg &\longmapsto \psi(f) d'\psi(g) \end{align} is surjective. The claim I am having trouble proving is that the kernel of this map is generated by terms $da$ with $\psi(a) \in \tau'(R)$. Questions have been asked about this before, see here, here, and here for example. In the first, an answer is given in terms of the functor of points, which is fine, but I would really like to be able to do this via a diagram chase. The other two questions refer to the same problem, but I still haven't been able to see how a diagram chase concludes this. Is someone able to spell it out to me?

Suppose we had $fdg \mapsto \psi(f) d' \psi(g)$. I'm really not sure what we can conclude about $f$ or $g$ here. The answerer in the third question I linked above claims that we can conclude that either $\psi(f)$ or $\psi(g)$ must be zero, but I see no reason why this must be the case. Can anyone shed some light on this?

Luke
  • 3,563

1 Answers1

1

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
  • 2,436