Mumford in his Red Book gives on page 144 an example of computation of a module of Kähler differential forms.
Namely, he lets $k$ be a field and considers the quotient algebra $B=k[X,Y]/(XY)$.
He defines $\omega=XdY=-YdX\in \Omega_{B/k} $ and claims there is a short exact sequence $$0\to k\cdot\omega \to \Omega_{B/k}\to k[X]dX\oplus k[Y]dY\to 0 $$ I more or less understand the idea behind this exact sequence (using the standard computation of the Kähler differential module of a quotient of a polynomial ring) , but I would like to see this exact sequence in a broader, more general context rather than the ad hoc explanations given by Mumford.

Also, he claims that his exact sequence (of $B$-modules I guess) does not split. Why is that?

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    $\begingroup$ This is covered in chapter 16 of Eisenbud's commutative algebra book. The key is the conormal sequence. There are also some notes by Ravi Vakil here. $\endgroup$ – Richard D. James May 28 at 18:39
  • $\begingroup$ @Richard D. James: The conormal sequence is certainly not what I'm asking about, since in that sequence the module $\Omega_{B/k}$ would be on the right and in the question it sits in the middle. Where in your references can I find the above exact sequence ? Where does it say that it doesn't split? I'd rather you wrote a precise, detailed answer to my question rather than pointing to a chapter called "Modules of Differentials" in a completely standard reference or directing me to an 800 pages well-known online book. $\endgroup$ – lefuneste May 28 at 19:42
  • $\begingroup$ One can combine the relative cotangent sequence with the conormal sequence I mentioned above. If you actually bother to click the link I gave, I think you will find considerably fewer than 800 pages of notes. Best of luck with your question. $\endgroup$ – Richard D. James May 28 at 20:04
  • $\begingroup$ @Richard D.James First of all thank you for your good wishes. Vakil explains the basic notions on Kähler differentials but definitely does not address my question . I don't think that "one" can combine those two well-known exact sequences to obtain what I want, and certainly no combination will prove that the exact sequence does not split. If you can do that "combination", please write it as an answer. If you can't, don't. $\endgroup$ – lefuneste May 28 at 21:32

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