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I was wondering if I am in the right track here. Let $A := k[x,y]/(x^2 - y^3)$, the cuspidal curve. Obviously this isn't etale or smooth over $k$ so its cotangent complex is not contractible. Now, I wish to compute $\mathbb{L}_{A/k}$.

So first we have maps $k \rightarrow k[x,y] \rightarrow A$ which induces the transitivity sequence $$\mathbb{L}_{k[x,y]/k} \otimes_{k[x,y]} A \rightarrow \mathbb{L}_{A/k} \rightarrow \mathbb{L}_{A/k[x,y]}.$$ By Iyengar's notes (http://arxiv.org/pdf/math/0609151.pdf) we know that:

  1. $\mathbb{L}_{A/k[x,y]} \simeq \Sigma A$
  2. $\mathbb{L}_{k[x,y]/k} \simeq \Omega_{k[x,y]/k}$

Hence am I right to conclude that Andre-Quillen homology is just $\Omega_{k[x,y]/k} \otimes_{k[x,y]} A$ in degree $0$ and $A$ in degree 1? Thank you!

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  • $\begingroup$ The transitivity sequence should give a long exact sequence in homology, and you seem to be claiming that it collapses- why is the boundary zero? (probably I've confused my shifts somehow and the map is obviously zero?) $\endgroup$ Dec 11, 2014 at 2:06
  • $\begingroup$ right it's not! as Johan pointed out on fb $\endgroup$ Dec 11, 2014 at 2:09
  • $\begingroup$ whoops, guess I should be more up on my math-social-media feeds... $\endgroup$ Dec 11, 2014 at 2:11
  • $\begingroup$ should have just gone to one of you guys for this in the first place lol $\endgroup$ Dec 11, 2014 at 2:14

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