# Andre-Quillen Homology of the cuspidal curve $k[x,y]/(x^2 - y^3)$

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!

• 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?) Dec 11, 2014 at 2:06
• right it's not! as Johan pointed out on fb Dec 11, 2014 at 2:09
• whoops, guess I should be more up on my math-social-media feeds... Dec 11, 2014 at 2:11
• should have just gone to one of you guys for this in the first place lol Dec 11, 2014 at 2:14