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Let f:X -> Y be a morphism of schemes over a field k. Can one check that f is formally smooth using only Artin rings of the form k'[t]/t^n, where k' is also a field?

Considering cuspidal curves one can show that you do at least need arbitrarily large n.

edited Oct 11 2009 at 20:20

asked Oct 6 2009 at 12:14

David Zureick-Brown♦
6,064●10●41

I misunderstood the original question: I didn't see the "over a field k" part, so I thought you were asking if it was sufficient to use square-zero extensions of the form k[t]/t^n --> k[t]/t^m where k could be an arbitrary field. – Anton Geraschenko♦ Oct 10 2009 at 20:26

I meant to let the k in the Artin rings vary so edited the question. – David Zureick-Brown♦ Oct 11 2009 at 20:21

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Is there more left to do in this question? – Greg Kuperberg Jan 23 2010 at 2:57

S. Carnahan · Answer 1 · 2009-10-10 04:14:24Z UTC

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I'm not sure what you mean by "using" but I think the answer is no. X can be nonempty and singular over Y with X(k) empty.

answered Oct 10 2009 at 4:14

S. Carnahan♦
25.6k●41●90

To check formal smoothness, you have to check that for any square-zero extension of Artin rings A'-->A, and any maps Spec(A)-->X and Spec(A')-->Y (making the obvious diagram commute), there's a compatible map Spec(A')-->X. I think that "using k[t]/t^n" means only checking the cases where A'=k[t]/t^n. If I understood this right, then your answer is correct. – Anton Geraschenko♦ Oct 10 2009 at 20:34

Can one check formal smoothness using only one-variable Artin rings?

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