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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.

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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 '09 at 20:26
    
I meant to let the k in the Artin rings vary so edited the question. –  David Zureick-Brown Oct 11 '09 at 20:21
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Is there more left to do in this question? –  Greg Kuperberg Jan 23 '10 at 2:57
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1 Answer

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.

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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 '09 at 20:34
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