Let $f:X\rightarrow Y$ be a morphism of schemes over a field $k$. Can one check that $f$ is formally smooth using only Artin rings of the form $k^{\prime}\left[t\right]/t^{n}$, where $k^{\prime}$ is also a field?

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

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

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.

  • $\begingroup$ 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. $\endgroup$ – Anton Geraschenko Oct 10 '09 at 20:34

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