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Apr 17, 2010 at 19:17 history edited user2330 CC BY-SA 2.5
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Apr 16, 2010 at 15:58 history edited user2330 CC BY-SA 2.5
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Apr 16, 2010 at 15:52 comment added BCnrd Although in general one doesn't expect much to come out of linear dual $T_ {X/S}$ when $\Omega^1_ {X/S}$ not locally free (of finite rank), as it generally fails to commute with base change, there is (at least) one situation where it has some interest beyond the smooth case: group schemes (not nec. smooth). For finite flat comm. group schemes is this useful (cf. papers of Fontaine), and in general get Lie alg. structure even w/o smoothness (see sga3 or Appendix A.7 of "Pseudo-reductive groups"). Beware: without smoothness, the notion of "adjoint repn" on Lie algebra is somewhat delicate.
Apr 16, 2010 at 15:25 comment added BCnrd The headline for the question is odd: one shouldn't say "tangent bundle" unless the sheaf is locally free, and it usually is not when $X$ is not $S$-smooth. The premise of the question seems a bit naive, especially killing nilpotents on X without doing the same for $S$. Probably the only interesting thing to say is that if $X$ is $S$-smooth then $X_ {\rm{red}} = X \times_S S_ {\rm{red}}$ and tangent/cotangent bundles for $X_ {\rm{red}}$ over $S_ {\rm{red}}$ are pullback of those for $X$ over $S$.
Apr 16, 2010 at 14:47 comment added user2330 Yes, I meant injective ! I've changed it in the question. Hence, with your example, we see that $T_x \ \pi$ is not surjective in general.
Apr 16, 2010 at 14:47 comment added Martin Brandenburg this is just intuition: tangent spaces should detect singularities, and reducing a scheme can destroy them. thus there won't be a good comparison.
Apr 16, 2010 at 14:45 history edited user2330 CC BY-SA 2.5
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Apr 16, 2010 at 14:18 comment added Thomas Nevins I don't think you mean "surjective" in your second bullet point. For example, consider $S = \operatorname{Spec}({\mathbb C})$ and $X = \operatorname{Spec}({\mathbb C}[t]/(t^2))$ (i.e. Spec of the dual numbers), where $x\in X_{red} \cong \operatorname{Spec}({\mathbb C})$ is the point.
Apr 16, 2010 at 14:12 history edited user2330 CC BY-SA 2.5
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Apr 16, 2010 at 14:02 history asked user2330 CC BY-SA 2.5