# Is there a relation between the tangent bundle of a scheme and the tangent bundle of the associated reduced scheme ?

My question is vague and general, and, if you want, naive.

$\bullet$ Given $X/S$ a scheme. We denote by $X_{red}$ the reduced scheme associated to $X$.

Grothendieck defines in EGA 4 (16.5.12.1) the tangent bundle $T_{X/S}$ of $X$ relatively to $S$. Is there any relation between $T_{X_{red}/S}$ and $T_{X/S}$ ? More precisely, we know that there is a morphism $$f : T_{X_{red}/S} \to T_{X/S}\times_{X} X_{red} ;$$ what can be said of $f$ ? I am interested, in particular, in the case when $S=Spec \ k$, with $k$ a field.

$\bullet$ A related question (but in which I am less interested) is the comparison of the Zariski tangent space at one point (as defined in Hartshorne p. 80) of $X$ and $X_{red}$. In general, if $\pi : X_{red} \to X$ denotes the canonical morphism, and if $x\in X_{red}$, the linear map $$T_x \ \pi : T_x \ X_{red} \to T_{\pi(x)} \ X$$ is injective, since $\pi$ is a closed immersion. Is it an isomorphism ?

$\bullet$ Finally, a related question is the comparison of $\Omega^1_{X/S}$ and $\Omega^1_{X_{red}/S}$.

$\bullet$ Ideally, I would like to transport sections of $T_{X_{red}/S}$ to sections of $T_{X/S}$.

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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. –  Thomas Nevins Apr 16 '10 at 14:18
this is just intuition: tangent spaces should detect singularities, and reducing a scheme can destroy them. thus there won't be a good comparison. –  Martin Brandenburg Apr 16 '10 at 14:47
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. –  user2330 Apr 16 '10 at 14:47
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 $S$. –  BCnrd Apr 16 '10 at 15:25
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. –  BCnrd Apr 16 '10 at 15:52