First order and infinitesimal deformations Terminology. Let $X$ be an algebraic variety over some algebraically closed field $k$. By an infinitesimal deformation of $X$, I mean a flat surjective map $\mathfrak X\to S=\textrm{Spec }A$, where $A$ is a local Artin $k$-algebra, such that $X$ is isomorphic to the closed fiber of this map. An isomorphism of deformations is an $S$-isomorphism between the total spaces, compatible with the inclusions of $X$ inside them. By a trivial deformation, I mean a deformation which is isomorphic to the product deformation $X\times \textrm{Spec }A\to \textrm{Spec }A$.
An algebraic variety $X$ is said to be rigid if it has no nontrivial infinitesimal deformations. One can show that if $X$ is nonsingular, then it is rigid of and only if $H^1(X,T_X)=0$, where $T_X$ is tha tangent bundle.
But $H^1(X,T_X)$ parameterizes, for any nonsingular $X$, isomorphism classes of first order deformations of $X$: those deformations defined over the dual numbers $D=k[t]/t^2$.
So, a posteriori, there should be something special in nonsingular varieties: if they have no nontrivial first order deformations, they have no infinitesimal deformations at all.
Question: can you show me an example of a (necessarily singular) variety with no nontrivial deformation over $D$, but with a nontrivial deformation over some other local Artin ring $A$?
Thank you!
 A: This is an interesting question! Here is my take on it.
The nonsingularity hypothesis is a simplification rather than a true necessity. If $X$ is nonsingular, then every first-order family is locally trivial, thus, by the Kodaira-Spencer equivalence the set of all first-order families for $X$ is parametrized by $H^1(X, \mathscr T_X)$. Moreover, the obstruction calculus is simplified, since the set of (small) liftings of a family for $X$ has a transitive $H^1(X,\mathscr T_X)$-action, which Sernesi takes advantage of to prove this theorem.
In the general case, not all first-order families for $X$ are locally trivial, and this is captured in the following comparison sequence:
$$0\to H^1(X,\mathscr T_X)\to \mathsf D_X(D)\to H^0(X,\mathscr T^1_X)\to H^2(X,\mathscr T_X)$$
where I've denoted the set of all first-order families for $X$ by $\mathsf D_X(D)$. In fact, Hartshorne's "Deformation Theory" extends this comparison sequence to the following sequence, for any local artinian ring $R$, useful for this lifting problem (Thm 10.2(b)):
$$0\to H^1(X,\mathscr T_X)\to \mathsf E(\mathcal X/R)\to H^0(X,\mathscr T^1_X)\to H^2(X,\mathscr T_X)$$
where $\mathsf E(\mathcal X/R)$ is the set of liftings of a chosen family $\mathcal X\to\operatorname{Spec}(R)$ for $X$ (actually, his version is slightly more general than this). Since $\mathscr T^1_X$ is supported on the singular locus of $X$, the nonsingularity hypothesis above simply ensures that $\mathscr T^1_X = 0$, so that $H^1(X,\mathscr T_X) = \mathsf E(\mathcal X/R)$. Thus, $X$ is rigid iff $H^1(X,\mathscr T_X)=0$ since all liftings are trivial, which is what Sernesi's proof boils down to. However in the general case, the theorem is something like the following:
$X$ is rigid iff $H^1(X,\mathscr T_X)=0$ and the obstruction map $H^0(X,\mathscr T^1_X)\to H^2(X,\mathscr T_X)$ is injective.
Now to answer your question. Suppose that $X$ has no nontrivial family over $D$. This means that $\mathsf D_X(D)=0$. By the first-order comparison sequence, this says exactly that $H^1(X,\mathscr T_X) = 0$ and $H^0(X,\mathscr T^1_X)\to H^2(X,\mathscr T_X)$ is injective, so that there is never a nontrivial lifting of any family for $X$ over any artinian local ring. Thus, $X$ is rigid.
