Let $f:X\to Y$ be a morphism of finite type between finite type schemes $X,Y$ over a field $k$. By the infinitesimal criterion for (formal) smoothness, f is smooth if given a commutative diagram

$$\begin{array}[c]{ccc} T& {\rightarrow}&X\\ \downarrow &&\downarrow\scriptstyle{f}\\ T'& {\rightarrow}&Y \end{array}$$

where $T⊂T′$ is a first order thickening of affine schemes, there exists a morphism $T'\to X$ that fits in the above diagram.

Question: Suppose that for any pair ($T', T$), where $T'$ is a trivial extension of $T$, i.e. $T'=T\times_{Spec k} Spec\ k[\epsilon]$ such that $\epsilon^2=0$, and any commutative diagram as above, there exists an arrow $T'\to X$ that fits in the diagram. Is it true that $f$ is smooth?

  • $\begingroup$ Trivially no? Say $Y=\operatorname{Spec} k$, then the condition asks for an extension of $f: T\to X$ to $f':T'\to X$. But if $T' = \operatorname{Spec} R[\varepsilon]/(\varepsilon^2)$ is a trivial extension of $T=\operatorname{Spec} R$, then $i:T\to T'$ has a retraction $r: T'\to T$ (defined by sending $\varepsilon$ to zero), and we can take $f' = f\circ r$. So the condition is satisfied, but $X$ is arbitrary. See also the edgarlorp's comment below. $\endgroup$ – Piotr Achinger Aug 31 '19 at 9:21
  • $\begingroup$ Yes, of course, thank you. Can you post it as an answer so that I can accept it? $\endgroup$ – sky223 Aug 31 '19 at 9:36

(Comment posted as answer)

The answer is no, in fact for $Y = \operatorname{Spec} k$, the condition is always satisfied, even if $X$ is not smooth. The point is that a trivial extension $$ i \colon T = \operatorname{Spec} R \to \operatorname{Spec} R[\varepsilon]/(\varepsilon^2) = T' $$ admits a retraction $r \colon T'\to T$ defined by the natural $R$-algebra structure on $R[\varepsilon]/(\varepsilon^2)$. Thus if $g\colon T\to X$ is a map, we can always extend it along $i$ to a map $g'\colon T'\to X$ by composing with $r$, i.e. $g' = g\circ r$.


(Can't post this as a comment, however:) The $\operatorname{spec}k[\epsilon]$ point $(\epsilon, \epsilon)$ of the variety $xy=0$ doesn't lift to a $\operatorname{spec}k[x]/(x^3)$ point, (whereas it does lift to a $k[\epsilon_1,\epsilon_2]$ point). Perhaps you want to mean something else by trivial extension?

  • $\begingroup$ I modified the question for clarity. I would like to know if there exists a non-smooth f which satisfies the criterion on all pairs of trivial extensions ($T',T$) as in the post, or if any morphism that satisfies the lifting property for all such pairs must then satisfy the property on all pairs of 1st order thickenings (and thus must be smooth). As you point out in your example, the singular variety (xy=0) over k doesn't satisfy the lifting property on all pairs of trivial extensions. $\endgroup$ – sky223 Aug 31 '19 at 9:26

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