
> 
> **Definition.** A locally finitely presented morphism of schemes $f\colon X\to Y$ is *smooth* (resp. *unramified*, resp. *étale*) if for any **affine** scheme $T$, any closed subscheme $T\_0$ defined by a square zero ideal $I$, and any morphisms $T\_0\to X$ and $T\to Y$ making the following diagram commute
> 
> 
> 
> ```
> 
>     g
> T0 --> X
> |      |
> |      |f
> v      v
> T ---> Y
> 
> ```
> 
> there exists (resp. exists at most one, resp. exists exactly one) morphism $T\to X$ which fills the diagram in so that it still commutes.
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> 
> 


For checking that $f$ is unramified or étale, it doesn't matter that I required $T$ to be affine. The reason is that for an arbitrary $T$, I can cover $T$ by affines, check if there exists (a unique) morphism on each affine, and then "glue the result". If there's at most one morphism locally, then there's at most one globally. If there's a unique morphism locally, then there's a unique morphism globally (uniqueness allows you to glue on overlaps).


But for checking that $f$ is smooth, it's really important to require $T$ to be affine in the definition, because it could be that there exist morphisms $T\to X$ locally on $T$, but it's impossible to find these local morphisms in such a way that they glue to give a global morphism.



> 
> **Question:** What is an example of a smooth morphism $f\colon X\to Y$, a square zero nilpotent thickening $T\_0\subseteq T$ and a commutative square as above so that there **does not** exist a morphism $T\to X$ filling in the diagram?
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> 
> 


I'm sure I worked out such an example with somebody years ago, but I can't seem to reproduce it now (and maybe it was wrong). One thing that may be worth noting is that the set of such filling morphisms $T\to X$, if it is non-empty, is a torsor under $Hom\_{\mathcal O\_{T\_0}}(g^\*\Omega\_{X/Y},I)=\Gamma(T\_0,g^\*\mathcal T\_{X/Y}\otimes I)$, where $\mathcal T\_{X/Y}$ is the relative tangent bundle. So the obstruction to finding such a lift will represent an element of $H^1(T\_0,g^\*\mathcal T\_{X/Y}\otimes I)$ (you can see this with Čech cocycles if you want). So in any example, this group will have to be non-zero.



Example of a smooth morphism where you can't lift a map from a nilpotent thickening?
Using some of BCnrd's ideas together with a different construction, I'll give a positive answer to Kevin Buzzard's stronger question; i.e., there is a counterexample for *any* non-etale smooth morphism.


Call a morphism $X \to Y$ *wicked smooth* if it is locally of finite presentation and for every (square-zero) nilpotent thickening $T\_0 \subseteq T$ of $Y$-schemes, every $Y$-morphism $T\_0 \to X$ lifts to a $Y$-morphism $T \to X$.



> 
> **Theorem:** A morphism is wicked smooth if and only if it is etale.
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**Proof:**
Anton already explained why etale implies wicked smooth. 


Now suppose that $X \to Y$ is wicked smooth. In particular, $X \to Y$ is smooth, so it remains to show that the geometric fibers are $0$-dimensional. Wicked smooth morphisms are preserved by base change, so by base extending by each $y \colon \operatorname{Spec} k \to Y$ with $k$ an algebraically closed field, we reduce to the case $Y=\operatorname{Spec} k$. Moreover, we may replace $X$ by an open subscheme to assume that $X$ is etale over $\mathbb{A}^n\_k$ for some $n \ge 0$.


Fix a projective variety $P$ and a surjection $\mathcal{F} \to \mathcal{G}$ of coherent sheaves on $P$ such that some $g \in \Gamma(P,\mathcal{G})$ is not in the image of $\Gamma(P,\mathcal{F})$. (For instance, take $P = \mathbb{P}^1$, let $\mathcal{F} = \mathcal{O}\_P$, and let $\mathcal{G}$ be the quotient corresponding to a subscheme consisting of two $k$-points.) Make $\mathcal{O}\_P \oplus \mathcal{F}$ an $\mathcal{O}\_P$-algebra by declaring that $\mathcal{F} \cdot \mathcal{F} = 0$, and let $T = \operatorname{\bf Spec}(\mathcal{O}\_P \oplus \mathcal{F})$. Similarly, define $T\_0 = \operatorname{\bf Spec}(\mathcal{O}\_P \oplus \mathcal{G})$, which is a closed subscheme of $T$ defined by a nilpotent ideal sheaf. We then may view $g = 0+g \in \Gamma(P,\mathcal{O}\_P \oplus \mathcal{G}) = \Gamma(T\_0,\mathcal{O}\_{T\_0})$.


Choose $x \in X(k)$; without loss of generality its image in $\mathbb{A}^n(k)$ is the origin. Using the infinitesimal lifting property for the etale morphism $X \to \mathbb{A}^n$ and the nilpotent thickening $P \subseteq T\_0$, we lift the point $(g,g,\ldots,g) \in \mathcal{A}^n(T\_0)$ mapping to $(0,0,\ldots,0) \in \mathbb{A}^n(P)$ to some $x\_0 \in X(T\_0)$ mapping to $x \in X(k) \subseteq X(P)$. By wicked smoothness, $x\_0$ lifts to some $x\_T \in X(T)$. The image of $x\_T$ in $\mathbb{A}^n(T)$ lifts $(g,g,\ldots,g)$, so each coordinate of $x\_T$ is a global section of $\mathcal{F}$ mapping to $g$, which is a contradiction unless $n=0$. Thus $X \to Y$ is etale.


