Lemma 76.20.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $Y$ is locally Noetherian and $f$ locally of finite type. The following are equivalent:

1. $f$ is smooth,

2. for every solid commutative diagram

$\xymatrix{ X \ar[d]_ f & \mathop{\mathrm{Spec}}(B) \ar[d]^ i \ar[l]^-\alpha \\ Y & \mathop{\mathrm{Spec}}(B') \ar[l]_-{\beta } \ar@{-->}[lu] }$

where $B' \to B$ is a small extension of Artinian local rings and $\beta$ of finite type (!) there exists a dotted arrow making the diagram commute.

Proof. If $f$ is smooth, then the infinitesimal lifting criterion (Lemma 76.19.6) says $f$ is formally smooth and (2) holds.

Assume $f$ is not smooth. The set of points $x \in X$ where $f$ is not smooth forms a closed subset $T$ of $|X|$. By Morphisms of Spaces, Lemma 67.25.6, there exists a point $x \in T \subset X$ with $x \in X_{\text{ft-pts}}$. Choose a commutative diagram

$\xymatrix{ X \ar[d] & U \ar[l] \ar[d] & u \ar@{|->}[d] \\ Y & V \ar[l] & v }$

with $U$ and $V$ affine, horizontal arrows étale and such that there is a point $u \in U$ mapping to $x$. Then $u$ is a finite type point of $U$. Since $U \to V$ is not smooth at the point $u$, by More on Morphisms, Lemma 37.12.1 there is a diagram

$\xymatrix{ X \ar[d] & U \ar[l] \ar[d] & \mathop{\mathrm{Spec}}(B) \ar[d]^ i \ar[l]^-\alpha \\ Y & V \ar[l] & \mathop{\mathrm{Spec}}(B') \ar[l]_-{\beta } \ar@{-->}[lu] }$

with $B' \to B$ a small extension of (Artinian) local rings such that the residue field of $B$ is equal to $\kappa (v)$ and such that the dotted arrow does not exist. Since $U \to V$ is of finite type, we see that $v$ is a finite type point of $V$. By Morphisms, Lemma 29.16.2 the morphism $\beta$ is of finite type, hence the composition $\mathop{\mathrm{Spec}}(B) \to Y$ is of finite type also. Arguing exactly as in the proof of Lemma 76.20.1 (using that $U \to X$ and $V \to Y$ are étale hence formally étale) we see that there cannot be an arrow $\mathop{\mathrm{Spec}}(B) \to X$ fitting into the outer rectangle of the last displayed diagram. In other words, (2) doesn't hold and the proof is complete. $\square$

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