## 76.20 Smoothness over a Noetherian base

This section is the analogue of More on Morphisms, Section 37.12.

Lemma 76.20.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $x \in |X|$. Assume that $Y$ is locally Noetherian and $f$ locally of finite type. The following are equivalent:

1. $f$ is smooth at $x$,

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 surjection of local rings with $\mathop{\mathrm{Ker}}(B' \to B)$ of square zero, and $\alpha$ mapping the closed point of $\mathop{\mathrm{Spec}}(B)$ to $x$ there exists a dotted arrow making the diagram commute, and

3. same as in (2) but with $B' \to B$ ranging over small extensions (see Algebra, Definition 10.141.1).

Proof. Condition (1) means there is an open subspace $X' \subset X$ such that $X' \to Y$ is smooth. Hence (1) implies conditions (2) and (3) by Lemma 76.19.6. Condition (2) implies condition (3) trivially. Assume (3). Choose a commutative diagram

$\xymatrix{ X \ar[d] & U \ar[l] \ar[d] \\ Y & V \ar[l] }$

with $U$ and $V$ affine, horizontal arrows étale and such that there is a point $u \in U$ mapping to $x$. Next, consider a diagram

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

as in (3) but for $u \in U \to V$. Let $\gamma : \mathop{\mathrm{Spec}}(B') \to X$ be the arrow we get from our assumption that (3) holds for $X$. Because $U \to X$ is étale and hence formally étale (Lemma 76.16.8) the morphism $\gamma$ has a unique lift to $U$ compatible with $\alpha$. Then because $V \to Y$ is étale hence formally étale this lift is compatible with $\beta$. Hence (3) holds for $u \in U \to V$ and we conclude that $U \to V$ is smooth at $u$ by More on Morphisms, Lemma 37.12.1. This proves that $X \to Y$ is smooth at $x$, thereby finishing the proof. $\square$

Sometimes it is useful to know that one only needs to check the lifting criterion for small extensions “centered” at points of finite type (see Morphisms of Spaces, Section 67.25).

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. $\square$

Here is a useful application.

Lemma 76.20.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is locally of finite type and $Y$ locally Noetherian. Let $Z \subset Y$ be a closed subspace with $n$th infinitesimal neighbourhood $Z_ n \subset Y$. Set $X_ n = Z_ n \times _ Y X$.

1. If $X_ n \to Z_ n$ is smooth for all $n$, then $f$ is smooth at every point of $f^{-1}(Z)$.

2. If $X_ n \to Z_ n$ is étale for all $n$, then $f$ is étale at every point of $f^{-1}(Z)$.

Proof. Assume $X_ n \to Z_ n$ is smooth for all $n$. Let $x \in X$ be a point lying over a point of $Z$. Given a small extension $B' \to B$ and morphisms $\alpha$, $\beta$ as in Lemma 76.20.1 part (3) the maximal ideal of $B'$ is nilpotent (as $B'$ is Artinian) and hence the morphism $\beta$ factors through $Z_ n$ and $\alpha$ factors through $X_ n$ for a suitable $n$. Thus the lifting property for $X_ n \to Z_ n$ kicks in to get the desired dotted arrow in the diagram. This proves (1). Part (2) follows from (1) and the fact that a morphism is étale if and only if it is smooth of relative dimension $0$. $\square$

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