Definition 76.19.1. Let $S$ be a scheme. A morphism $f : X \to Y$ of algebraic spaces over $S$ is said to be *formally smooth* if it is formally smooth as a transformation of functors as in Definition 76.13.1.

## 76.19 Formally smooth morphisms

In this section we introduce the notion of a formally smooth morphism $X \to Y$ of algebraic spaces. Such a morphism is characterized by the property that $T$-valued points of $X$ lift to infinitesimal thickenings of $T$ provided $T$ is affine. The main result is that a morphism which is formally smooth and locally of finite presentation is smooth, see Lemma 76.19.6. It turns out that this criterion is often easier to use than the Jacobian criterion.

In the cases of formally unramified and formally étale morphisms the condition that $T'$ be affine could be dropped, see Lemmas 76.14.3 and 76.16.3. This is no longer true in the case of formally smooth morphisms. In fact, a slightly more natural condition would be that we should be able to fill in the dotted arrow étale locally on $T'$. In fact, analyzing the proof of Lemma 76.19.6 shows that this would be equivalent to the definition as it currently stands. It is also true that requiring the existence of the dotted arrow fppf locally on $T'$ would be sufficient, but that is slightly more difficult to prove.

We will not restate the results proved in the more general setting of formally smooth transformations of functors in Section 76.13.

Lemma 76.19.2. A composition of formally smooth morphisms is formally smooth.

**Proof.**
Omitted.
$\square$

Lemma 76.19.3. A base change of a formally smooth morphism is formally smooth.

**Proof.**
Omitted, but see Algebra, Lemma 10.138.2 for the algebraic version.
$\square$

Lemma 76.19.4. Let $f : X \to S$ be a morphism of schemes. Then $f$ is formally étale if and only if $f$ is formally smooth and formally unramified.

**Proof.**
Omitted.
$\square$

Here is a helper lemma which will be superseded by Lemma 76.19.10.

Lemma 76.19.5. Let $S$ be a scheme. Let

be a commutative diagram of morphisms of algebraic spaces over $S$. If the vertical arrows are étale and $f$ is formally smooth, then $\psi $ is formally smooth.

**Proof.**
By Lemma 76.13.5 the morphisms $U \to X$ and $V \to Y$ are formally étale. By Lemma 76.13.3 the composition $U \to Y$ is formally smooth. By Lemma 76.13.8 we see $\psi : U \to V$ is formally smooth.
$\square$

The following lemma is the main result of this section. It implies, combined with Limits of Spaces, Proposition 70.3.10, that we can recognize whether a morphism of algebraic spaces $f : X \to Y$ is smooth in terms of “simple” properties of the transformation of functors $X \to Y$.

Lemma 76.19.6 (Infinitesimal lifting criterion). Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:

The morphism $f$ is smooth.

The morphism $f$ is locally of finite presentation, and formally smooth.

**Proof.**
Assume $f : X \to S$ is locally of finite presentation and formally smooth. Consider a commutative diagram

where $U$ and $V$ are schemes and the vertical arrows are étale and surjective. By Lemma 76.19.5 we see $\psi : U \to V$ is formally smooth. By Morphisms of Spaces, Lemma 67.28.4 the morphism $\psi $ is locally of finite presentation. Hence by the case of schemes the morphism $\psi $ is smooth, see More on Morphisms, Lemma 37.11.7. Hence $f$ is smooth, see Morphisms of Spaces, Lemma 67.37.4.

Conversely, assume that $f : X \to Y$ is smooth. Consider a solid commutative diagram

as in Definition 76.19.1. We will show the dotted arrow exists thereby proving that $f$ is formally smooth. Let $\mathcal{F}$ be the sheaf of sets on $(T')_{spaces, {\acute{e}tale}}$ of Lemma 76.17.4 as in the special case discussed in Remark 76.17.6. Let

be the sheaf of $\mathcal{O}_ T$-modules on $T_{spaces, {\acute{e}tale}}$ with action $\mathcal{H} \times \mathcal{F} \to \mathcal{F}$ as in Lemma 76.17.5. The action $\mathcal{H} \times \mathcal{F} \to \mathcal{F}$ turns $\mathcal{F}$ into a pseudo $\mathcal{H}$-torsor, see Cohomology on Sites, Definition 21.4.1. Our goal is to show that $\mathcal{F}$ is a trivial $\mathcal{H}$-torsor. There are two steps: (I) To show that $\mathcal{F}$ is a torsor we have to show that $\mathcal{F}$ has étale locally a section. (II) To show that $\mathcal{F}$ is the trivial torsor it suffices to show that $H^1(T_{\acute{e}tale}, \mathcal{H}) = 0$, see Cohomology on Sites, Lemma 21.4.3.

First we prove (I). To see this choose a commutative diagram

where $U$ and $V$ are schemes and the vertical arrows are étale and surjective. As $f$ is assumed smooth we see that $\psi $ is smooth and hence formally smooth by Lemma 76.13.5. By the same lemma the morphism $V \to Y$ is formally étale. Thus by Lemma 76.13.3 the composition $U \to Y$ is formally smooth. Then (I) follows from Lemma 76.13.6 part (4).

Finally we prove (II). By Lemma 76.7.15 we see that $\Omega _{X/S}$ is of finite presentation. Hence $a^*\Omega _{X/S}$ is of finite presentation (see Properties of Spaces, Section 66.30). Hence the sheaf $\mathcal{H} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ T}(a^*\Omega _{X/Y}, \mathcal{C}_{T/T'})$ is quasi-coherent by Properties of Spaces, Lemma 66.29.7. Thus by Descent, Proposition 35.9.3 and Cohomology of Schemes, Lemma 30.2.2 we have

as desired. $\square$

Smooth morphisms satisfy strong local lifting property, see Lemma 76.19.7. If in the lemma we assume $T'$ is affine, then we do not know if it is necessary to take an étale covering. More precisely, if we have a commutative diagram

of algebraic spaces where $X \to Y$ is smooth and $T \to T'$ is a thickening of affine schemes, the does a dotted arrow making the diagram commute always exist? If you know the answer, or if you have a reference, please email stacks.project@gmail.com.

Lemma 76.19.7. Let $S$ be a scheme. Consider a commutative diagram

of algebraic spaces over $S$ where $X \to Y$ is smooth and $T \to T'$ is a thickening. Then there exists an étale covering $\{ T'_ i \to T'\} $ such that we can find the dotted arrow in

making the diagram commute (for all $i$).

**Proof.**
Choose an étale covering $\{ Y_ i \to Y\} $ with each $Y_ i$ affine. After replacing $T'$ by the induced étale covering we may assume $Y$ is affine.

Assume $Y$ is affine. Choose an étale covering $\{ X_ i \to X\} $. This gives rise to an étale covering of $T$. This étale covering of $T$ comes from an étale covering of $T'$ (by Theorem 76.8.1, see discussion in Section 76.9). Hence we may assume $X$ is affine.

Assume $X$ and $Y$ are affine. We can do one more étale covering of $T'$ and assume $T'$ is affine. In this case the lemma follows from Algebra, Lemma 10.138.17. $\square$

We do a bit more work to show that being formally smooth is étale local on the source. To begin we show that a formally smooth morphism has a nice sheaf of differentials. The notion of a locally projective quasi-coherent module is defined in Properties of Spaces, Section 66.31.

Lemma 76.19.8. Let $S$ be a scheme. Let $f : X \to Y$ be a formally smooth morphism of algebraic spaces over $S$. Then $\Omega _{X/Y}$ is locally projective on $X$.

**Proof.**
Choose a diagram

where $U$ and $V$ are affine(!) schemes and the vertical arrows are étale. By Lemma 76.19.5 we see $\psi : U \to V$ is formally smooth. Hence $\Gamma (V, \mathcal{O}_ V) \to \Gamma (U, \mathcal{O}_ U)$ is a formally smooth ring map, see More on Morphisms, Lemma 37.11.6. Hence by Algebra, Lemma 10.138.7 the $\Gamma (U, \mathcal{O}_ U)$-module $\Omega _{\Gamma (U, \mathcal{O}_ U)/\Gamma (V, \mathcal{O}_ V)}$ is projective. Hence $\Omega _{U/V}$ is locally projective, see Properties, Section 28.21. Since $\Omega _{X/Y}|_ U = \Omega _{U/V}$ we see that $\Omega _{X/Y}$ is locally projective too. (Because we can find an étale covering of $X$ by the affine $U$'s fitting into diagrams as above – details omitted.) $\square$

Lemma 76.19.9. Let $T$ be an affine scheme. Let $\mathcal{F}$, $\mathcal{G}$ be quasi-coherent $\mathcal{O}_ T$-modules on $T_{\acute{e}tale}$. Consider the internal hom sheaf $\mathcal{H} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ T}(\mathcal{F}, \mathcal{G})$ on $T_{\acute{e}tale}$. If $\mathcal{F}$ is locally projective, then $H^1(T_{\acute{e}tale}, \mathcal{H}) = 0$.

**Proof.**
By the definition of a locally projective sheaf on an algebraic space (see Properties of Spaces, Definition 66.31.2) we see that $\mathcal{F}_{Zar} = \mathcal{F}|_{T_{Zar}}$ is a locally projective sheaf on the scheme $T$. Thus $\mathcal{F}_{Zar}$ is a direct summand of a free $\mathcal{O}_{T_{Zar}}$-module. Whereupon we conclude (as $\mathcal{F} = (\mathcal{F}_{Zar})^ a$, see Descent, Proposition 35.8.9) that $\mathcal{F}$ is a direct summand of a free $\mathcal{O}_ T$-module on $T_{\acute{e}tale}$. Hence we may assume that $\mathcal{F} = \bigoplus _{i \in I} \mathcal{O}_ T$ is a free module. In this case $\mathcal{H} = \prod _{i \in I} \mathcal{G}$ is a product of quasi-coherent modules. By Cohomology on Sites, Lemma 21.12.5 we conclude that $H^1 = 0$ because the cohomology of a quasi-coherent sheaf on an affine scheme is zero, see Descent, Proposition 35.9.3 and Cohomology of Schemes, Lemma 30.2.2.
$\square$

Lemma 76.19.10. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:

$f$ is formally smooth,

for every diagram

\[ \xymatrix{ U \ar[d] \ar[r]_\psi & V \ar[d] \\ X \ar[r]^ f & Y } \]where $U$ and $V$ are schemes and the vertical arrows are étale the morphism of schemes $\psi $ is formally smooth (as in More on Morphisms, Definition 37.6.1), and

for one such diagram with surjective vertical arrows the morphism $\psi $ is formally smooth.

**Proof.**
We have seen that (1) implies (2) and (3) in Lemma 76.19.5. Assume (3). The proof that $f$ is formally smooth is entirely similar to the proof of (1) $\Rightarrow $ (2) of Lemma 76.19.6.

Consider a solid commutative diagram

as in Definition 76.19.1. We will show the dotted arrow exists thereby proving that $f$ is formally smooth. Let $\mathcal{F}$ be the sheaf of sets on $(T')_{spaces, {\acute{e}tale}}$ of Lemma 76.17.4 as in the special case discussed in Remark 76.17.6. Let

be the sheaf of $\mathcal{O}_ T$-modules on $T_{spaces, {\acute{e}tale}}$ with action $\mathcal{H} \times \mathcal{F} \to \mathcal{F}$ as in Lemma 76.17.5. The action $\mathcal{H} \times \mathcal{F} \to \mathcal{F}$ turns $\mathcal{F}$ into a pseudo $\mathcal{H}$-torsor, see Cohomology on Sites, Definition 21.4.1. Our goal is to show that $\mathcal{F}$ is a trivial $\mathcal{H}$-torsor. There are two steps: (I) To show that $\mathcal{F}$ is a torsor we have to show that $\mathcal{F}$ has étale locally a section. (II) To show that $\mathcal{F}$ is the trivial torsor it suffices to show that $H^1(T_{\acute{e}tale}, \mathcal{H}) = 0$, see Cohomology on Sites, Lemma 21.4.3.

First we prove (I). To see this consider a diagram (which exists because we are assuming (3))

where $U$ and $V$ are schemes, the vertical arrows are étale and surjective, and $\psi $ is formally smooth. By Lemma 76.13.5 the morphism $V \to Y$ is formally étale. Thus by Lemma 76.13.3 the composition $U \to Y$ is formally smooth. Then (I) follows from Lemma 76.13.6 part (4).

Finally we prove (II). By Lemma 76.19.8 we see that $\Omega _{U/V}$ locally projective. Hence $\Omega _{X/Y}$ is locally projective, see Descent on Spaces, Lemma 74.6.5. Hence $a^*\Omega _{X/Y}$ is locally projective, see Properties of Spaces, Lemma 66.31.3. Hence

by Lemma 76.19.9 as desired. $\square$

Lemma 76.19.11. The property $\mathcal{P}(f) =$“$f$ is formally smooth” is fpqc local on the base.

**Proof.**
Let $f : X \to Y$ be a morphism of algebraic spaces over a scheme $S$. Choose an index set $I$ and diagrams

with étale vertical arrows and $U_ i$, $V_ i$ affine schemes. Moreover, assume that $\coprod U_ i \to X$ and $\coprod V_ i \to Y$ are surjective, see Properties of Spaces, Lemma 66.6.1. By Lemma 76.19.10 we see that $f$ is formally smooth if and only if each of the morphisms $\psi _ i$ are formally smooth. Hence we reduce to the case of a morphism of affine schemes. In this case the result follows from Algebra, Lemma 10.138.16. Some details omitted. $\square$

Lemma 76.19.12. Let $S$ be a scheme. Let $f : X \to Y$, $g : Y \to Z$ be morphisms of algebraic spaces over $S$. Assume $f$ is formally smooth. Then

Lemma 76.7.8 is short exact.

**Proof.**
Follows from the case of schemes, see More on Morphisms, Lemma 37.11.11, by étale localization, see Lemmas 76.19.10 and 76.7.3.
$\square$

Lemma 76.19.13. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $h : Z \to X$ be a formally unramified morphism of algebraic spaces over $B$. Assume that $Z$ is formally smooth over $B$. Then the canonical exact sequence

of Lemma 76.15.13 is short exact.

**Proof.**
Let $Z \to Z'$ be the universal first order thickening of $Z$ over $X$. From the proof of Lemma 76.15.13 we see that our sequence is identified with the sequence

Since $Z \to S$ is formally smooth we can étale locally on $Z'$ find a left inverse $Z' \to Z$ over $B$ to the inclusion map $Z \to Z'$. Thus the sequence is étale locally split, see Lemma 76.7.11. $\square$

Lemma 76.19.14. Let $S$ be a scheme. Let

be a commutative diagram of algebraic spaces over $S$ where $i$ and $j$ are formally unramified and $f$ is formally smooth. Then the canonical exact sequence

of Lemma 76.15.14 is exact and locally split.

**Proof.**
Denote $Z \to Z'$ the universal first order thickening of $Z$ over $X$. Denote $Z \to Z''$ the universal first order thickening of $Z$ over $Y$. By Lemma 76.15.13 here is a canonical morphism $Z' \to Z''$ so that we have a commutative diagram

The sequence above is identified with the sequence

via our definitions concerning conormal sheaves of formally unramified morphisms. Let $U'' \to Z''$ be an étale morphism with $U''$ affine. Denote $U \to Z$ and $U' \to Z'$ the corresponding affine schemes étale over $Z$ and $Z'$. As $f$ is formally smooth there exists a morphism $h : U'' \to X$ which agrees with $i$ on $U$ and such that $f \circ h$ equals $b|_{U''}$. Since $Z'$ is the universal first order thickening we obtain a unique morphism $g : U'' \to Z'$ such that $g = a \circ h$. The universal property of $Z''$ implies that $k \circ g$ is the inclusion map $U'' \to Z''$. Hence $g$ is a left inverse to $k$. Picture

Thus $g$ induces a map $\mathcal{C}_{Z/Z'}|_ U \to \mathcal{C}_{Z/Z''}|_ U$ which is a left inverse to the map $\mathcal{C}_{Z/Z''} \to \mathcal{C}_{Z/Z'}$ over $U$. $\square$

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