The Stacks project

75.19 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 75.19.6. It turns out that this criterion is often easier to use than the Jacobian criterion.

Definition 75.19.1. 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 75.13.1.

In the cases of formally unramified and formally étale morphisms the condition that $T'$ be affine could be dropped, see Lemmas 75.14.3 and 75.16.3. 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 75.19.6 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 75.13.

Lemma 75.19.2. A composition of formally smooth morphisms is formally smooth.

Proof. Omitted. $\square$

Lemma 75.19.3. 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 75.19.4. 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Here is a helper lemma which will be superseded by Lemma 75.19.10.

Lemma 75.19.5. Let $S$ be a scheme. Let

\[ \xymatrix{ U \ar[d] \ar[r]_\psi & V \ar[d] \\ X \ar[r]^ f & Y } \]

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 75.13.5 the morphisms $U \to X$ and $V \to Y$ are formally étale. By Lemma 75.13.3 the composition $U \to Y$ is formally smooth. By Lemma 75.13.8 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 69.3.10, 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 75.19.6 (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:

  1. The morphism $f$ is smooth.

  2. 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

\[ \xymatrix{ U \ar[d] \ar[r]_\psi & V \ar[d] \\ X \ar[r]^ f & Y } \]

where $U$ and $V$ are schemes and the vertical arrows are étale and surjective. By Lemma 75.19.5 we see $\psi : U \to V$ is formally smooth. By Morphisms of Spaces, Lemma 66.28.4 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 66.37.4.

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

\[ \xymatrix{ X \ar[d]_ f & T \ar[d]^ i \ar[l]^ a \\ Y & T' \ar[l] \ar@{-->}[lu] } \]

as in Definition 75.19.1. 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 75.17.4 as in the special case discussed in Remark 75.17.6. Let

\[ \mathcal{H} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ T}(a^*\Omega _{X/Y}, \mathcal{C}_{T/T'}) \]

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 75.17.5. 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

\[ \xymatrix{ U \ar[d] \ar[r]_\psi & V \ar[d] \\ X \ar[r]^ f & Y } \]

where $U$ and $V$ are schemes and the vertical arrows are étale and surjective. As $f$ is assumed smooth we see that $\psi $ is smooth and hence formally smooth by Lemma 75.13.5. By the same lemma the morphism $V \to Y$ is formally étale. Thus by Lemma 75.13.3 the composition $U \to Y$ is formally smooth. Then (I) follows from Lemma 75.13.6 part (4).

Finally we prove (II). By Lemma 75.7.15 we see that $\Omega _{X/S}$ is of finite presentation. Hence $a^*\Omega _{X/S}$ is of finite presentation (see Properties of Spaces, Section 65.30). 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 65.29.7. Thus by Descent, Proposition 35.9.3 and Cohomology of Schemes, Lemma 30.2.2 we have

\[ H^1(T_{spaces, {\acute{e}tale}}, \mathcal{H}) = H^1(T_{\acute{e}tale}, \mathcal{H}) = H^1(T, \mathcal{H}) = 0 \]

as desired. $\square$

Smooth morphisms satisfy strong local lifting property, see Lemma 75.19.7. 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

\[ \xymatrix{ X \ar[d] & T \ar[l] \ar[d] \\ Y & T' \ar[l] \ar@{..>}[lu] } \]

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 75.19.7. Let $S$ be a scheme. Consider a commutative diagram

\[ \xymatrix{ X \ar[d] & T \ar[l] \ar[d] \\ Y & T' \ar[l] } \]

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

\[ \xymatrix{ X \ar[d] & T \ar[l] \ar[d] & T \times _{T'} T'_ i \ar[l] \ar[d] \\ Y & T' \ar[l] & T'_ i \ar[l] \ar@{..>}[llu] } \]

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 75.8.1, see discussion in Section 75.9). 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 65.31.

Lemma 75.19.8. 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

\[ \xymatrix{ U \ar[d] \ar[r]_\psi & V \ar[d] \\ X \ar[r]^ f & Y } \]

where $U$ and $V$ are affine(!) schemes and the vertical arrows are étale. By Lemma 75.19.5 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 75.19.9. 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 65.31.2) 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 75.19.10. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:

  1. $f$ is formally smooth,

  2. 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

  3. 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 75.19.5. Assume (3). The proof that $f$ is formally smooth is entirely similar to the proof of (1) $\Rightarrow $ (2) of Lemma 75.19.6.

Consider a solid commutative diagram

\[ \xymatrix{ X \ar[d]_ f & T \ar[d]^ i \ar[l]^ a \\ Y & T' \ar[l] \ar@{-->}[lu] } \]

as in Definition 75.19.1. 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 75.17.4 as in the special case discussed in Remark 75.17.6. Let

\[ \mathcal{H} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ T}(a^*\Omega _{X/Y}, \mathcal{C}_{T/T'}) \]

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 75.17.5. 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))

\[ \xymatrix{ U \ar[d] \ar[r]_\psi & V \ar[d] \\ X \ar[r]^ f & Y } \]

where $U$ and $V$ are schemes, the vertical arrows are étale and surjective, and $\psi $ is formally smooth. By Lemma 75.13.5 the morphism $V \to Y$ is formally étale. Thus by Lemma 75.13.3 the composition $U \to Y$ is formally smooth. Then (I) follows from Lemma 75.13.6 part (4).

Finally we prove (II). By Lemma 75.19.8 we see that $\Omega _{U/V}$ locally projective. Hence $\Omega _{X/Y}$ is locally projective, see Descent on Spaces, Lemma 73.6.5. Hence $a^*\Omega _{X/Y}$ is locally projective, see Properties of Spaces, Lemma 65.31.3. Hence

\[ H^1(T_{\acute{e}tale}, \mathcal{H}) = H^1(T_{\acute{e}tale}, \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ T}(a^*\Omega _{X/Y}, \mathcal{C}_{T/T'}) = 0 \]

by Lemma 75.19.9 as desired. $\square$

Lemma 75.19.11. 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

\[ \xymatrix{ U_ i \ar[d] \ar[r]_{\psi _ i} & V_ i \ar[d] \\ X \ar[r]^ f & Y } \]

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 65.6.1. By Lemma 75.19.10 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 75.19.12. 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

\[ 0 \to f^*\Omega _{Y/Z} \to \Omega _{X/Z} \to \Omega _{X/Y} \to 0 \]

Lemma 75.7.8 is short exact.

Proof. Follows from the case of schemes, see More on Morphisms, Lemma 37.11.11, by étale localization, see Lemmas 75.19.10 and 75.7.3. $\square$

Lemma 75.19.13. 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

\[ 0 \to \mathcal{C}_{Z/X} \to i^*\Omega _{X/B} \to \Omega _{Z/B} \to 0 \]

of Lemma 75.15.13 is short exact.

Proof. Let $Z \to Z'$ be the universal first order thickening of $Z$ over $X$. From the proof of Lemma 75.15.13 we see that our sequence is identified with the sequence

\[ \mathcal{C}_{Z/Z'} \to \Omega _{Z'/B} \otimes \mathcal{O}_ Z \to \Omega _{Z/B} \to 0. \]

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 75.7.11. $\square$

Lemma 75.19.14. Let $S$ be a scheme. Let

\[ \xymatrix{ Z \ar[r]_ i \ar[rd]_ j & X \ar[d]^ f \\ & Y } \]

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

\[ 0 \to \mathcal{C}_{Z/Y} \to \mathcal{C}_{Z/X} \to i^*\Omega _{X/Y} \to 0 \]

of Lemma 75.15.14 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 75.15.13 here is a canonical morphism $Z' \to Z''$ so that we have a commutative diagram

\[ \xymatrix{ Z \ar[r]_{i'} \ar[rd]_{j'} & Z' \ar[r]_ a \ar[d]^ k & X \ar[d]^ f \\ & Z'' \ar[r]^ b & Y } \]

The sequence above is identified with the sequence

\[ \mathcal{C}_{Z/Z''} \to \mathcal{C}_{Z/Z'} \to (i')^*\Omega _{Z'/Z''} \to 0 \]

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

\[ \xymatrix{ U \ar[d] \ar[r] & Z' \ar[d]^ k \\ U'' \ar[r] \ar[ru]^ g & Z'' } \]

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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