The Stacks project

Proof. Omitted. $\square$

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

Proof. Omitted. $\square$

Proof. Consider a solid diagram

as in Definition 37.11.1. If $f$ is formally smooth, then there exists an $S$-morphism $a' : T' \to X$ such that $a'|_ T = a$. Since the underlying sets of $T$ and $T'$ are the same we see that $a'$ is a morphism into $U$ (see Schemes, Section 26.3). And it clearly is a $V$-morphism as well. Hence the dotted arrow above as desired. $\square$

Proof. This is immediate from the definitions (Definition 37.11.1 and Algebra, Definition 10.138.1) by the equivalence of categories of rings and affine schemes, see Schemes, Lemma 26.6.5. $\square$

Proof. Assume $f : X \to S$ is locally of finite presentation and formally smooth. Consider a pair of affine opens $\mathop{\mathrm{Spec}}(A) = U \subset X$ and $\mathop{\mathrm{Spec}}(R) = V \subset S$ such that $f(U) \subset V$. By Lemma 37.11.5 we see that $U \to V$ is formally smooth. By Lemma 37.11.6 we see that $R \to A$ is formally smooth. By Morphisms, Lemma 29.21.2 we see that $R \to A$ is of finite presentation. By Algebra, Proposition 10.138.13 we see that $R \to A$ is smooth. Hence by the definition of a smooth morphism we see that $X \to S$ is smooth.

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

as in Definition 37.11.1. We will show the dotted arrow exists thereby proving that $f$ is formally smooth.

Let $\mathcal{F}$ be the sheaf of sets on $T'$ of Lemma 37.9.4 in the special case discussed in Remark 37.9.6. Let

be the sheaf of $\mathcal{O}_ T$-modules with action $\mathcal{H} \times \mathcal{F} \to \mathcal{F}$ as in Lemma 37.9.5. Our goal is simply to show that $\mathcal{F}(T) \not= \emptyset $. In other words we are trying to show that $\mathcal{F}$ is a trivial $\mathcal{H}$-torsor on $T$ (see Cohomology, Section 20.4). There are two steps: (I) To show that $\mathcal{F}$ is a torsor we have to show that $\mathcal{F}_ t \not= \emptyset $ for all $t \in T$ (see Cohomology, Definition 20.4.1). (II) To show that $\mathcal{F}$ is the trivial torsor it suffices to show that $H^1(T, \mathcal{H}) = 0$ (see Cohomology, Lemma 20.4.3 – we may use either cohomology of $\mathcal{H}$ as an abelian sheaf or as an $\mathcal{O}_ T$-module, see Cohomology, Lemma 20.13.3).

First we prove (I). To see this, for every $t \in T$ we can choose an affine open $U \subset T$ neighbourhood of $t$ such that $a(U)$ is contained in an affine open $\mathop{\mathrm{Spec}}(A) = W \subset X$ which maps to an affine open $\mathop{\mathrm{Spec}}(R) = V \subset S$. By Morphisms, Lemma 29.34.2 the ring map $R \to A$ is smooth. Hence by Algebra, Proposition 10.138.13 the ring map $R \to A$ is formally smooth. Lemma 37.11.6 in turn implies that $W \to V$ is formally smooth. Hence we can lift $a|_ U : U \to W$ to a $V$-morphism $a' : U' \to W \subset X$ showing that $\mathcal{F}(U) \not= \emptyset $.

Finally we prove (II). By Morphisms, Lemma 29.32.13 we see that $\Omega _{X/S}$ is of finite presentation (it is even finite locally free by Morphisms, Lemma 29.34.12). Hence $a^*\Omega _{X/S}$ is of finite presentation (see Modules, Lemma 17.11.4). Hence the sheaf $\mathcal{H} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ T}(a^*\Omega _{X/S}, \mathcal{C}_{T/T'})$ is quasi-coherent by the discussion in Schemes, Section 26.24. Thus by Cohomology of Schemes, Lemma 30.2.2 we have $H^1(T, \mathcal{H}) = 0$ as desired. $\square$

Proof. Choose $U \subset X$ and $V \subset Y$ affine open such that $f(U) \subset V$. By Lemma 37.11.5 $f|_ U : U \to V$ is formally smooth. Hence $\Gamma (V, \mathcal{O}_ V) \to \Gamma (U, \mathcal{O}_ U)$ is a formally smooth ring map, see 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. $\square$

Proof. By the definition of a locally projective sheaf on a scheme (see Properties, Definition 28.21.1) we see that $\mathcal{F}$ is a direct summand of a free $\mathcal{O}_ T$-module. 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, Lemma 20.11.12 we conclude that $H^1 = 0$ because the cohomology of a quasi-coherent sheaf on an affine scheme is zero, see Cohomology of Schemes, Lemma 30.2.2. $\square$

Proof. The implications (1) $\Rightarrow $ (2), (1) $\Rightarrow $ (3), and (2) $\Rightarrow $ (4) follow from Lemma 37.11.5. The implication (3) $\Rightarrow $ (4) is immediate.

Assume (4). The proof that $f$ is formally smooth is the same as the second part of the proof of Lemma 37.11.7. Consider a solid commutative diagram

as in Definition 37.11.1. We will show the dotted arrow exists thereby proving that $f$ is formally smooth. Let $\mathcal{F}$ be the sheaf of sets on $T'$ of Lemma 37.9.4 as in the special case discussed in Remark 37.9.6. Let

be the sheaf of $\mathcal{O}_ T$-modules on $T$ with action $\mathcal{H} \times \mathcal{F} \to \mathcal{F}$ as in Lemma 37.9.5. The action $\mathcal{H} \times \mathcal{F} \to \mathcal{F}$ turns $\mathcal{F}$ into a pseudo $\mathcal{H}$-torsor, see Cohomology, Definition 20.4.1. 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 $\mathcal{F}$ locally has a section. (II) To show that $\mathcal{F}$ is the trivial torsor it suffices to show that $H^1(T, \mathcal{H}) = 0$, see Cohomology, Lemma 20.4.3.

First we prove (I). To see this, for every $t \in T$ we can choose an affine open $W \subset T$ neighbourhood of $t$ such that $a(W)$ is contained in $U_{ji}$ for some $i, j$. Let $W' \subset T'$ be the corresponding open subscheme. By assumption (4) we can lift $a|_ W : W \to U_{ji}$ to a $V_ j$-morphism $a' : W' \to U_{ji}$ showing that $\mathcal{F}(W')$ is nonempty.

Finally we prove (II). By Lemma 37.11.8 we see that $\Omega _{U_{ji}/V_ j}$ locally projective. Hence $\Omega _{X/Y}$ is locally projective, see Properties, Lemma 28.21.2. Hence $a^*\Omega _{X/Y}$ is locally projective, see Properties, Lemma 28.21.3. Hence

by Lemma 37.11.9 as desired. $\square$

Proof. The algebraic version of this lemma is the following: Given ring maps $A \to B \to C$ with $B \to C$ formally smooth, then the sequence

of Algebra, Lemma 10.131.7 is exact. This is Algebra, Lemma 10.138.9. $\square$

Proof. Let $Z \to Z'$ be the universal first order thickening of $Z$ over $X$. From the proof of Lemma 37.7.10 we see that our sequence is identified with the sequence

Since $Z \to S$ is formally smooth we can locally on $Z'$ find a left inverse $Z' \to Z$ over $S$ to the inclusion map $Z \to Z'$. Thus the sequence is locally split, see Morphisms, Lemma 29.32.16. $\square$

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 37.7.10 here is a canonical morphism $Z' \to Z''$ so that we have a commutative diagram

In the proof of Lemma 37.7.11 we identified the sequence above with the sequence

Let $U'' \subset Z''$ be an affine open. Denote $U \subset Z$ and $U' \subset Z'$ the corresponding affine open subschemes. 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 $U$. $\square$