Michael Artin's position on differential criteria of smoothness (e.g., Morphisms, Lemma 29.34.14) is that they are basically useless (in practice). In this section we introduce the notion of a formally smooth morphism $X \to S$. 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 37.11.7. It turns out that this criterion is often easier to use than the differential criteria mentioned above.

Recall that a ring map $R \to A$ is called *formally smooth* (see Algebra, Definition 10.138.1) if for every commutative solid diagram

\[ \xymatrix{ A \ar[r] \ar@{-->}[rd] & B/I \\ R \ar[r] \ar[u] & B \ar[u] } \]

where $I \subset B$ is an ideal of square zero, a dotted arrow exists which makes the diagram commute. This motivates the following analogue for morphisms of schemes.

Definition 37.11.1. Let $f : X \to S$ be a morphism of schemes. We say $f$ is *formally smooth* if given any solid commutative diagram

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

where $T \subset T'$ is a first order thickening of affine schemes over $S$ there exists a dotted arrow making the diagram commute.

In the cases of formally unramified and formally étale morphisms the condition that $T'$ be affine could be dropped, see Lemmas 37.6.2 and 37.8.2. 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 Zariski locally on $T'$. In fact, analyzing the proof of Lemma 37.11.10 shows that this would be equivalent to the definition as it currently stands. In particular, being formally smooth is Zariski local on the source (and in fact it is smooth local on the source, insert future reference here).

Lemma 37.11.2. A composition of formally smooth morphisms is formally smooth.

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

Lemma 37.11.3. 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 37.11.4. 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.
$\square$

Lemma 37.11.5. Let $f : X \to S$ be a morphism of schemes. Let $U \subset X$ and $V \subset S$ be open subschemes such that $f(U) \subset V$. If $f$ is formally smooth, so is $f|_ U : U \to V$.

**Proof.**
Consider a solid diagram

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

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.
$\square$

Lemma 37.11.6. Let $f : X \to S$ be a morphism of schemes. Assume $X$ and $S$ are affine. Then $f$ is formally smooth if and only if $\mathcal{O}_ S(S) \to \mathcal{O}_ X(X)$ is a formally smooth ring map.

**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.
$\square$

The following lemma is the main result of this section. It is a victory of the functorial point of view in that it implies (combined with Limits, Proposition 32.6.1) that we can recognize whether a morphism $f : X \to S$ is smooth in terms of “simple” properties of the functor $h_ X : \mathit{Sch}/S \to \textit{Sets}$.

Lemma 37.11.7 (Infinitesimal lifting criterion). Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

The morphism $f$ is smooth, and

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 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

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

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

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

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.
$\square$

Locally projective quasi-coherent modules are defined in Properties, Section 28.21.

Lemma 37.11.8. Let $f : X \to Y$ be a formally smooth morphism of schemes. Then $\Omega _{X/Y}$ is locally projective on $X$.

**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.
$\square$

Lemma 37.11.9. Let $T$ be an affine scheme. Let $\mathcal{F}$, $\mathcal{G}$ be quasi-coherent $\mathcal{O}_ T$-modules. Consider $\mathcal{H} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ T}(\mathcal{F}, \mathcal{G})$. If $\mathcal{F}$ is locally projective, then $H^1(T, \mathcal{H}) = 0$.

**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.
$\square$

Lemma 37.11.10. Let $f : X \to Y$ be a morphism of schemes. The following are equivalent:

$f$ is formally smooth,

for every $x \in X$ there exist opens $x \in U \subset X$ and $f(x) \in V \subset Y$ with $f(U) \subset V$ such that $f|_ U : U \to V$ is formally smooth,

for every pair of affine opens $U \subset X$ and $V \subset Y$ with $f(U) \subset V$ the ring map $\mathcal{O}_ Y(V) \to \mathcal{O}_ X(U)$ is formally smooth, and

there exists an affine open covering $Y = \bigcup V_ j$ and for each $j$ an affine open covering $f^{-1}(V_ j) = \bigcup U_{ji}$ such that $\mathcal{O}_ Y(V) \to \mathcal{O}_ X(U)$ is a formally smooth ring map for all $j$ and $i$.

**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

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

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

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

be the sheaf of $\mathcal{O}_ T$-modules on $T$ 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

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

by Lemma 37.11.9 as desired.
$\square$

Lemma 37.11.11. Let $f : X \to Y$, $g : Y \to S$ be morphisms of schemes. Assume $f$ is formally smooth. Then

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

(see Morphisms, Lemma 29.32.9) is short exact.

**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

\[ 0 \to C \otimes _ B \Omega _{B/A} \to \Omega _{C/A} \to \Omega _{C/B} \to 0 \]

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

Lemma 37.11.12. Let $h : Z \to X$ be a formally unramified morphism of schemes over $S$. Assume that $Z$ is formally smooth over $S$. Then the canonical exact sequence

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

of Lemma 37.7.10 is short exact.

**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

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

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.
$\square$

Lemma 37.11.13. Let

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

be a commutative diagram of schemes 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 37.7.11 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 37.7.10 here is a canonical morphism $Z' \to Z''$ so that we have a commutative diagram

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

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

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

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

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

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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