## 37.8 Formally étale morphisms

Recall that a ring map $R \to A$ is called formally étale (see Algebra, Definition 10.150.1) 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, there exists exactly one dotted arrow which makes the diagram commute. This motivates the following analogue for morphisms of schemes.

Definition 37.8.1. Let $f : X \to S$ be a morphism of schemes. We say $f$ is formally étale 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 exactly one dotted arrow making the diagram commute.

It is clear that a formally étale morphism is formally unramified. Hence if $f : X \to S$ is formally étale, then $\Omega _{X/S}$ is zero, see Lemma 37.6.7.

Lemma 37.8.2. If $f : X \to S$ is a formally étale morphism, then 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 schemes over $S$ there exists exactly one dotted arrow making the diagram commute. In other words, in Definition 37.8.1 the condition that $T$ be affine may be dropped.

Proof. Let $T' = \bigcup T'_ i$ be an affine open covering, and let $T_ i = T \cap T'_ i$. Then we get morphisms $a'_ i : T'_ i \to X$ fitting into the diagram. By uniqueness we see that $a'_ i$ and $a'_ j$ agree on any affine open subscheme of $T'_ i \cap T'_ j$. Hence $a'_ i$ and $a'_ j$ agree on $T'_ i \cap T'_ j$. Thus we see that the morphisms $a'_ i$ glue to a global morphism $a' : T' \to X$. The uniqueness of $a'$ we have seen in Lemma 37.6.2. $\square$

Lemma 37.8.3. A composition of formally étale morphisms is formally étale.

Proof. This is formal. $\square$

Lemma 37.8.4. A base change of a formally étale morphism is formally étale.

Proof. This is formal. $\square$

Lemma 37.8.5. 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 étale, 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 37.8.1. If $f$ is formally ramified, then there exists exactly one $S$-morphism $a' : T' \to X$ such that $a'|_ T = a$. Since $|T'| = |T|$ we conclude that $a'(T') \subset U$ which gives our unique morphism from $T'$ into $U$. $\square$

Lemma 37.8.6. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

1. $f$ is formally étale,

2. $f$ is formally unramified and the universal first order thickening of $X$ over $S$ is equal to $X$,

3. $f$ is formally unramified and $\mathcal{C}_{X/S} = 0$, and

4. $\Omega _{X/S} = 0$ and $\mathcal{C}_{X/S} = 0$.

Proof. Actually, the last assertion only make sense because $\Omega _{X/S} = 0$ implies that $\mathcal{C}_{X/S}$ is defined via Lemma 37.6.7 and Definition 37.7.2. This also makes it clear that (3) and (4) are equivalent.

Either of the assumptions (1), (2), and (3) imply that $f$ is formally unramified. Hence we may assume $f$ is formally unramified. The equivalence of (1), (2), and (3) follow from the universal property of the universal first order thickening $X'$ of $X$ over $S$ and the fact that $X = X' \Leftrightarrow \mathcal{C}_{X/S} = 0$ since after all by definition $\mathcal{C}_{X/S} = \mathcal{C}_{X/X'}$ is the ideal sheaf of $X$ in $X'$. $\square$

Proof. Say $X \to S$ is unramified and flat. Then $\Delta : X \to X \times _ S X$ is an open immersion, see Morphisms, Lemma 29.35.13. We have to show that $\mathcal{C}_{X/S}$ is zero. Consider the two projections $p, q : X \times _ S X \to X$. As $f$ is formally unramified (see Lemma 37.6.8), $q$ is formally unramified (see Lemma 37.6.4). As $f$ is flat, $p$ is flat, see Morphisms, Lemma 29.25.8. Hence $p^*\mathcal{C}_{X/S} = \mathcal{C}_ q$ by Lemma 37.7.7 where $\mathcal{C}_ q$ denotes the conormal sheaf of the formally unramified morphism $q : X \times _ S X \to X$. But $\Delta (X) \subset X \times _ S X$ is an open subscheme which maps isomorphically to $X$ via $q$. Hence by Lemma 37.7.8 we see that $\mathcal{C}_ q|_{\Delta (X)} = \mathcal{C}_{X/X} = 0$. In other words, the pullback of $\mathcal{C}_{X/S}$ to $X$ via the identity morphism is zero, i.e., $\mathcal{C}_{X/S} = 0$. $\square$

Lemma 37.8.8. Let $f : X \to S$ be a morphism of schemes. Assume $X$ and $S$ are affine. Then $f$ is formally étale if and only if $\mathcal{O}_ S(S) \to \mathcal{O}_ X(X)$ is a formally étale ring map.

Proof. This is immediate from the definitions (Definition 37.8.1 and Algebra, Definition 10.150.1) by the equivalence of categories of rings and affine schemes, see Schemes, Lemma 26.6.5. $\square$

Lemma 37.8.9. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

1. The morphism $f$ is étale, and

2. the morphism $f$ is locally of finite presentation and formally étale.

Proof. Assume $f$ is étale. An étale morphism is locally of finite presentation, flat and unramified, see Morphisms, Section 29.36. Hence $f$ is locally of finite presentation and formally étale, see Lemma 37.8.7.

Conversely, suppose that $f$ is locally of finite presentation and formally étale. Being étale is local in the Zariski topology on $X$ and $S$, see Morphisms, Lemma 29.36.2. By Lemma 37.8.5 we can cover $X$ by affine opens $U$ which map into affine opens $V$ such that $U \to V$ is formally étale (and of finite presentation, see Morphisms, Lemma 29.21.2). By Lemma 37.8.8 we see that the ring maps $\mathcal{O}(V) \to \mathcal{O}(U)$ are formally étale (and of finite presentation). We win by Algebra, Lemma 10.150.2. (We will give another proof of this implication when we discuss formally smooth morphisms.) $\square$

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