## 75.16 Formally étale morphisms

In this section we work out what it means that a morphism of algebraic spaces is formally étale.

Definition 75.16.1. Let $S$ be a scheme. A morphism $f : X \to Y$ of algebraic spaces over $S$ is said to be formally étale if it is formally étale as a transformation of functors as in Definition 75.13.1.

We will not restate the results proved in the more general setting of formally étale transformations of functors in Section 75.13.

Lemma 75.16.2. 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 étale,

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 étale (as in More on Morphisms, Definition 37.8.1), and

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

Proof. Assume $f$ is formally étale. By Lemma 75.13.5 the morphisms $U \to X$ and $V \to Y$ are formally étale. Thus by Lemma 75.13.3 the composition $U \to Y$ is formally étale. Then it follows from Lemma 75.13.8 that $U \to V$ is formally étale. Thus (1) implies (2). And (2) implies (3) trivially

Assume given a diagram as in (3). 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 étale. Then it follows from Lemma 75.13.6 that $X \to Y$ is formally étale, i.e., (1) holds. $\square$

Lemma 75.16.3. Let $S$ be a scheme. Let $f : X \to Y$ be a formally étale morphism of algebraic spaces over $S$. Then given any solid commutative diagram

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

where $T \subset T'$ is a first order thickening of algebraic spaces over $Y$ there exists exactly one dotted arrow making the diagram commute. In other words, in Definition 75.16.1 the condition that $T$ be affine may be dropped.

Proof. Let $U' \to T'$ be a surjective étale morphism where $U' = \coprod U'_ i$ is a disjoint union of affine schemes. Let $U_ i = T \times _{T'} U'_ i$. Then we get morphisms $a'_ i : U'_ i \to X$ such that $a'_ i|_{U_ i}$ equals the composition $U_ i \to T \to X$. By uniqueness (see Lemma 75.14.3) we see that $a'_ i$ and $a'_ j$ agree on the fibre product $U'_ i \times _{T'} U'_ j$. Hence $\coprod a'_ i : U' \to X$ descends to give a unique morphism $a' : T' \to X$. $\square$

Lemma 75.16.4. A composition of formally étale morphisms is formally étale.

Proof. This is formal. $\square$

Lemma 75.16.5. A base change of a formally étale morphism is formally étale.

Proof. This is formal. $\square$

Lemma 75.16.6. 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 étale,

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

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

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

Proof. Actually, the last assertion only make sense because $\Omega _{X/Y} = 0$ implies that $\mathcal{C}_{X/Y}$ is defined via Lemma 75.14.6 and Definition 75.15.5. 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 $X = X' \Leftrightarrow \mathcal{C}_{X/Y} = 0$ since after all by definition $\mathcal{C}_{X/Y} = \mathcal{C}_{X/X'}$ is the ideal sheaf of $X$ in $X'$. $\square$

Proof. Follows from the case of schemes, see More on Morphisms, Lemma 37.8.7 and étale localization, see Lemmas 75.14.2 and 75.16.2 and Morphisms of Spaces, Lemma 66.30.5. $\square$

Lemma 75.16.8. 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 étale, and

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

Proof. Follows from the case of schemes, see More on Morphisms, Lemma 37.8.9 and étale localization, see Lemma 75.16.2 and Morphisms of Spaces, Lemmas 66.28.4 and 66.39.2. $\square$

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