Definition 72.16.1. 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 72.13.1.
72.16 Formally étale morphisms
In this section we work out what it means that a morphism of algebraic spaces is formally étale.
We will not restate the results proved in the more general setting of formally étale transformations of functors in Section 72.13.
Lemma 72.16.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:
$f$ is formally étale,
for every diagram
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 36.8.1), and
for one such diagram with surjective vertical arrows the morphism $\psi $ is formally étale.
\[ \xymatrix{ U \ar[d] \ar[r]_\psi & V \ar[d] \\ X \ar[r]^ f & Y } \]
Proof. Assume $f$ is formally étale. By Lemma 72.13.5 the morphisms $U \to X$ and $V \to Y$ are formally étale. Thus by Lemma 72.13.3 the composition $U \to Y$ is formally étale. Then it follows from Lemma 72.13.8 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 72.13.5 the morphism $V \to Y$ is formally étale. Thus by Lemma 72.13.3 the composition $U \to Y$ is formally étale. Then it follows from Lemma 72.13.6 that $X \to Y$ is formally étale, i.e., (1) holds. $\square$
Lemma 72.16.3. 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 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 72.16.1 the condition that $T$ be affine may be dropped.
\[ \xymatrix{ X \ar[d]_ f & T \ar[d]^ i \ar[l]_ a \\ Y & T' \ar[l] \ar@{-->}[lu] } \]
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 72.14.3) 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 72.16.4. A composition of formally étale morphisms is formally étale.
Proof. This is formal. $\square$
Lemma 72.16.5. A base change of a formally étale morphism is formally étale.
Proof. This is formal. $\square$
Lemma 72.16.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ The following are equivalent:
$f$ is formally étale,
$f$ is formally unramified and the universal first order thickening of $X$ over $Y$ is equal to $X$,
$f$ is formally unramified and $\mathcal{C}_{X/Y} = 0$, and
$\Omega _{X/Y} = 0$ 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 72.14.6 and Definition 72.15.5. 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 $X'$. $\square$
Lemma 72.16.7. An unramified flat morphism is formally étale.
Proof. Follows from the case of schemes, see More on Morphisms, Lemma 36.8.7 and étale localization, see Lemmas 72.14.2 and 72.16.2 and Morphisms of Spaces, Lemma 63.30.5. $\square$
Lemma 72.16.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:
The morphism $f$ is étale, and
the morphism $f$ is locally of finite presentation and formally étale.
Proof. Follows from the case of schemes, see More on Morphisms, Lemma 36.8.9 and étale localization, see Lemma 72.16.2 and Morphisms of Spaces, Lemmas 63.28.4 and 63.39.2. $\square$
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