Definition 76.16.1. 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 76.13.1.
76.16 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 76.13.
Lemma 76.16.2. 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 37.8.1), 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 76.13.5 the morphisms $U \to X$ and $V \to Y$ are formally étale. Thus by Lemma 76.13.3 the composition $U \to Y$ is formally étale. Then it follows from Lemma 76.13.8 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 76.13.5 the morphism $V \to Y$ is formally étale. Thus by Lemma 76.13.3 the composition $U \to Y$ is formally étale. Then it follows from Lemma 76.13.6 that $X \to Y$ is formally étale, i.e., (1) holds. $\square$
Lemma 76.16.3. 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 76.16.1 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 76.14.3) 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 76.16.4. A composition of formally étale morphisms is formally étale.
Proof. This is formal. $\square$
Lemma 76.16.5. A base change of a formally étale morphism is formally étale.
Proof. This is formal. $\square$
Lemma 76.16.6. 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 76.14.6 and Definition 76.15.5. 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 76.16.7. An unramified flat morphism is formally étale.
Proof. Follows from the case of schemes, see More on Morphisms, Lemma 37.8.7 and étale localization, see Lemmas 76.14.2 and 76.16.2 and Morphisms of Spaces, Lemma 67.30.5. $\square$
Lemma 76.16.8. 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 37.8.9 and étale localization, see Lemma 76.16.2 and Morphisms of Spaces, Lemmas 67.28.4 and 67.39.2. $\square$
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