Definition 76.14.1. Let $S$ be a scheme. A morphism $f : X \to Y$ of algebraic spaces over $S$ is said to be *formally unramified* if it is formally unramified as a transformation of functors as in Definition 76.13.1.

## 76.14 Formally unramified morphisms

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

We will not restate the results proved in the more general setting of formally unramified transformations of functors in Section 76.13. It turns out we can characterize this property in terms of vanishing of the module of relative differentials, see Lemma 76.14.6.

Lemma 76.14.2. 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 unramified,

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 unramified (as in More on Morphisms, Definition 37.6.1), and

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

**Proof.**
Assume $f$ is formally unramified. By Lemma 76.13.5 the morphisms $U \to X$ and $V \to Y$ are formally unramified. Thus by Lemma 76.13.3 the composition $U \to Y$ is formally unramified. Then it follows from Lemma 76.13.8 that $U \to V$ is formally unramified. 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 unramified. Thus by Lemma 76.13.3 the composition $U \to Y$ is formally unramified. Then it follows from Lemma 76.13.6 that $X \to Y$ is formally unramified, i.e., (1) holds. $\square$

Lemma 76.14.3. Let $S$ be a scheme. If $f : X \to Y$ is a formally unramified 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 $S$ there exists at most one dotted arrow making the diagram commute. In other words, in Definition 76.14.1 the condition that $T$ be an affine scheme may be dropped.

**Proof.**
This is true because there exists a surjective étale morphism $U' \to T'$ where $U'$ is a disjoint union of affine schemes (see Properties of Spaces, Lemma 66.6.1) and a morphism $T' \to X$ is determined by its restriction to $U'$.
$\square$

Lemma 76.14.4. A composition of formally unramified morphisms is formally unramified.

**Proof.**
This is formal.
$\square$

Lemma 76.14.5. A base change of a formally unramified morphism is formally unramified.

**Proof.**
This is formal.
$\square$

Lemma 76.14.6. 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 unramified, and

$\Omega _{X/Y} = 0$.

**Proof.**
This is a combination of Lemma 76.14.2, More on Morphisms, Lemma 37.6.7, and Lemma 76.7.3.
$\square$

Lemma 76.14.7. 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 unramified,

the morphism $f$ is locally of finite type and $\Omega _{X/Y} = 0$, and

the morphism $f$ is locally of finite type and formally unramified.

**Proof.**
Choose a diagram

where $U$ and $V$ are schemes and the vertical arrows are étale and surjective. Then we see

Here we have used Morphisms, Lemma 29.35.2 and Lemma 76.14.6. $\square$

Lemma 76.14.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:

$f$ is unramified and a monomorphism,

$f$ is unramified and universally injective,

$f$ is locally of finite type and a monomorphism,

$f$ is universally injective, locally of finite type, and formally unramified.

Moreover, in this case $f$ is also representable, separated, and locally quasi-finite.

**Proof.**
We have seen in Lemma 76.14.7 that being formally unramified and locally of finite type is the same thing as being unramified. Hence (4) is equivalent to (2). A monomorphism is certainly formally unramified hence (3) implies (4). It is clear that (1) implies (3). Finally, if (2) holds, then $\Delta : X \to X \times _ Y X$ is both an open immersion (Morphisms of Spaces, Lemma 67.38.9) and surjective (Morphisms of Spaces, Lemma 67.19.2) hence an isomorphism, i.e., $f$ is a monomorphism. In this way we see that (2) implies (1). Finally, we see that $f$ is representable, separated, and locally quasi-finite by Morphisms of Spaces, Lemmas 67.27.10 and 67.51.1.
$\square$

Lemma 76.14.9. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:

$f$ is a closed immersion,

$f$ is universally closed, unramified, and a monomorphism,

$f$ is universally closed, unramified, and universally injective,

$f$ is universally closed, locally of finite type, and a monomorphism,

$f$ is universally closed, universally injective, locally of finite type, and formally unramified.

**Proof.**
The equivalence of (2) – (5) follows immediately from Lemma 76.14.8. Moreover, if (2) – (5) are satisfied then $f$ is representable. Similarly, if (1) is satisfied then $f$ is representable. Hence the result follows from the case of schemes, see Étale Morphisms, Lemma 41.7.2.
$\square$

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