## 100.36 Unramified morphisms

For a justification of our choice of definition of unramified morphisms we refer the reader to the discussion in the section on étale morphisms Section 100.35.

In Lemma 100.36.9 we will characterize unramified morphisms of algebraic stacks as morphisms which are locally of finite type and have étale diagonal.

The property “unramified” of morphisms of algebraic spaces is étale-smooth local on the source-and-target, see Descent on Spaces, Remark 73.21.5. It is also stable under base change and fpqc local on the target, see Morphisms of Spaces, Lemma 66.38.4 and Descent on Spaces, Lemma 73.11.27. Hence, by Lemma 100.34.1 above, we may define what it means for a morphism of algebraic spaces to be unramified as follows and it agrees with the already existing notion defined in Properties of Stacks, Section 99.3 when the morphism is representable by algebraic spaces because such a morphism is automatically DM by Lemma 100.4.3.

Definition 100.36.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. We say $f$ is unramified if $f$ is DM and the equivalent conditions of Lemma 100.34.1 hold with $\mathcal{P} =$“unramified”.

We will use without further mention that this agrees with the already existing notion of unramified morphisms in case $f$ is representable by algebraic spaces or if $\mathcal{X}$ and $\mathcal{Y}$ are representable by algebraic spaces.

Proof. Combine Remark 100.34.3 with Morphisms of Spaces, Lemma 66.38.3. $\square$

Proof. Combine Remark 100.34.4 with Morphisms of Spaces, Lemma 66.38.4. $\square$

Proof. Follows from Remark 100.34.5 and Morphisms of Spaces, Lemma 66.39.10. $\square$

Proof. Let $j : \mathcal{Z} \to \mathcal{X}$ be an immersion of algebraic stacks. Since $j$ is representable, it is DM by Lemma 100.4.3. On the other hand, if $X \to \mathcal{X}$ is a smooth and surjective morphism where $X$ is a scheme, then $Z = \mathcal{Z} \times _\mathcal {X} X$ is a locally closed subscheme of $X$. Hence $Z \to X$ is unramified (Morphisms, Lemmas 29.35.7 and 29.35.8) and we conclude that $j$ is unramified from the definition. $\square$

Lemma 100.36.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent

1. $f$ is unramified,

2. $f$ is DM and for any morphism $V \to \mathcal{Y}$ where $V$ is an algebraic space and any étale morphism $U \to V \times _\mathcal {Y} \mathcal{X}$ where $U$ is an algebraic space, the morphism $U \to V$ is unramified,

3. there exists some surjective, locally of finite presentation, and flat morphism $W \to \mathcal{Y}$ where $W$ is an algebraic space and some surjective étale morphism $T \to W \times _\mathcal {Y} \mathcal{X}$ where $T$ is an algebraic space such that the morphism $T \to W$ is unramified.

Proof. Assume (1). Then $f$ is DM and since being unramified is preserved by base change, we see that (2) holds.

Assume (2). Choose a scheme $V$ and a surjective étale morphism $V \to \mathcal{Y}$. Choose a scheme $U$ and a surjective étale morphism $U \to V \times _\mathcal {Y} \mathcal{X}$ (Lemma 100.21.7). Thus we see that (3) holds.

Assume $W \to \mathcal{Y}$ and $T \to W \times _\mathcal {Y} \mathcal{X}$ are as in (3). We first check $f$ is DM. Namely, it suffices to check $W \times _\mathcal {Y} \mathcal{X} \to W$ is DM, see Lemma 100.4.5. By Lemma 100.4.12 it suffices to check $W \times _\mathcal {Y} \mathcal{X}$ is DM. This follows from the existence of $T \to W \times _\mathcal {Y} \mathcal{X}$ by (the easy direction of) Theorem 100.21.6.

Assume $f$ is DM and $W \to \mathcal{Y}$ and $T \to W \times _\mathcal {Y} \mathcal{X}$ are as in (3). Let $V$ be an algebraic space, let $V \to \mathcal{Y}$ be surjective smooth, let $U$ be an algebraic space, and let $U \to V \times _\mathcal {Y} \mathcal{X}$ is surjective and étale (Lemma 100.21.7). We have to check that $U \to V$ is unramified. It suffices to prove $U \times _\mathcal {Y} W \to V \times _\mathcal {Y} W$ is unramified by Descent on Spaces, Lemma 73.11.27. We may replace $\mathcal{X}, \mathcal{Y}, W, T, U, V$ by $\mathcal{X} \times _\mathcal {Y} W, W, W, T, U \times _\mathcal {Y} W, V \times _\mathcal {Y} W$ (small detail omitted). Thus we may assume that $Y = \mathcal{Y}$ is an algebraic space, there exists an algebraic space $T$ and a surjective étale morphism $T \to \mathcal{X}$ such that $T \to Y$ is unramified, and $U$ and $V$ are as before. In this case we know that

$U \to V\text{ is unramified} \Leftrightarrow \mathcal{X} \to Y\text{ is unramified} \Leftrightarrow T \to Y\text{ is unramified}$

by the equivalence of properties (1) and (2) of Lemma 100.34.1 and Definition 100.36.1. This finishes the proof. $\square$

Lemma 100.36.7. An unramified morphism of algebraic stacks is locally quasi-finite.

Proof. This follows from Lemma 100.36.6 (characterizing unramified morphisms), Lemma 100.23.7 (characterizing locally quasi-finite morphisms), and Morphisms of Spaces, Lemma 66.38.7 (the corresponding result for algebraic spaces). $\square$

Lemma 100.36.8. Let $\mathcal{X} \to \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks. If $\mathcal{X} \to \mathcal{Z}$ is unramified and $\mathcal{Y} \to \mathcal{Z}$ is DM, then $\mathcal{X} \to \mathcal{Y}$ is unramified.

Proof. Assume $\mathcal{X} \to \mathcal{Z}$ is unramified. By Lemma 100.4.12 the morphism $\mathcal{X} \to \mathcal{Y}$ is DM. Choose a commutative diagram

$\xymatrix{ U \ar[d] \ar[r] & V \ar[d] \ar[r] & W \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{Y} \ar[r] & \mathcal{Z} }$

with $U, V, W$ algebraic spaces, with $W \to \mathcal{Z}$ surjective smooth, $V \to \mathcal{Y} \times _\mathcal {Z} W$ surjective étale, and $U \to \mathcal{X} \times _\mathcal {Y} V$ surjective étale (see Lemma 100.21.7). Then also $U \to \mathcal{X} \times _\mathcal {Z} W$ is surjective and étale. Hence we know that $U \to W$ is unramified and we have to show that $U \to V$ is unramified. This follows from Morphisms of Spaces, Lemma 66.38.11. $\square$

Lemma 100.36.9. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent

1. $f$ is unramified, and

2. $f$ is locally of finite type and its diagonal is étale.

Proof. Assume $f$ is unramified. Then $f$ is DM hence we can choose algebraic spaces $U$, $V$, a smooth surjective morphism $V \to \mathcal{Y}$ and a surjective étale morphism $U \to \mathcal{X} \times _\mathcal {Y} V$ (Lemma 100.21.7). Since $f$ is unramified the induced morphism $U \to V$ is unramified. Thus $U \to V$ is locally of finite type (Morphisms of Spaces, Lemma 66.38.6) and we conclude that $f$ is locally of finite type. The diagonal $\Delta : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ is a morphism of algebraic stacks over $\mathcal{Y}$. The base change of $\Delta$ by the surjective smooth morphism $V \to \mathcal{Y}$ is the diagonal of the base change of $f$, i.e., of $\mathcal{X}_ V = \mathcal{X} \times _\mathcal {Y} V \to V$. In other words, the diagram

$\xymatrix{ \mathcal{X}_ V \ar[r] \ar[d] & \mathcal{X}_ V \times _ V \mathcal{X}_ V \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{X} \times _\mathcal {Y} \mathcal{X} }$

is cartesian. Since the right vertical arrow is surjective and smooth it suffices to show that the top horizontal arrow is étale by Properties of Stacks, Lemma 99.3.4. Consider the commutative diagram

$\xymatrix{ U \ar[d] \ar[r] & U \times _ V U \ar[d] \\ \mathcal{X}_ V \ar[r] & \mathcal{X}_ V \times _ V \mathcal{X}_ V }$

All arrows are representable by algebraic spaces, the vertical arrows are étale, the left one is surjective, and the top horizontal arrow is an open immersion by Morphisms of Spaces, Lemma 66.38.9. This implies what we want: first we see that $U \to \mathcal{X}_ V \times _ V \mathcal{X}_ V$ is étale as a composition of étale morphisms, and then we can use Properties of Stacks, Lemma 99.3.5 to see that $\mathcal{X}_ V \to \mathcal{X}_ V \times _ V \mathcal{X}_ V$ is étale because being étale (for morphisms of algebraic spaces) is local on the source in the étale topology (Descent on Spaces, Lemma 73.19.1).

Assume $f$ is locally of finite type and that its diagonal is étale. Then $f$ is DM by definition (as étale morphisms of algebraic spaces are unramified). As above this means we can choose algebraic spaces $U$, $V$, a smooth surjective morphism $V \to \mathcal{Y}$ and a surjective étale morphism $U \to \mathcal{X} \times _\mathcal {Y} V$ (Lemma 100.21.7). To finish the proof we have to show that $U \to V$ is unramified. We already know that $U \to V$ is locally of finite type. Arguing as above we find a commutative diagram

$\xymatrix{ U \ar[d] \ar[r] & U \times _ V U \ar[d] \\ \mathcal{X}_ V \ar[r] & \mathcal{X}_ V \times _ V \mathcal{X}_ V }$

where all arrows are representable by algebraic spaces, the vertical arrows are étale, and the lower horizontal one is étale as a base change of $\Delta$. It follows that $U \to U \times _ V U$ is étale for example by Lemma 100.35.61. Thus $U \to U \times _ V U$ is an étale monomorphism hence an open immersion (Morphisms of Spaces, Lemma 66.51.2). Then $U \to V$ is unramified by Morphisms of Spaces, Lemma 66.38.9. $\square$

Lemma 100.36.10. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent

1. $f$ is étale, and

2. $f$ is locally of finite presentation, flat, and unramified,

3. $f$ is locally of finite presentation, flat, and its diagonal is étale.

Proof. The equivalence of (2) and (3) follows immediately from Lemma 100.36.9. Thus in each case the morphism $f$ is DM. Then we can choose Then we can choose algebraic spaces $U$, $V$, a smooth surjective morphism $V \to \mathcal{Y}$ and a surjective étale morphism $U \to \mathcal{X} \times _\mathcal {Y} V$ (Lemma 100.21.7). To finish the proof we have to show that $U \to V$ is étale if and only if it is locally of finite presentation, flat, and unramified. This follows from Morphisms of Spaces, Lemma 66.39.12 (and the more trivial Morphisms of Spaces, Lemmas 66.39.10, 66.39.8, and 66.39.7). $\square$

[1] It is quite easy to deduce this directly from Morphisms of Spaces, Lemma 66.39.11.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).