The Stacks project

66.38 Unramified morphisms

The property “unramified” (resp. “G-unramified”) of morphisms of schemes is étale local on the source-and-target, see Descent, Remark 35.32.7. It is also stable under base change and fpqc local on the target, see Morphisms, Lemma 29.35.5 and Descent, Lemma 35.23.28. Hence, by Lemma 66.22.1 above, we may define the notion of an unramified morphism (resp. G-unramified morphism) of algebraic spaces as follows and it agrees with the already existing notion defined in Section 66.3 when the morphism is representable.

Definition 66.38.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.

  1. We say $f$ is unramified if the equivalent conditions of Lemma 66.22.1 hold with $\mathcal{P} = \text{unramified}$.

  2. Let $x \in |X|$. We say $f$ is unramified at $x$ if there exists an open neighbourhood $X' \subset X$ of $x$ such that $f|_{X'} : X' \to Y$ is unramified.

  3. We say $f$ is G-unramified if the equivalent conditions of Lemma 66.22.1 hold with $\mathcal{P} = \text{G-unramified}$.

  4. Let $x \in |X|$. We say $f$ is G-unramified at $x$ if there exists an open neighbourhood $X' \subset X$ of $x$ such that $f|_{X'} : X' \to Y$ is G-unramified.

Because of the following lemma, from here on we will only develop theory for unramified morphisms, and whenever we want to use a G-unramified morphism we will simply say “an unramified morphism locally of finite presentation”.

Lemma 66.38.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Then $f$ is G-unramified if and only if $f$ is unramified and locally of finite presentation.

Proof. Consider any diagram as in Lemma 66.22.1. Then all we are saying is that the morphism $h$ is G-unramified if and only if it is unramified and locally of finite presentation. This is clear from Morphisms, Definition 29.35.1. $\square$

Lemma 66.38.3. The composition of unramified morphisms is unramified.

Lemma 66.38.4. The base change of an unramified morphism is unramified.

Lemma 66.38.5. 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 unramified,

  2. for every $x \in |X|$ the morphism $f$ is unramified at $x$,

  3. for every scheme $Z$ and any morphism $Z \to Y$ the morphism $Z \times _ Y X \to Z$ is unramified,

  4. for every affine scheme $Z$ and any morphism $Z \to Y$ the morphism $Z \times _ Y X \to Z$ is unramified,

  5. there exists a scheme $V$ and a surjective étale morphism $V \to Y$ such that $V \times _ Y X \to V$ is an unramified morphism,

  6. there exists a scheme $U$ and a surjective étale morphism $\varphi : U \to X$ such that the composition $f \circ \varphi $ is unramified,

  7. for every commutative diagram

    \[ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y } \]

    where $U$, $V$ are schemes and the vertical arrows are étale the top horizontal arrow is unramified,

  8. there exists a commutative diagram

    \[ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y } \]

    where $U$, $V$ are schemes, the vertical arrows are étale, and $U \to X$ is surjective such that the top horizontal arrow is unramified, and

  9. there exist Zariski coverings $Y = \bigcup _{i \in I} Y_ i$, and $f^{-1}(Y_ i) = \bigcup X_{ij}$ such that each morphism $X_{ij} \to Y_ i$ is unramified.

Proof. Omitted. $\square$

Lemma 66.38.6. An unramified morphism of algebraic spaces is locally of finite type.

Proof. Via a diagram as in Lemma 66.22.1 this translates into Morphisms, Lemma 29.35.9. $\square$

Lemma 66.38.7. If $f$ is unramified at $x$ then $f$ is quasi-finite at $x$. In particular, an unramified morphism is locally quasi-finite.

Proof. Via a diagram as in Lemma 66.22.1 this translates into Morphisms, Lemma 29.35.10. $\square$

Proof. Let $i : X \to Y$ be an immersion of algebraic spaces. Choose a scheme $V$ and a surjective étale morphism $V \to Y$. Then $V \times _ Y X \to V$ is an immersion of schemes, hence unramified (see Morphisms, Lemmas 29.35.7 and 29.35.8). Thus by definition $i$ is unramified. $\square$

Lemma 66.38.9. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.

  1. If $f$ is unramified, then the diagonal morphism $\Delta _{X/Y} : X \to X \times _ Y X$ is an open immersion.

  2. If $f$ is locally of finite type and $\Delta _{X/Y}$ is an open immersion, then $f$ is unramified.

Proof. We know in any case that $\Delta _{X/Y}$ is a representable monomorphism, see Lemma 66.4.1. Choose a scheme $V$ and a surjective étale morphism $V \to Y$. Choose a scheme $U$ and a surjective étale morphism $U \to X \times _ Y V$. Consider the commutative diagram

\[ \xymatrix{ U \ar[d] \ar[rr]_-{\Delta _{U/V}} & & U \times _ V U \ar[d] \ar[r] & V \ar[d]^{\Delta _{V/Y}} \\ X \ar[rr]^-{\Delta _{X/Y}} & & X \times _ Y X \ar[r] & V \times _ Y V } \]

with cartesian right square. The left vertical arrow is surjective étale. The right vertical arrow is étale as a morphism between schemes étale over $Y$, see Properties of Spaces, Lemma 65.16.6. Hence the middle vertical arrow is étale too (but it need not be surjective).

Assume $f$ is unramified. Then $U \to V$ is unramified, hence $\Delta _{U/V}$ is an open immersion by Morphisms, Lemma 29.35.13. Looking at the left square of the diagram above we conclude that $\Delta _{X/Y}$ is an étale morphism, see Properties of Spaces, Lemma 65.16.3. Hence $\Delta _{X/Y}$ is a representable étale monomorphism, which implies that it is an open immersion by Étale Morphisms, Theorem 41.14.1. (See also Spaces, Lemma 64.5.8 for the translation from schemes language into the language of functors.)

Assume that $f$ is locally of finite type and that $\Delta _{X/Y}$ is an open immersion. This implies that $U \to V$ is locally of finite type too (by definition of a morphism of algebraic spaces which is locally of finite type). Looking at the displayed diagram above we conclude that $\Delta _{U/V}$ is étale as a morphism between schemes étale over $X \times _ Y X$, see Properties of Spaces, Lemma 65.16.6. But since $\Delta _{U/V}$ is the diagonal of a morphism between schemes we see that it is in any case an immersion, see Schemes, Lemma 26.21.2. Hence it is an open immersion, and we conclude that $U \to V$ is unramified by Morphisms, Lemma 29.35.13. This in turn means that $f$ is unramified by definition. $\square$

Lemma 66.38.10. Let $S$ be a scheme. Consider a commutative diagram

\[ \xymatrix{ X \ar[rr]_ f \ar[rd]_ p & & Y \ar[ld]^ q \\ & Z } \]

of algebraic spaces over $S$. Assume that $X \to Z$ is locally of finite type. Then there exists an open subspace $U(f) \subset X$ such that $|U(f)| \subset |X|$ is the set of points where $f$ is unramified. Moreover, for any morphism of algebraic spaces $Z' \to Z$, if $f' : X' \to Y'$ is the base change of $f$ by $Z' \to Z$, then $U(f')$ is the inverse image of $U(f)$ under the projection $X' \to X$.

Proof. This lemma is the analogue of Morphisms, Lemma 29.35.15 and in fact we will deduce the lemma from it. By Definition 66.38.1 the set $\{ x \in |X| : f \text{ is unramified at }x\} $ is open in $X$. Hence we only need to prove the final statement. By Lemma 66.23.6 the morphism $X \to Y$ is locally of finite type. By Lemma 66.23.3 the morphism $X' \to Y'$ is locally of finite type.

Choose a scheme $W$ and a surjective étale morphism $W \to Z$. Choose a scheme $V$ and a surjective étale morphism $V \to W \times _ Z Y$. Choose a scheme $U$ and a surjective étale morphism $U \to V \times _ Y X$. Finally, choose a scheme $W'$ and a surjective étale morphism $W' \to W \times _ Z Z'$. Set $V' = W' \times _ W V$ and $U' = W' \times _ W U$, so that we obtain surjective étale morphisms $V' \to Y'$ and $U' \to X'$. We will use without further mention an étale morphism of algebraic spaces induces an open map of associated topological spaces (see Properties of Spaces, Lemma 65.16.7). This combined with Lemma 66.38.5 implies that $U(f)$ is the image in $|X|$ of the set $T$ of points in $U$ where the morphism $U \to V$ is unramified. Similarly, $U(f')$ is the image in $|X'|$ of the set $T'$ of points in $U'$ where the morphism $U' \to V'$ is unramified. Now, by construction the diagram

\[ \xymatrix{ U' \ar[r] \ar[d] & U \ar[d] \\ V' \ar[r] & V } \]

is cartesian (in the category of schemes). Hence the aforementioned Morphisms, Lemma 29.35.15 applies to show that $T'$ is the inverse image of $T$. Since $|U'| \to |X'|$ is surjective this implies the lemma. $\square$

Lemma 66.38.11. Let $S$ be a scheme. Let $X \to Y \to Z$ be morphisms of algebraic spaces over $S$. If $X \to Z$ is unramified, then $X \to Y$ is unramified.

Proof. Choose a commutative diagram

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

with vertical arrows étale and surjective. (See Spaces, Lemma 64.11.6.) Apply Morphisms, Lemma 29.35.16 to the top row. $\square$

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