The Stacks project

66.12 Immersions

Open, closed and locally closed immersions of algebraic spaces were defined in Spaces, Section 64.12. Namely, a morphism of algebraic spaces is a closed immersion (resp. open immersion, resp. immersion) if it is representable and a closed immersion (resp. open immersion, resp. immersion) in the sense of Section 66.3.

In particular these types of morphisms are stable under base change and compositions of morphisms in the category of algebraic spaces over $S$, see Spaces, Lemmas 64.12.2 and 64.12.3.

Lemma 66.12.1. 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 a closed immersion (resp. open immersion, resp. immersion),

  2. for every scheme $Z$ and any morphism $Z \to Y$ the morphism $Z \times _ Y X \to Z$ is a closed immersion (resp. open immersion, resp. immersion),

  3. for every affine scheme $Z$ and any morphism $Z \to Y$ the morphism $Z \times _ Y X \to Z$ is a closed immersion (resp. open immersion, resp. immersion),

  4. there exists a scheme $V$ and a surjective étale morphism $V \to Y$ such that $V \times _ Y X \to V$ is a closed immersion (resp. open immersion, resp. immersion), and

  5. there exists a Zariski covering $Y = \bigcup Y_ i$ such that each of the morphisms $f^{-1}(Y_ i) \to Y_ i$ is a closed immersion (resp. open immersion, resp. immersion).

Proof. Using that a base change of a closed immersion (resp. open immersion, resp. immersion) is another one it is clear that (1) implies (2) and (2) implies (3). Also (3) implies (4) since we can take $V$ to be a disjoint union of affines, see Properties of Spaces, Lemma 65.6.1.

Assume $V \to Y$ is as in (4). Let $\mathcal{P}$ be the property closed immersion (resp. open immersion, resp. immersion) of morphisms of schemes. Note that property $\mathcal{P}$ is preserved under any base change and fppf local on the base (see Section 66.3). Moreover, morphisms of type $\mathcal{P}$ are separated and locally quasi-finite (in each of the three cases, see Schemes, Lemma 26.23.8, and Morphisms, Lemma 29.20.16). Hence by More on Morphisms, Lemma 37.55.1 the morphisms of type $\mathcal{P}$ satisfy descent for fppf covering. Thus Spaces, Lemma 64.11.5 applies and we see that $X \to Y$ is representable and has property $\mathcal{P}$, in other words (1) holds.

The equivalence of (1) and (5) follows from the fact that $\mathcal{P}$ is Zariski local on the target (since we saw above that $\mathcal{P}$ is in fact fppf local on the target). $\square$

Lemma 66.12.2. Let $S$ be a scheme. Let $Z \to Y \to X$ be morphisms of algebraic spaces over $S$.

  1. If $Z \to X$ is representable, locally of finite type, locally quasi-finite, separated, and a monomorphism, then $Z \to Y$ is representable, locally of finite type, locally quasi-finite, separated, and a monomorphism.

  2. If $Z \to X$ is an immersion and $Y \to X$ is locally separated, then $Z \to Y$ is an immersion.

  3. If $Z \to X$ is a closed immersion and $Y \to X$ is separated, then $Z \to Y$ is a closed immersion.

Proof. In each case the proof is to contemplate the commutative diagram

\[ \xymatrix{ Z \ar[r] \ar[rd] & Y \times _ X Z \ar[r] \ar[d] & Z \ar[d] \\ & Y \ar[r] & X } \]

where the composition of the top horizontal arrows is the identity. Let us prove (1). The first horizontal arrow is a section of $Y \times _ X Z \to Z$, whence representable, locally of finite type, locally quasi-finite, separated, and a monomorphism by Lemma 66.4.7. The arrow $Y \times _ X Z \to Y$ is a base change of $Z \to X$ hence is representable, locally of finite type, locally quasi-finite, separated, and a monomorphism (as each of these properties of morphisms of schemes is stable under base change, see Spaces, Remark 64.4.1). Hence the same is true for the composition (as each of these properties of morphisms of schemes is stable under composition, see Spaces, Remark 64.4.2). This proves (1). The other results are proved in exactly the same manner. $\square$

Lemma 66.12.3. Let $S$ be a scheme. Let $i : Z \to X$ be an immersion of algebraic spaces over $S$. Then $|i| : |Z| \to |X|$ is a homeomorphism onto a locally closed subset, and $i$ is a closed immersion if and only if the image $|i|(|Z|) \subset |X|$ is a closed subset.

Proof. The first statement is Properties of Spaces, Lemma 65.12.1. Let $U$ be a scheme and let $U \to X$ be a surjective étale morphism. By assumption $T = U \times _ X Z$ is a scheme and the morphism $j : T \to U$ is an immersion of schemes. By Lemma 66.12.1 the morphism $i$ is a closed immersion if and only if $j$ is a closed immersion. By Schemes, Lemma 26.10.4 this is true if and only if $j(T)$ is closed in $U$. However, the subset $j(T) \subset U$ is the inverse image of $|i|(|Z|) \subset |X|$, see Properties of Spaces, Lemma 65.4.3. This finishes the proof. $\square$

Remark 66.12.4. Let $S$ be a scheme. Let $i : Z \to X$ be an immersion of algebraic spaces over $S$. Since $i$ is a monomorphism we may think of $|Z|$ as a subset of $|X|$; in the rest of this remark we do so. Let $\partial |Z|$ be the boundary of $|Z|$ in the topological space $|X|$. In a formula

\[ \partial |Z| = \overline{|Z|} \setminus |Z|. \]

Let $\partial Z$ be the reduced closed subspace of $X$ with $|\partial Z| = \partial |Z|$ obtained by taking the reduced induced closed subspace structure, see Properties of Spaces, Definition 65.12.5. By construction we see that $|Z|$ is closed in $|X| \setminus |\partial Z| = |X \setminus \partial Z|$. Hence it is true that any immersion of algebraic spaces can be factored as a closed immersion followed by an open immersion (but not the other way in general, see Morphisms, Example 29.3.4).

Remark 66.12.5. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $T \subset |X|$ be a locally closed subset. Let $\partial T$ be the boundary of $T$ in the topological space $|X|$. In a formula

\[ \partial T = \overline{T} \setminus T. \]

Let $U \subset X$ be the open subspace of $X$ with $|U| = |X| \setminus \partial T$, see Properties of Spaces, Lemma 65.4.8. Let $Z$ be the reduced closed subspace of $U$ with $|Z| = T$ obtained by taking the reduced induced closed subspace structure, see Properties of Spaces, Definition 65.12.5. By construction $Z \to U$ is a closed immersion of algebraic spaces and $U \to X$ is an open immersion, hence $Z \to X$ is an immersion of algebraic spaces over $S$ (see Spaces, Lemma 64.12.2). Note that $Z$ is a reduced algebraic space and that $|Z| = T$ as subsets of $|X|$. We sometimes say $Z$ is the reduced induced subspace structure on $T$.

Lemma 66.12.6. Let $S$ be a scheme. Let $Z \to X$ be an immersion of algebraic spaces over $S$. Assume $Z \to X$ is quasi-compact. There exists a factorization $Z \to \overline{Z} \to X$ where $Z \to \overline{Z}$ is an open immersion and $\overline{Z} \to X$ is a closed immersion.

Proof. Let $U$ be a scheme and let $U \to X$ be surjective étale. As usual denote $R = U \times _ X U$ with projections $s, t : R \to U$. Set $T = Z \times _ U X$. Let $\overline{T} \subset U$ be the scheme theoretic image of $T \to U$. Note that $s^{-1}\overline{T} = t^{-1}\overline{T}$ as taking scheme theoretic images of quasi-compact morphisms commute with flat base change, see Morphisms, Lemma 29.25.16. Hence we obtain a closed subspace $\overline{Z} \subset X$ whose pullback to $U$ is $\overline{T}$, see Properties of Spaces, Lemma 65.12.2. By Morphisms, Lemma 29.7.7 the morphism $T \to \overline{T}$ is an open immersion. It follows that $Z \to \overline{Z}$ is an open immersion and we win. $\square$


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