The Stacks project

66.40 Proper morphisms

The notion of a proper morphism plays an important role in algebraic geometry. Here is the definition of a proper morphism of algebraic spaces.

Definition 66.40.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. We say $f$ is proper if $f$ is separated, finite type, and universally closed.

Lemma 66.40.2. 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 proper,

  2. for every scheme $Z$ and every morphism $Z \to Y$ the projection $Z \times _ Y X \to Z$ is proper,

  3. for every affine scheme $Z$ and every morphism $Z \to Y$ the projection $Z \times _ Y X \to Z$ is proper,

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

Lemma 66.40.5. A closed immersion of algebraic spaces is a proper morphism of algebraic spaces.

Proof. As a closed immersion is by definition representable this follows from Spaces, Lemma 64.5.8 and the corresponding result for morphisms of schemes, see Morphisms, Lemma 29.41.6. $\square$

Lemma 66.40.6. Let $S$ be a scheme. Consider a commutative diagram of algebraic spaces

\[ \xymatrix{ X \ar[rr] \ar[rd] & & Y \ar[ld] \\ & B & } \]

over $S$.

  1. If $X \to B$ is universally closed and $Y \to B$ is separated, then the morphism $X \to Y$ is universally closed. In particular, the image of $|X|$ in $|Y|$ is closed.

  2. If $X \to B$ is proper and $Y \to B$ is separated, then the morphism $X \to Y$ is proper.

Proof. Assume $X \to B$ is universally closed and $Y \to B$ is separated. We factor the morphism as $X \to X \times _ B Y \to Y$. The first morphism is a closed immersion, see Lemma 66.4.6 hence universally closed. The projection $X \times _ B Y \to Y$ is the base change of a universally closed morphism and hence universally closed, see Lemma 66.9.3. Thus $X \to Y$ is universally closed as the composition of universally closed morphisms, see Lemma 66.9.4. This proves (1). To deduce (2) combine (1) with Lemmas 66.4.10, 66.8.9, and 66.23.6. $\square$

Lemma 66.40.7. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $f : X \to Y$ be a morphism of algebraic spaces over $B$. If $X$ is universally closed over $B$ and $f$ is surjective then $Y$ is universally closed over $B$. In particular, if also $Y$ is separated and of finite type over $B$, then $Y$ is proper over $B$.

Proof. Assume $X$ is universally closed and $f$ surjective. Denote $p : X \to B$, $q : Y \to B$ the structure morphisms. Let $B' \to B$ be a morphism of algebraic spaces over $S$. The base change $f' : X_{B'} \to Y_{B'}$ is surjective (Lemma 66.5.5), and the base change $p' : X_{B'} \to B'$ is closed. If $T \subset Y_{B'}$ is closed, then $(f')^{-1}(T) \subset X_{B'}$ is closed, hence $p'((f')^{-1}(T)) = q'(T)$ is closed. So $q'$ is closed. $\square$

Lemma 66.40.8. Let $S$ be a scheme. Let

\[ \xymatrix{ X \ar[rr]_ h \ar[rd]_ f & & Y \ar[ld]^ g \\ & B } \]

be a commutative diagram of morphism of algebraic spaces over $S$. Assume

  1. $X \to B$ is a proper morphism,

  2. $Y \to B$ is separated and locally of finite type,

Then the scheme theoretic image $Z \subset Y$ of $h$ is proper over $B$ and $X \to Z$ is surjective.

Proof. The scheme theoretic image of $h$ is constructed in Section 66.16. Observe that $h$ is quasi-compact (Lemma 66.8.10) hence $|h|(|X|) \subset |Z|$ is dense (Lemma 66.16.3). On the other hand $|h|(|X|)$ is closed in $|Y|$ (Lemma 66.40.6) hence $X \to Z$ is surjective. Thus $Z \to B$ is a proper (Lemma 66.40.7). $\square$

Lemma 66.40.9. 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 separated,

  2. $\Delta _{X/Y} : X \to X \times _ Y X$ is universally closed, and

  3. $\Delta _{X/Y} : X \to X \times _ Y X$ is proper.

Proof. The implication (1) $\Rightarrow $ (3) follows from Lemma 66.40.5. We will use Spaces, Lemma 64.5.8 without further mention in the rest of the proof. Recall that $\Delta _{X/Y}$ is a representable monomorphism which is locally of finite type, see Lemma 66.4.1. Since proper $\Rightarrow $ universally closed for morphisms of schemes we conclude that (3) implies (2). If $\Delta _{X/Y}$ is universally closed then √Čtale Morphisms, Lemma 41.7.2 implies that it is a closed immersion. Thus (2) $\Rightarrow $ (1) and we win. $\square$


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