The Stacks project

64.17 Scheme theoretic closure and density

This section is the analogue of Morphisms, Section 29.7.

Lemma 64.17.1. Let $S$ be a scheme. Let $W \subset S$ be a scheme theoretically dense open subscheme (Morphisms, Definition 29.7.1). Let $f : X \to S$ be a morphism of schemes which is flat, locally of finite presentation, and locally quasi-finite. Then $f^{-1}(W)$ is scheme theoretically dense in $X$.

Proof. We will use the characterization of Morphisms, Lemma 29.7.5. Assume $V \subset X$ is an open and $g \in \Gamma (V, \mathcal{O}_ V)$ is a function which restricts to zero on $f^{-1}(W) \cap V$. We have to show that $g = 0$. Assume $g \not= 0$ to get a contradiction. By More on Morphisms, Lemma 37.40.6 we may shrink $V$, find an open $U \subset S$ fitting into a commutative diagram

\[ \xymatrix{ V \ar[r] \ar[d]_\pi & X \ar[d]^ f \\ U \ar[r] & S, } \]

a quasi-coherent subsheaf $\mathcal{F} \subset \mathcal{O}_ U$, an integer $r > 0$, and an injective $\mathcal{O}_ U$-module map $\mathcal{F}^{\oplus r} \to \pi _*\mathcal{O}_ V$ whose image contains $g|_ V$. Say $(g_1, \ldots , g_ r) \in \Gamma (U, \mathcal{F}^{\oplus r})$ maps to $g$. Then we see that $g_ i|_{W \cap U} = 0$ because $g|_{f^{-1}W \cap V} = 0$. Hence $g_ i = 0$ because $\mathcal{F} \subset \mathcal{O}_ U$ and $W$ is scheme theoretically dense in $S$. This implies $g = 0$ which is the desired contradiction. $\square$

Lemma 64.17.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $U \subset X$ be an open subspace. The following are equivalent

  1. for every étale morphism $\varphi : V \to X$ (of algebraic spaces) the scheme theoretic closure of $\varphi ^{-1}(U)$ in $V$ is equal to $V$,

  2. there exists a scheme $V$ and a surjective étale morphism $\varphi : V \to X$ such that the scheme theoretic closure of $\varphi ^{-1}(U)$ in $V$ is equal to $V$,

Proof. Observe that if $V \to V'$ is a morphism of algebraic spaces étale over $X$, and $Z \subset V$, resp. $Z' \subset V'$ is the scheme theoretic closure of $U \times _ X V$, resp. $U \times _ X V'$ in $V$, resp. $V'$, then $Z$ maps into $Z'$. Thus if $V \to V'$ is surjective and étale then $Z = V$ implies $Z' = V'$. Next, note that an étale morphism is flat, locally of finite presentation, and locally quasi-finite (see Morphisms, Section 29.34). Thus Lemma 64.17.1 implies that if $V$ and $V'$ are schemes, then $Z' = V'$ implies $Z = V$. A formal argument using that every algebraic space has an étale covering by a scheme shows that (1) and (2) are equivalent. $\square$

It follows from Lemma 64.17.2 that the following definition is compatible with the definition in the case of schemes.

Definition 64.17.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $U \subset X$ be an open subspace.

  1. The scheme theoretic image of the morphism $U \to X$ is called the scheme theoretic closure of $U$ in $X$.

  2. We say $U$ is scheme theoretically dense in $X$ if the equivalent conditions of Lemma 64.17.2 are satisfied.

With this definition it is not the case that $U$ is scheme theoretically dense in $X$ if and only if the scheme theoretic closure of $U$ is $X$. This is somewhat inelegant. But with suitable finiteness conditions we will see that it does hold.

Lemma 64.17.4. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $U \subset X$ be an open subspace. If $U \to X$ is quasi-compact, then $U$ is scheme theoretically dense in $X$ if and only if the scheme theoretic closure of $U$ in $X$ is $X$.

Proof. Follows from Lemma 64.16.3 part (3). $\square$

Lemma 64.17.5. Let $S$ be a scheme. Let $j : U \to X$ be an open immersion of algebraic spaces over $S$. Then $U$ is scheme theoretically dense in $X$ if and only if $\mathcal{O}_ X \to j_*\mathcal{O}_ U$ is injective.

Proof. If $\mathcal{O}_ X \to j_*\mathcal{O}_ U$ is injective, then the same is true when restricted to any algebraic space $V$ étale over $X$. Hence the scheme theoretic closure of $U \times _ X V$ in $V$ is equal to $V$, see proof of Lemma 64.16.1. Conversely, assume the scheme theoretic closure of $U \times _ X V$ is equal to $V$ for all $V$ étale over $X$. Suppose that $\mathcal{O}_ X \to j_*\mathcal{O}_ U$ is not injective. Then we can find an affine, say $V = \mathop{\mathrm{Spec}}(A)$, étale over $X$ and a nonzero element $f \in A$ such that $f$ maps to zero in $\Gamma (V \times _ X U, \mathcal{O})$. In this case the scheme theoretic closure of $V \times _ X U$ in $V$ is clearly contained in $\mathop{\mathrm{Spec}}(A/(f))$ a contradiction. $\square$

Lemma 64.17.6. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. If $U$, $V$ are scheme theoretically dense open subspaces of $X$, then so is $U \cap V$.

Proof. Let $W \to X$ be any étale morphism. Consider the map $\mathcal{O}(W) \to \mathcal{O}(W \times _ X V) \to \mathcal{O}(W \times _ X (V \cap U))$. By Lemma 64.17.5 both maps are injective. Hence the composite is injective. Hence by Lemma 64.17.5 $U \cap V$ is scheme theoretically dense in $X$. $\square$

Lemma 64.17.7. Let $S$ be a scheme. Let $h : Z \to X$ be an immersion of algebraic spaces over $S$. Assume either $Z \to X$ is quasi-compact or $Z$ is reduced. Let $\overline{Z} \subset X$ be the scheme theoretic image of $h$. Then the morphism $Z \to \overline{Z}$ is an open immersion which identifies $Z$ with a scheme theoretically dense open subspace of $\overline{Z}$. Moreover, $Z$ is topologically dense in $\overline{Z}$.

Proof. In both cases the formation of $\overline{Z}$ commutes with étale localization, see Lemmas 64.16.3 and 64.16.4. Hence this lemma follows from the case of schemes, see Morphisms, Lemma 29.7.7. $\square$

Lemma 64.17.8. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $f, g : X \to Y$ be morphisms of algebraic spaces over $B$. Let $U \subset X$ be an open subspace such that $f|_ U = g|_ U$. If the scheme theoretic closure of $U$ in $X$ is $X$ and $Y \to B$ is separated, then $f = g$.

Proof. As $Y \to B$ is separated the fibre product $Y \times _{\Delta , Y \times _ B Y, (f, g)} X$ is a closed subspace $Z \subset X$. As $f|_ U = g|_ U$ we see that $U \subset Z$. Hence $Z = X$ as $U$ is assumed scheme theoretically dense in $X$. $\square$


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