## 64.12 Immersions and Zariski coverings of algebraic spaces

At this point an interesting phenomenon occurs. We have already defined the notion of an open immersion of algebraic spaces (through Definition 64.5.1) but we have yet to define the notion of a point1. Thus the Zariski topology of an algebraic space has already been defined, but there is no space yet!

Perhaps superfluously we formally introduce immersions as follows.

Definition 64.12.1. Let $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{fppf})$ be a scheme. Let $F$ be an algebraic space over $S$.

1. A morphism of algebraic spaces over $S$ is called an open immersion if it is representable, and an open immersion in the sense of Definition 64.5.1.

2. An open subspace of $F$ is a subfunctor $F' \subset F$ such that $F'$ is an algebraic space and $F' \to F$ is an open immersion.

3. A morphism of algebraic spaces over $S$ is called a closed immersion if it is representable, and a closed immersion in the sense of Definition 64.5.1.

4. A closed subspace of $F$ is a subfunctor $F' \subset F$ such that $F'$ is an algebraic space and $F' \to F$ is a closed immersion.

5. A morphism of algebraic spaces over $S$ is called an immersion if it is representable, and an immersion in the sense of Definition 64.5.1.

6. A locally closed subspace of $F$ is a subfunctor $F' \subset F$ such that $F'$ is an algebraic space and $F' \to F$ is an immersion.

We note that these definitions make sense since an immersion is in particular a monomorphism (see Schemes, Lemma 26.23.8 and Lemma 64.5.8), and hence the image of an immersion $G \to F$ of algebraic spaces is a subfunctor $F' \subset F$ which is (canonically) isomorphic to $G$. Thus some of the discussion of Schemes, Section 26.10 carries over to the setting of algebraic spaces.

Lemma 64.12.2. Let $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{fppf})$ be a scheme. A composition of (closed, resp. open) immersions of algebraic spaces over $S$ is a (closed, resp. open) immersion of algebraic spaces over $S$.

Proof. See Lemma 64.5.4 and Remarks 64.4.3 (see very last line of that remark) and 64.4.2. $\square$

Lemma 64.12.3. Let $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{fppf})$ be a scheme. A base change of a (closed, resp. open) immersion of algebraic spaces over $S$ is a (closed, resp. open) immersion of algebraic spaces over $S$.

Proof. See Lemma 64.5.5 and Remark 64.4.3 (see very last line of that remark). $\square$

Lemma 64.12.4. Let $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{fppf})$ be a scheme. Let $F$ be an algebraic space over $S$. Let $F_1$, $F_2$ be locally closed subspaces of $F$. If $F_1 \subset F_2$ as subfunctors of $F$, then $F_1$ is a locally closed subspace of $F_2$. Similarly for closed and open subspaces.

Proof. Let $T \to F_2$ be a morphism with $T$ a scheme. Since $F_2 \to F$ is a monomorphism, we see that $T \times _{F_2} F_1 = T \times _ F F_1$. The lemma follows formally from this. $\square$

Let us formally define the notion of a Zariski open covering of algebraic spaces. Note that in Lemma 64.8.5 we have already encountered such open coverings as a method for constructing algebraic spaces.

Definition 64.12.5. Let $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{fppf})$ be a scheme. Let $F$ be an algebraic space over $S$. A Zariski covering $\{ F_ i \subset F\} _{i \in I}$ of $F$ is given by a set $I$ and a collection of open subspaces $F_ i \subset F$ such that $\coprod F_ i \to F$ is a surjective map of sheaves.

Note that if $T$ is a schemes, and $a : T \to F$ is a morphism, then each of the fibre products $T \times _ F F_ i$ is identified with an open subscheme $T_ i \subset T$. The final condition of the definition signifies exactly that $T = \bigcup _{i \in I} T_ i$.

It is clear that the collection $F_{Zar}$ of open subspaces of $F$ is a set (as $(\mathit{Sch}/S)_{fppf}$ is a site, hence a set). Moreover, we can turn $F_{Zar}$ into a category by letting the morphisms be inclusions of subfunctors (which are automatically open immersions by Lemma 64.12.4). Finally, Definition 64.12.5 provides the notion of a Zariski covering $\{ F_ i \to F'\} _{i \in I}$ in the category $F_{Zar}$. Hence, just as in the case of a topological space (see Sites, Example 7.6.4) by suitably choosing a set of coverings we may obtain a Zariski site of the algebraic space $F$.

Definition 64.12.6. Let $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{fppf})$ be a scheme. Let $F$ be an algebraic space over $S$. A small Zariski site $F_{Zar}$ of an algebraic space $F$ is one of the sites described above.

Hence this gives a notion of what it means for something to be true Zariski locally on an algebraic space, which is how we will use this notion. In general the Zariski topology is not fine enough for our purposes. For example we can consider the category of Zariski sheaves on an algebraic space. It will turn out that this is not the correct thing to consider, even for quasi-coherent sheaves. One only gets the desired result when using the étale or fppf site of $F$ to define quasi-coherent sheaves.

[1] We will associate a topological space to an algebraic space in Properties of Spaces, Section 65.4, and its opens will correspond exactly to the open subspaces defined below.

Comment #2241 by clarification on

In the first parenthetical of the 3rd paragraph of Definition 53.12.5 is the statement "as $(Sch/S)_{fppf}$ is a site, hence a set". This seems inconsistent with the notation or logic of section 53.2. General Remarks (025T), which uses $(Sch/S)_{fppf}$ to represent the "big fppf site". If interpreted as objects in a category of (pre)sheaves, the underlying category stills seem to be "big", per 4.2.2 (0015).

Comment #2244 by dumb question on

Following the definition of Zariski covering, it says that $\coprod F_i \to F$ surjective as sheaves "signifies exactly that $T=\bigcup T_i$". Can you add reference or argument for why? If it said the morphism was surjective following the convention of Tag 02YN, then I would understand - that $T = \cup T_i$ is exacty what $\coprod F_i \to F$ surjective means, if i understand correctly. But surjective as sheaves means that for any $s:T \to F$, there exists an fppf cover $f: T' \to T$ and $s':T' \to F_i$ for some $i$ such that $sf = s'$. We get an induced map $T' \to T_i$, which seems to imply $T_i \to T$ is surjective as schemes (for a single $i$) since $T' \to T$ is surjective I'm obviously being stupid. Basically, it seems you are using the converse of tag 00WT.

Comment #2276 by on

Dear Clarification. Please consult Chapter 34 (and in particular Section 34.2) for our conventions regarding sites of schemes. In particular the "big" does not refer to what you think it refers to (and had you followed the links in 64.2 you would have found this out).

Comment #2277 by on

Dear dumb question, I'd prefer not to give the details in the text, but the underlying reason is simple. In Definition 64.12.5 we are assuming that the maps $F_i \to F$ are representable by open immersions. This means that the possible confusion mentioned in Remark 64.5.2 does not happen. Namely, given $T \to F$ let $T_i \subset T$ be the open subscheme corresponding to $T \times_F F_i$. Then $\coprod T_i \to T$ is surjective as a map of sheaves if and only if $\coprod T_i \to T$ is surjective as a map of schemes by the very fact that Zariski coverings are coverings for the topology we are considering (for me this is usually fppf).

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).