## 97.16 Algebraic spaces

The following is our first main result on algebraic spaces.

Proposition 97.16.1. Let $S$ be a locally Noetherian scheme. Let $F : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$ be a functor. Assume that

1. $\Delta : F \to F \times F$ is representable by algebraic spaces,

2. $F$ satisfies axioms [-1], , , , , ,  (see Section 97.15), and

3. $\mathcal{O}_{S, s}$ is a G-ring for all finite type points $s$ of $S$.

Then $F$ is an algebraic space.

Proof. Lemma 97.13.8 applies to $F$. Using this we choose, for every finite type field $k$ over $S$ and $x_0 \in F(\mathop{\mathrm{Spec}}(k))$, an affine scheme $U_{k, x_0}$ of finite type over $S$ and a smooth morphism $U_{k, x_0} \to F$ such that there exists a finite type point $u_{k, x_0} \in U_{k, x_0}$ with residue field $k$ such that $x_0$ is the image of $u_{k, x_0}$. Then

$U = \coprod \nolimits _{k, x_0} U_{k, x_0} \longrightarrow F$

is smooth1. To finish the proof it suffices to show this map is surjective, see Bootstrap, Lemma 79.12.3 (this is where we use axiom ). By Criteria for Representability, Lemma 96.5.6 it suffices to show that $U \times _ F V \to V$ is surjective for those $V \to F$ where $V$ is an affine scheme locally of finite presentation over $S$. Since $U \times _ F V \to V$ is smooth the image is open. Hence it suffices to show that the image of $U \times _ F V \to V$ contains all finite type points of $V$, see Morphisms, Lemma 29.16.7. Let $v_0 \in V$ be a finite type point. Then $k = \kappa (v_0)$ is a finite type field over $S$. Denote $x_0$ the composition $\mathop{\mathrm{Spec}}(k) \xrightarrow {v_0} V \to F$. Then $(u_{k, x_0}, v_0) : \mathop{\mathrm{Spec}}(k) \to U \times _ F V$ is a point mapping to $v_0$ and we win. $\square$

Lemma 97.16.2. Let $S$ be a locally Noetherian scheme. Let $a : F \to G$ be a transformation of functors $(\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$. Assume that

1. $a$ is injective,

2. $F$ satisfies axioms , , , , and ,

3. $\mathcal{O}_{S, s}$ is a G-ring for all finite type points $s$ of $S$,

4. $G$ is an algebraic space locally of finite type over $S$,

Then $F$ is an algebraic space.

Proof. By Lemma 97.8.1 the functor $G$ satisfies . As $F \to G$ is injective, we conclude that $F$ also satisfies . Moreover, as $F \to G$ is injective, we see that given schemes $U$, $V$ and morphisms $U \to F$ and $V \to F$, then $U \times _ F V = U \times _ G V$. Hence $\Delta : F \to F \times F$ is representable (by schemes) as this holds for $G$ by assumption. Thus Proposition 97.16.1 applies2. $\square$

 Set theoretical remark: This coproduct is (isomorphic) to an object of $(\mathit{Sch}/S)_{fppf}$ as we have a bound on the index set by axiom [-1], see Sets, Lemma 3.9.9.
 The set theoretic condition [-1] holds for $F$ as it holds for $G$. Details omitted.

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).