## 64.6 Reasonable and decent algebraic spaces

In Lemma 64.5.1 we have seen a number of conditions on algebraic spaces related to the behaviour of étale morphisms from affine schemes into $X$ and related to the existence of special étale coverings of $X$ by schemes. We tabulate the different types of conditions here:

\[ \boxed { \begin{matrix} (\alpha )
& \text{fibres of étale morphisms from affines are finite}
\\ (\beta )
& \text{points come from monomorphisms of spectra of fields}
\\ (\gamma )
& \text{points come from quasi-compact monomorphisms of spectra of fields}
\\ (\delta )
& \text{fibres of étale morphisms from affines are universally bounded}
\\ (\epsilon )
& \text{cover by étale morphisms from schemes quasi-compact onto their image}
\end{matrix} } \]

The conditions in the following definition are not exactly conditions on the diagonal of $X$, but they are in some sense separation conditions on $X$.

Definition 64.6.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.

We say $X$ is *decent* if for every point $x \in X$ the equivalent conditions of Lemma 64.4.5 hold, in other words property $(\gamma )$ of Lemma 64.5.1 holds.

We say $X$ is *reasonable* if the equivalent conditions of Lemma 64.4.6 hold, in other words property $(\delta )$ of Lemma 64.5.1 holds.

We say $X$ is *very reasonable* if the equivalent conditions of Lemma 64.4.7 hold, i.e., property $(\epsilon )$ of Lemma 64.5.1 holds.

We have the following implications among these conditions on algebraic spaces:

\[ \xymatrix{ \text{representable} \ar@{=>}[rd] & & & \\ & \text{very reasonable} \ar@{=>}[r] & \text{reasonable} \ar@{=>}[r] & \text{decent} \\ \text{quasi-separated} \ar@{=>}[ru] & & & } \]

The notion of a very reasonable algebraic space is obsolete. It was introduced because the assumption was needed to prove some results which are now proven for the class of decent spaces. The class of decent spaces is the largest class of spaces $X$ where one has a good relationship between the topology of $|X|$ and properties of $X$ itself.

Example 64.6.2. The algebraic space $\mathbf{A}^1_{\mathbf{Q}}/\mathbf{Z}$ constructed in Spaces, Example 61.14.8 is not decent as its “generic point” cannot be represented by a monomorphism from the spectrum of a field.

Lemma 64.6.4. Let $S$ be a scheme. Let $X$ be a quasi-compact reasonable algebraic space. Then there exists a directed system of quasi-compact and quasi-separated algebraic spaces $X_ i$ such that $X = \mathop{\mathrm{colim}}\nolimits _ i X_ i$ (colimit in the category of sheaves).

**Proof.**
We sketch the proof. By Properties of Spaces, Lemma 62.6.3 we have $X = U/R$ with $U$ affine. In this case, reasonable means $U \to X$ is universally bounded. Hence there exists an integer $N$ such that the “fibres” of $U \to X$ have degree at most $N$, see Definition 64.3.1. Denote $s, t : R \to U$ and $c : R \times _{s, U, t} R \to R$ the groupoid structural maps.

Claim: for every quasi-compact open $A \subset R$ there exists an open $R' \subset R$ such that

$A \subset R'$,

$R'$ is quasi-compact, and

$(U, R', s|_{R'}, t|_{R'}, c|_{R' \times _{s, U, t} R'})$ is a groupoid scheme.

Note that $e : U \to R$ is open as it is a section of the étale morphism $s : R \to U$, see Étale Morphisms, Proposition 40.6.1. Moreover $U$ is affine hence quasi-compact. Hence we may replace $A$ by $A \cup e(U) \subset R$, and assume that $A$ contains $e(U)$. Next, we define inductively $A^1 = A$, and

\[ A^ n = c(A^{n - 1} \times _{s, U, t} A) \subset R \]

for $n \geq 2$. Arguing inductively, we see that $A^ n$ is quasi-compact for all $n \geq 2$, as the image of the quasi-compact fibre product $A^{n - 1} \times _{s, U, t} A$. If $k$ is an algebraically closed field over $S$, and we consider $k$-points then

\[ A^ n(k) = \left\{ (u, u') \in U(k) : \begin{matrix} \text{there exist } u = u_1, u_2, \ldots , u_ n \in U(k)\text{ with}
\\ (u_ i , u_{i + 1}) \in A \text{ for all }i = 1, \ldots , n - 1.
\end{matrix} \right\} \]

But as the fibres of $U(k) \to X(k)$ have size at most $N$ we see that if $n > N$ then we get a repeat in the sequence above, and we can shorten it proving $A^ N = A^ n$ for all $n \geq N$. This implies that $R' = A^ N$ gives a groupoid scheme $(U, R', s|_{R'}, t|_{R'}, c|_{R' \times _{s, U, t} R'})$, proving the claim above.

Consider the map of sheaves on $(\mathit{Sch}/S)_{fppf}$

\[ \mathop{\mathrm{colim}}\nolimits _{R' \subset R} U/R' \longrightarrow U/R \]

where $R' \subset R$ runs over the quasi-compact open subschemes of $R$ which give étale equivalence relations as above. Each of the quotients $U/R'$ is an algebraic space (see Spaces, Theorem 61.10.5). Since $R'$ is quasi-compact, and $U$ affine the morphism $R' \to U \times _{\mathop{\mathrm{Spec}}(\mathbf{Z})} U$ is quasi-compact, and hence $U/R'$ is quasi-separated. Finally, if $T$ is a quasi-compact scheme, then

\[ \mathop{\mathrm{colim}}\nolimits _{R' \subset R} U(T)/R'(T) \longrightarrow U(T)/R(T) \]

is a bijection, since every morphism from $T$ into $R$ ends up in one of the open subrelations $R'$ by the claim above. This clearly implies that the colimit of the sheaves $U/R'$ is $U/R$. In other words the algebraic space $X = U/R$ is the colimit of the quasi-separated algebraic spaces $U/R'$.
$\square$

Lemma 64.6.5. Let $S$ be a scheme. Let $X$, $Y$ be algebraic spaces over $S$. Let $X \to Y$ be a representable morphism. If $Y$ is decent (resp. reasonable), then so is $X$.

**Proof.**
Translation of Lemma 64.5.3.
$\square$

Lemma 64.6.6. Let $S$ be a scheme. Let $X \to Y$ be an étale morphism of algebraic spaces over $S$. If $Y$ is decent, resp. reasonable, then so is $X$.

**Proof.**
Let $U$ be an affine scheme and $U \to X$ an étale morphism. Set $R = U \times _ X U$ and $R' = U \times _ Y U$. Note that $R \to R'$ is a monomorphism.

Let $x \in |X|$. To show that $X$ is decent, we have to show that the fibres of $|U| \to |X|$ and $|R| \to |X|$ over $x$ are finite. But if $Y$ is decent, then the fibres of $|U| \to |Y|$ and $|R'| \to |Y|$ are finite. Hence the result for “decent”.

To show that $X$ is reasonable, we have to show that the fibres of $U \to X$ are universally bounded. However, if $Y$ is reasonable, then the fibres of $U \to Y$ are universally bounded, which immediately implies the same thing for the fibres of $U \to X$. Hence the result for “reasonable”.
$\square$

## Comments (2)

Comment #2253 by Antoine Chambert-Loir on

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