Proposition 66.50.2. Let $S$ be a scheme. Let $f : X \to T$ be a morphism of algebraic spaces over $S$. Assume

$T$ is representable,

$f$ is locally quasi-finite, and

$f$ is separated.

Then $X$ is representable.

Proposition 66.50.2. Let $S$ be a scheme. Let $f : X \to T$ be a morphism of algebraic spaces over $S$. Assume

$T$ is representable,

$f$ is locally quasi-finite, and

$f$ is separated.

Then $X$ is representable.

**Proof.**
Let $T = \bigcup T_ i$ be an affine open covering of the scheme $T$. If we can show that the open subspaces $X_ i = f^{-1}(T_ i)$ are representable, then $X$ is representable, see Properties of Spaces, Lemma 65.13.1. Note that $X_ i = T_ i \times _ T X$ and that locally quasi-finite and separated are both stable under base change, see Lemmas 66.4.4 and 66.27.4. Hence we may assume $T$ is an affine scheme.

By Properties of Spaces, Lemma 65.6.2 there exists a Zariski covering $X = \bigcup X_ i$ such that each $X_ i$ has a surjective étale covering by an affine scheme. By Properties of Spaces, Lemma 65.13.1 again it suffices to prove the proposition for each $X_ i$. Hence we may assume there exists an affine scheme $U$ and a surjective étale morphism $U \to X$. This reduces us to the situation in the next paragraph.

Assume we have

\[ U \longrightarrow X \longrightarrow T \]

where $U$ and $T$ are affine schemes, $U \to X$ is étale surjective, and $X \to T$ is separated and locally quasi-finite. By Lemmas 66.39.5 and 66.27.3 the morphism $U \to T$ is locally quasi-finite. Since $U$ and $T$ are affine it is quasi-finite. Set $R = U \times _ X U$. Then $X = U/R$, see Spaces, Lemma 64.9.1. As $X \to T$ is separated the morphism $R \to U \times _ T U$ is a closed immersion, see Lemma 66.4.5. In particular $R$ is an affine scheme also. As $U \to X$ is étale the projection morphisms $t, s : R \to U$ are étale as well. In particular $s$ and $t$ are quasi-finite, flat and of finite presentation (see Morphisms, Lemmas 29.36.6, 29.36.12 and 29.36.11).

Let $(U, R, s, t, c)$ be the groupoid associated to the étale equivalence relation $R$ on $U$. Let $u \in U$ be a point, and denote $p \in T$ its image. We are going to use More on Groupoids, Lemma 40.13.2 for the groupoid $(U, R, s, t, c)$ over the scheme $T$ with points $p$ and $u$ as above. By the discussion in the previous paragraph all the assumptions (1) – (7) of that lemma are satisfied. Hence we get an étale neighbourhood $(T', p') \to (T, p)$ and disjoint union decompositions

\[ U_{T'} = U' \amalg W, \quad R_{T'} = R' \amalg W' \]

and $u' \in U'$ satisfying conclusions (a), (b), (c), (d), (e), (f), (g), and (h) of the aforementioned More on Groupoids, Lemma 40.13.2. We may and do assume that $T'$ is affine (after possibly shrinking $T'$). Conclusion (h) implies that $R' = U' \times _{X_{T'}} U'$ with projection mappings identified with the restrictions of $s'$ and $t'$. Thus $(U', R', s'|_{R'}, t'|_{R'}, c'|_{R' \times _{t', U', s'} R'})$ of conclusion (g) is an étale equivalence relation. By Spaces, Lemma 64.10.2 we conclude that $U'/R'$ is an open subspace of $X_{T'}$. By conclusion (d) the schemes $U'$, $R'$ are affine and the morphisms $s'|_{R'}, t'|_{R'}$ are finite étale. Hence Groupoids, Proposition 39.23.9 kicks in and we see that $U'/R'$ is an affine scheme.

We conclude that for every pair of points $(u, p)$ as above we can find an étale neighbourhood $(T', p') \to (T, p)$ with $\kappa (p) = \kappa (p')$ and a point $u' \in U_{T'}$ mapping to $u$ such that the image $x'$ of $u'$ in $|X_{T'}|$ has an open neighbourhood $V'$ in $X_{T'}$ which is an affine scheme. We apply Lemma 66.50.1 to obtain an open subspace $W \subset X$ which is a scheme, and which contains $x$ (the image of $u$ in $|X|$). Since this works for every $x$ we see that $X$ is a scheme by Properties of Spaces, Lemma 65.13.1. This ends the proof. $\square$

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## Comments (1)

Comment #688 by Kestutis Cesnavicius on