Lemma 75.4.1 (David Rydh). A flat monomorphism of algebraic spaces is representable by schemes.

## 75.4 Monomorphisms

This section is the continuation of Morphisms of Spaces, Section 66.10. We would like to know whether or not every monomorphism of algebraic spaces is representable. If you can prove this is true or have a counterexample, please email stacks.project@gmail.com. For the moment this is known in the following cases

for monomorphisms which are locally of finite type (more generally any separated, locally quasi-finite morphism is representable by Morphisms of Spaces, Lemma 66.51.1 and a monomorphism which is locally of finite type is locally quasi-finite by Morphisms of Spaces, Lemma 66.27.10),

if the target is a disjoint union of spectra of zero dimensional local rings (Decent Spaces, Lemma 67.19.1), and

for flat monomorphisms (see below).

**Proof.**
Let $f : X \to Y$ be a flat monomorphism of algebraic spaces. To prove $f$ is representable, we have to show $X \times _ Y V$ is a scheme for every scheme $V$ mapping to $Y$. Since being a scheme is local (Properties of Spaces, Lemma 65.13.1), we may assume $V$ is affine. Thus we may assume $Y = \mathop{\mathrm{Spec}}(B)$ is an affine scheme. Next, we can assume that $X$ is quasi-compact by replacing $X$ by a quasi-compact open. The space $X$ is separated as $X \to X \times _{\mathop{\mathrm{Spec}}(B)} X$ is an isomorphism. Applying Limits of Spaces, Lemma 69.17.3 we reduce to the case where $B$ is local, $X \to \mathop{\mathrm{Spec}}(B)$ is a flat monomorphism, and there exists a point $x \in X$ mapping to the closed point of $\mathop{\mathrm{Spec}}(B)$. Then $X \to \mathop{\mathrm{Spec}}(B)$ is surjective as generalizations lift along flat morphisms of separated algebraic spaces, see Decent Spaces, Lemma 67.7.4. Hence we see that $\{ X \to \mathop{\mathrm{Spec}}(B)\} $ is an fpqc cover. Then $X \to \mathop{\mathrm{Spec}}(B)$ is a morphism which becomes an isomorphism after base change by $X \to \mathop{\mathrm{Spec}}(B)$. Hence it is an isomorphism by fpqc descent, see Descent on Spaces, Lemma 73.11.15.
$\square$

The following is (in some sense) a variant of the lemma above.

Lemma 75.4.2. Let $S$ be a scheme. Let $f : X \to Y$ be a quasi-compact monomorphism of algebraic spaces such that for every $T \to Y$ the map

is injective. Then $f$ is an isomorphism (and hence representable by schemes).

**Proof.**
The question is étale local on $Y$, hence we may assume $Y = \mathop{\mathrm{Spec}}(A)$ is affine. Then $X$ is quasi-compact and we may choose an affine scheme $U = \mathop{\mathrm{Spec}}(B)$ and a surjective étale morphism $U \to X$ (Properties of Spaces, Lemma 65.6.3). Note that $U \times _ X U = \mathop{\mathrm{Spec}}(B \otimes _ A B)$. Hence the category of quasi-coherent $\mathcal{O}_ X$-modules is equivalent to the category $DD_{B/A}$ of descent data on modules for $A \to B$. See Properties of Spaces, Proposition 65.32.1, Descent, Definition 35.3.1, and Descent, Subsection 35.4.14. On the other hand,

is a universally injective ring map. Namely, given an $A$-module $M$ we see that $A \oplus M \to B \otimes _ A (A \oplus M)$ is injective by the assumption of the lemma. Hence $DD_{B/A}$ is equivalent to the category of $A$-modules by Descent, Theorem 35.4.22. Thus pullback along $f : X \to \mathop{\mathrm{Spec}}(A)$ determines an equivalence of categories of quasi-coherent modules. In particular $f^*$ is exact on quasi-coherent modules and we see that $f$ is flat (small detail omitted). Moreover, it is clear that $f$ is surjective (for example because $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is surjective). Hence we see that $\{ X \to \mathop{\mathrm{Spec}}(A)\} $ is an fpqc cover. Then $X \to \mathop{\mathrm{Spec}}(A)$ is a morphism which becomes an isomorphism after base change by $X \to \mathop{\mathrm{Spec}}(A)$. Hence it is an isomorphism by fpqc descent, see Descent on Spaces, Lemma 73.11.15. $\square$

Lemma 75.4.3. A quasi-compact flat surjective monomorphism of algebraic spaces is an isomorphism.

**Proof.**
Such a morphism satisfies the assumptions of Lemma 75.4.2.
$\square$

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