Lemma 67.13.4. Let $S$ be a scheme. Let $Z' \to Z$ be a monomorphism of algebraic spaces over $S$. Assume there exists a field $k$ and a locally finitely presented, surjective, flat morphism $\mathop{\mathrm{Spec}}(k) \to Z$. Then either $Z'$ is empty or $Z' = Z$.

Proof. We may assume that $Z'$ is nonempty. In this case the fibre product $T = Z' \times _ Z \mathop{\mathrm{Spec}}(k)$ is nonempty, see Properties of Spaces, Lemma 65.4.3. Now $T$ is an algebraic space and the projection $T \to \mathop{\mathrm{Spec}}(k)$ is a monomorphism. Hence $T = \mathop{\mathrm{Spec}}(k)$, see Morphisms of Spaces, Lemma 66.10.8. We conclude that $\mathop{\mathrm{Spec}}(k) \to Z$ factors through $Z'$. But as $\mathop{\mathrm{Spec}}(k) \to Z$ is surjective, flat and locally of finite presentation, we see that $\mathop{\mathrm{Spec}}(k) \to Z$ is surjective as a map of sheaves on $(\mathit{Sch}/S)_{fppf}$ (see Spaces, Remark 64.5.2) and we conclude that $Z' = Z$. $\square$

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).