The Stacks project

Proof. It suffices to prove the following: For any field $k$ and object $y$ of $\mathcal{Y}$ over $\mathop{\mathrm{Spec}}(k)$ there exists an integer $d \geq 1$ and an object $(U, Z, y, x, \alpha )$ of $\mathcal{H}_{d, lci}(\mathcal{X}/\mathcal{Y})$ with $U = \mathop{\mathrm{Spec}}(k)$. Namely, in this case we see that $p$ is surjective on objects in the strong sense that an extension of the field is not needed.

Denote $X_ y$ the algebraic space over $U = \mathop{\mathrm{Spec}}(k)$ representing the $2$-fibre product $(\mathit{Sch}/U')_{fppf} \times _{y', \mathcal{Y}, F} \mathcal{X}$. By assumption the morphism $X_ y \to \mathop{\mathrm{Spec}}(k)$ is surjective and locally of finite presentation (and flat). In particular $X_ y$ is nonempty. Choose a nonempty affine scheme $V$ and an étale morphism $V \to X_ y$. Note that $V \to \mathop{\mathrm{Spec}}(k)$ is (flat), surjective, and locally of finite presentation (by Morphisms of Spaces, Definition 67.28.1). Pick a closed point $v \in V$ where $V \to \mathop{\mathrm{Spec}}(k)$ is Cohen-Macaulay (i.e., $V$ is Cohen-Macaulay at $v$), see More on Morphisms, Lemma 37.22.7. Applying More on Morphisms, Lemma 37.23.4 we find a regular immersion $Z \to V$ with $Z = \{ v\}$. This implies $Z \to V$ is a closed immersion. Moreover, it follows that $Z \to \mathop{\mathrm{Spec}}(k)$ is finite (for example by Algebra, Lemma 10.122.1). Hence $Z \to \mathop{\mathrm{Spec}}(k)$ is finite locally free of some degree $d$. Now $Z \to X_ y$ is unramified as the composition of a closed immersion followed by an étale morphism (see Morphisms of Spaces, Lemmas 67.38.3, 67.39.10, and 67.38.8). Finally, $Z \to X_ y$ is a local complete intersection morphism as a composition of a regular immersion of schemes and an étale morphism of algebraic spaces (see More on Morphisms, Lemma 37.62.9 and Morphisms of Spaces, Lemmas 67.39.6 and 67.37.8 and More on Morphisms of Spaces, Lemmas 76.48.6 and 76.48.5). The morphism $Z \to X_ y$ corresponds to an object $x$ of $\mathcal{X}$ over $Z$ together with an isomorphism $\alpha : y|_ Z \to F(x)$. We obtain an object $(U, Z, y, x, \alpha )$ of $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$. By what was said above about the morphism $Z \to X_ y$ we see that it actually is an object of the subcategory $\mathcal{H}_{d, lci}(\mathcal{X}/\mathcal{Y})$ and we win. $\square$