The Stacks project

Proof. Choose a scheme $W$ and a surjective smooth morphism $W \to \mathcal{X}$. Choose an extension of fields $k'/k$ and a morphism $w : \mathop{\mathrm{Spec}}(k') \to W$ such that $\mathop{\mathrm{Spec}}(k') \to W \to \mathcal{X}$ is $2$-isomorphic to $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$. This is possible as $W \to \mathcal{X}$ is surjective. Consider the commutative diagram

both of whose squares are cartesian. By our choice of $w$ we see that $F' = F \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(k')$. Thus $F' \to F$ is surjective and we can choose a point $z' \in |F'|$ mapping to $z$. Since $F' \to F$ is flat we see that $\mathcal{O}_{F, \overline{z}} \to \mathcal{O}_{F', \overline{z}'}$ is flat, see Morphisms of Spaces, Lemma 67.30.8. Hence $f_1, \ldots , f_ r$ map to a regular sequence in $\mathcal{O}_{F', \overline{z}'}$, see Algebra, Lemma 10.68.5. Note that $U \times _\mathcal {X} W \to W$ is a morphism of algebraic spaces which is flat and locally of finite presentation. Hence by More on Morphisms of Spaces, Lemma 76.28.1 we see that there exists an open subspace $U'$ of $U \times _\mathcal {X} W$ containing $p(z')$ such that the intersection $U' \cap (V(f_1, \ldots , f_ r) \times _\mathcal {X} W)$ is flat and locally of finite presentation over $W$. Note that $\text{pr}_0(U')$ is an open subspace of $U$ containing $p(z)$ as $\text{pr}_0$ is smooth hence open. Now we see that $U' \cap (V(f_1, \ldots , f_ r) \times _\mathcal {X} W) \to \mathcal{X}$ is flat and locally of finite presentation as the composition

Hence Properties of Stacks, Lemma 100.3.5 implies $\text{pr}_0(U') \cap V(f_1, \ldots , f_ r) \to \mathcal{X}$ is flat and locally of finite presentation as desired. $\square$