Lemma 76.28.1. Let S be a scheme. Consider a cartesian diagram
\xymatrix{ X \ar[d] & F \ar[l]^ p \ar[d] \\ Y & \mathop{\mathrm{Spec}}(k) \ar[l] }
where X \to Y is a morphism of algebraic spaces over S which is flat and locally of finite presentation, and where k is a field over S. Let f_1, \ldots , f_ r \in \Gamma (X, \mathcal{O}_ X) and z \in |F| such that f_1, \ldots , f_ r map to a regular sequence in the local ring \mathcal{O}_{F, \overline{z}}. Then, after replacing X by an open subspace containing p(z), the morphism
V(f_1, \ldots , f_ r) \longrightarrow Y
is flat and locally of finite presentation.
Proof.
Set Z = V(f_1, \ldots , f_ r). It is clear that Z \to X is locally of finite presentation, hence the composition Z \to Y is locally of finite presentation, see Morphisms of Spaces, Lemma 67.28.2. Hence it suffices to show that Z \to Y is flat in a neighbourhood of p(z). Let k'/k be an extension field. Then F' = F \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(k') is surjective and flat over F, hence we can find a point z' \in |F'| mapping to z and the local ring map \mathcal{O}_{F, \overline{z}} \to \mathcal{O}_{F', \overline{z}'} is flat, see Morphisms of Spaces, Lemma 67.30.8. Hence the image of f_1, \ldots , f_ r in \mathcal{O}_{F', \overline{z}'} is a regular sequence too, see Algebra, Lemma 10.68.5. Thus, during the proof we may replace k by an extension field. In particular, we may assume that z \in |F| comes from a section z : \mathop{\mathrm{Spec}}(k) \to F of the structure morphism F \to \mathop{\mathrm{Spec}}(k).
Choose a scheme V and a surjective étale morphism V \to Y. Choose a scheme U and a surjective étale morphism U \to X \times _ Y V. After possibly enlarging k once more we may assume that \mathop{\mathrm{Spec}}(k) \to F \to X factors through U (as U \to X is surjective). Let u : \mathop{\mathrm{Spec}}(k) \to U be such a factorization and denote v \in V the image of u. Note that the morphisms
U_ v \times _{\mathop{\mathrm{Spec}}(\kappa (v))} \mathop{\mathrm{Spec}}(k) = U \times _ V \mathop{\mathrm{Spec}}(k) \to U \times _ Y \mathop{\mathrm{Spec}}(k) \to F
are étale (the first as the base change of V \to V \times _ Y V and the second as the base change of U \to X). Moreover, by construction the point u : \mathop{\mathrm{Spec}}(k) \to U gives a point of the left most space which maps to z on the right. Hence the elements f_1, \ldots , f_ r map to a regular sequence in the local ring on the right of the following map
\mathcal{O}_{U_ v, u} \longrightarrow \mathcal{O}_{U_ v \times _{\mathop{\mathrm{Spec}}(\kappa (v)} \mathop{\mathrm{Spec}}(k), \overline{u}} = \mathcal{O}_{U \times _ V \mathop{\mathrm{Spec}}(k), \overline{u}}.
But since the displayed arrow is flat (combine More on Flatness, Lemma 38.2.5 and Morphisms of Spaces, Lemma 67.30.8) we see from Algebra, Lemma 10.68.5 that f_1, \ldots , f_ r maps to a regular sequence in \mathcal{O}_{U_ v, u}. By More on Morphisms, Lemma 37.23.2 we conclude that the morphism of schemes
V(f_1, \ldots , f_ r) \times _ X U = V(f_1|_ U, \ldots , f_ r|_ U) \to V
is flat in an open neighbourhood U' of u. Let X' \subset X be the open subspace corresponding to the image of |U'| \to |X| (see Properties of Spaces, Lemmas 66.4.6 and 66.4.8). We conclude that V(f_1, \ldots , f_ r) \cap X' \to Y is flat (see Morphisms of Spaces, Definition 67.30.1) as we have the commutative diagram
\xymatrix{ V(f_1, \ldots , f_ r) \times _ X U' \ar[d]_ a \ar[r] & V \ar[d]^ b \\ V(f_1, \ldots , f_ r) \cap X' \ar[r] & Y }
with a, b étale and a surjective.
\square
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