Lemma 49.11.7. Let f : Y \to X be a morphism of schemes. If f satisfies the equivalent conditions of Lemma 49.11.1 then for every x \in X there exist a d and a commutative diagram
with the following properties
Lemma 49.11.7. Let f : Y \to X be a morphism of schemes. If f satisfies the equivalent conditions of Lemma 49.11.1 then for every x \in X there exist a d and a commutative diagram
with the following properties
Proof. Choose an affine open neighbourhood U = \mathop{\mathrm{Spec}}(A) \subset X of x. Write V = f^{-1}(U) = \mathop{\mathrm{Spec}}(B). Then B is a finite locally free A-module and the inclusion A \subset B is a locally direct summand. Thus after shrinking U we can choose a basis 1 = e_1, e_2, \ldots , e_ d of B as an A-module. Write e_ i e_ j = \sum \alpha _{ij}^ l e_ l for unique elements \alpha _{ij}^ l \in A which satisfy the relations \sum _ l \alpha _{ij}^ l \alpha _{lk}^ m = \sum _ l \alpha _{il}^ m \alpha _{jk}^ l and \alpha _{ij}^ k = \alpha _{ji}^ k and \alpha _{i1}^ j - \delta _{ij} in A. This determines a morphism \mathop{\mathrm{Spec}}(A) \to X_ d by sending a_{ij}^ l \in A_ d to \alpha _{ij}^ l \in A. By construction V \cong \mathop{\mathrm{Spec}}(A) \times _{X_ d} Y_ d. By the definition of U_ d we see that \mathop{\mathrm{Spec}}(A) \to X_ d factors through U_ d. This finishes the proof. \square
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Comment #7809 by Peng Du on
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