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. $\square$
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #7809 by Peng Du on
Comment #8039 by Stacks Project on