Lemma 107.4.2. In the situation of Definition 107.4.1 there is a canonical surjection from the set of formal branches of $\mathcal{X}$ through $x_0$ to the set of irreducible components of $|\mathcal{X}|$ containing $x_0$ in $|\mathcal{X}|$.
Proof. Let $A$ be as in Definition 107.4.1 and let $\mathop{\mathrm{Spec}}(A) \to \mathcal{X}$ be as in Remark 107.2.9. We claim that the generic point of an irreducible component of $\mathop{\mathrm{Spec}}(A)$ maps to a generic point of an irreducible component of $|\mathcal{X}|$. Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Consider the diagram
\[ \xymatrix{ \mathop{\mathrm{Spec}}(A) \times _\mathcal {X} U \ar[d]_ p \ar[r]_-q & U \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar[r]^ j & \mathcal{X} } \]
By Lemma 107.2.10 we see that $j$ is flat. Hence $q$ is flat. On the other hand, $f$ is surjective smooth hence $p$ is surjective smooth. This implies that any generic point $\eta \in \mathop{\mathrm{Spec}}(A)$ of an irreducible component is the image of a codimension $0$ point $\eta '$ of the algebraic space $\mathop{\mathrm{Spec}}(A) \times _\mathcal {X} U$ (see Properties of Spaces, Section 66.11 for notation and use going down on étale local rings). Since $q$ is flat, $q(\eta ')$ is a codimension $0$ point of $U$ (same argument). Since $U$ is a scheme, $q(\eta ')$ is the generic point of an irreducible component of $U$. Thus the closure of the image of $q(\eta ')$ in $|\mathcal{X}|$ is an irreducible component by Lemma 107.3.1 as claimed.
Clearly the claim provides a mechanism for defining the desired map. To see that it is surjective, we choose $u_0 \in U$ mapping to $x_0$ in $|\mathcal{X}|$. Choose an affine open $U' \subset U$ neighbourhood of $u_0$. After shrinking $U'$ we may assume every irreducible component of $U'$ passes through $u_0$. Then we may replace $\mathcal{X}$ by the open substack corresponding to the image of $|U'| \to |\mathcal{X}|$. Thus we may assume $U$ is affine has a point $u_0$ mapping to $x_0 \in |\mathcal{X}|$ and every irreducible component of $U$ passes through $u_0$. By Properties of Stacks, Lemma 100.4.3 there is a point $t \in |\mathop{\mathrm{Spec}}(A) \times _\mathcal {X} U|$ mapping to the closed point of $\mathop{\mathrm{Spec}}(A)$ and to $u_0$. Using going down for the flat local ring homomorphisms
\[ A \longrightarrow \mathcal{O}_{\mathop{\mathrm{Spec}}(A) \times _\mathcal {X} U, \overline{t}} \longleftarrow \mathcal{O}_{U, u_0} \]
we see that every minimal prime of $\mathcal{O}_{U, u_0}$ is the image of a minimal prime of the local ring in the middle and such a minimal prime maps to a minimal prime of $A$. This proves the surjectivity. Some details omitted. $\square$
Post a comment
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 (0)