The Stacks project

Lemma 106.4.2. In the situation of Definition 106.4.1 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 106.4.1 and let $\mathop{\mathrm{Spec}}(A) \to \mathcal{X}$ be as in Remark 106.2.9. 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 106.2.10 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 65.11 for notation and use going down on ├ętale 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 106.3.1 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 99.4.3 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. $\square$


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