106.4 Formal branches and multiplicities

It will be convenient to have a comparison between the notion of multiplicity of an irreducible component given by Definition 106.3.4 and the related notion of multiplicities of irreducible components of (the spectra of) versal rings of $\mathcal{X}$ at finite type points.

In Situation 106.2.1 let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism, where $k$ is a finite type field over $S$. Let $A$, $A'$ be versal rings to $\mathcal{X}$ at $x_0$. After possibly swapping $A$ and $A'$, we know there is a formally smooth1 map $\varphi : A \to A'$ compatible with versal formal objects, see Lemma 106.2.4 and Remark 106.2.9. Moreover, $\varphi$ is well defined up to formal homotopy, see Formal Deformation Theory, Lemma 89.28.3. In particular, we find that $\varphi (\mathfrak p)A'$ is a well defined ideal of $A'$ by Formal Deformation Theory, Lemma 89.28.4. Since $A \to A'$ is formally smooth, in fact $\varphi (\mathfrak p)A'$ is a minimal prime of $A'$ and every minimal prime of $A'$ is of this form for a unique minimal prime $\mathfrak p \subset A$ (all of this is easy to prove by writing $A'$ as a power series ring over $A$). Therefore, recalling that minimal primes correspond to irreducible components, the following definition makes sense.

Definition 106.4.1. Let $\mathcal{X}$ be an algebraic stack locally of finite type over a locally Noetherian scheme $S$. Let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ is a morphism where $k$ is a field of finite type over $S$. The formal branches of $\mathcal{X}$ through $x_0$ is the set of irreducible components of $\mathop{\mathrm{Spec}}(A)$ for any choice of versal ring to $\mathcal{X}$ at $x_0$ identified for different choices of $A$ by the procedure described above.

Suppose in the situation of Definition 106.4.1 we are given a finite extension $l/k$. Set $x_{l, 0} : \mathop{\mathrm{Spec}}(l) \to \mathcal{X}$ equal to the composition of $\mathop{\mathrm{Spec}}(l) \to \mathop{\mathrm{Spec}}(k)$ with $x_0$. Let $A \to A'$ be as in Lemma 106.2.5. Since $A \to A'$ is faithfully flat, the morphism

$\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A)$

sends (generic points of) irreducible components to (generic points of) irreducible components. This will be a surjective map, but in general this map will not be a bijection. In other words, we obtain a surjective map

$\text{formal branches of }\mathcal{X}\text{ through }x_{l, 0} \longrightarrow \text{formal branches of }\mathcal{X}\text{ through }x_0$

It turns out that if $l/k$ is purely inseparable, then the map is injective as well (we'll add a precise statement and proof here if we ever need this).

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 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 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. $\square$

Let $A$ be a Noetherian complete local ring. Then the irreducible components of $\mathop{\mathrm{Spec}}(A)$ have multiplicities, see introduction to Section 106.3. If $A' = A[[t_1, \ldots , t_ r]]$, then the morphism $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A)$ induces a bijection on irreducible components preserving multiplicities (we omit the easy proof). This and the discussion preceding Definition 106.4.1 mean that the following definition makes sense.

Definition 106.4.3. Let $\mathcal{X}$ be an algebraic stack locally of finite type over a locally Noetherian scheme $S$. Let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ is a morphism where $k$ is a field of finite type over $S$. The multiplicity of a formal branch of $\mathcal{X}$ through $x_0$ is the multiplicity of the corresponding irreducible component of $\mathop{\mathrm{Spec}}(A)$ for any choice of versal ring to $\mathcal{X}$ at $x_0$ (see discussion above).

Lemma 106.4.4. Let $\mathcal{X}$ be an algebraic stack locally of finite type over a locally Noetherian scheme $S$. Let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ is a morphism where $k$ is a field of finite type over $S$ with image $s \in S$. If $\mathcal{O}_{S, s}$ is a G-ring, then the map of Lemma 106.4.2 preserves multiplicities.

Proof. By Lemma 106.2.8 we may assume there is a smooth morphism $U \to \mathcal{X}$ where $U$ is a scheme and a $k$-valued point $u_0$ of $U$ such that $\mathcal{O}_{U, u_0}^\wedge$ is a versal ring to $\mathcal{X}$ at $x_0$. By construction of our map in the proof of Lemma 106.4.2 (which simplifies greatly because $A = \mathcal{O}_{U, u_0}^\wedge$) we find that it suffices to show: the multiplicity of an irreducible component of $U$ passing through $u_0$ is the same as the multiplicity of any irreducible component of $\mathop{\mathrm{Spec}}(\mathcal{O}_{U, u_0}^\wedge )$ mapping into it.

Translated into commutative algebra we find the following: Let $C = \mathcal{O}_{U, u_0}$. This is essentially of finite type over $\mathcal{O}_{S, s}$ and hence is a G-ring (More on Algebra, Proposition 15.50.10). Then $A = C^\wedge$. Therefore $C \to A$ is a regular ring map. Let $\mathfrak q \subset C$ be a minimal prime and let $\mathfrak p \subset A$ be a minimal prime lying over $\mathfrak q$. Then

$R = C_\mathfrak p \longrightarrow A_\mathfrak p = R'$

is a regular ring map of Artinian local rings. For such a ring map it is always the case that

$\text{length}_ R R = \text{length}_{R'} R'$

This is what we have to show because the left hand side is the multiplicity of our component on $U$ and the right hand side is the multiplicity of our component on $\mathop{\mathrm{Spec}}(A)$. To see the equality, first we use that

$\text{length}_ R(R) \text{length}_{R'}(R'/\mathfrak m_ R R') = \text{length}_{R'}(R')$

by Algebra, Lemma 10.52.13. Thus it suffices to show $\mathfrak m_ R R' = \mathfrak m_{R'}$, which is a consequence of being a regular homomorphism of zero dimensional local rings. $\square$

[1] In the sense that $A'$ becomes isomorphic to a power series ring over $A$.

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