## Tag `0DR4`

## 96.3. Multiplicities of components of algebraic stacks

If $X$ is a locally Noetherian scheme, then we may write $X$ (thought of simply as a topological space) as a union of irreducible components, say $X = \bigcup T_i$. Each irreducible component is the closure of a unique generic point $\xi_i$, and the local ring $\mathcal O_{X,\xi_i}$ is a local Artin ring. We may define the

multiplicity of $X$ along $T_i$or themultiplicity of $T_i$ in $X$by $$ m_{T_i, X} = \text{length}_{\mathcal O_{X, \xi_i}} \mathcal O_{X, \xi_i} $$ In other words, it is the length of the local Artinian ring. Please compare with Chow Homology, Section 41.10.Our goal here is to generalise this definition to locally Noetherian algebraic stacks. If $\mathcal{X}$ is a stack, then its topological space $|\mathcal{X}|$ (see Properties of Stacks, Definition 89.4.8) is locally Noetherian (Morphisms of Stacks, Lemma 90.8.3). The irreducible components of $|\mathcal{X}|$ are sometimes referred to as the irreducible components of $\mathcal{X}$. If $\mathcal{X}$ is quasi-separated, then $|\mathcal{X}|$ is sober (Morphisms of Stacks, Lemma 90.29.3), but it need not be in the non-quasi-separated case. Consider for example the non-quasi-separated algebraic space $X = \mathbf{A}^1_\mathbf{C}/\mathbf{Z}$. Furthermore, there is no structure sheaf on $|\mathcal{X}|$ whose stalks can be used to define multiplicities.

Lemma 96.3.1. Let $f : U \to \mathcal{X}$ be a smooth morphism from a scheme to a locally Noetherian algebraic stack. The closure of the image of any irreducible component of $|U|$ is an irreducible component of $|\mathcal{X}|$. If $U \to \mathcal{X}$ is surjective, then all irreducible components of $|\mathcal{X}|$ are obtained in this way.

Proof.The map $|U| \to |\mathcal{X}|$ is continuous and open by Properties of Stacks, Lemma 89.4.7. Let $T \subset |U|$ be an irreducible component. Since $U$ is locally Noetherian, we can find a nonempty affine open $W \subset U$ contained in $T$. Then $f(T) \subset |\mathcal{X}|$ is irreducible and contains the nonempty open subset $f(W)$. Thus the closure of $f(T)$ is irreducible and contains a nonempty open. It follows that this closure is an irreducible component.Assume $U \to \mathcal{X}$ is surjective and let $Z \subset |\mathcal{X}|$ be an irreducible component. Choose a Noetherian open subset $V$ of $|\mathcal{X}|$ meeting $Z$. After removing the other irreducible components from $V$ we may assume that $V \subset Z$. Take an irreducible component of the nonempty open $f^{-1}(V) \subset |U|$ and let $T \subset |U|$ be its closure. This is an irreducible component of $|U|$ and the closure of $f(T)$ must agree with $Z$ by our choice of $T$. $\square$

The preceding lemma applies in particular in the case of smooth morphisms between locally Noetherian schemes. This particular case is implicitly invoked in the statement of the following lemma.

Lemma 96.3.2. Let $U \to X$ be a smooth morphism of locally Noetherian schemes. Let $T'$ is an irreducible component of $U$. Let $T$ be the irreducible component of $X$ obtained as the closure of the image of $T'$. Then $m_{T', U} = m_{T, X}$.

Proof.Write $\xi'$ for the generic point of $T'$, and $\xi$ for the generic point of $T$. Let $A = \mathcal{O}_{X, \xi}$ and $B = \mathcal{O}_{U, \xi'}$. We need to show that $\text{length}_A A = \text{length}_B B$. Since $A \to B$ is a flat local homomorphism of rings (since smooth morphisms are flat), we have $$ \text{length}_A(A) \text{length}_B(B/\mathfrak m_A B) = \text{length}_B(B) $$ by Algebra, Lemma 10.51.13. Thus it suffices to show $\mathfrak m_A B = \mathfrak m_B$, or equivalently, that $B/\mathfrak m_A B$ is reduced. Since $U \to X$ is smooth, so is its base change $U_{\xi} \to \mathop{\rm Spec} \kappa(\xi)$. As $U_{\xi}$ is a smooth scheme over a field, it is reduced, and thus so its local ring at any point (Varieties, Lemma 32.24.4). In particular, $$ B/\mathfrak m_A B = \mathcal{O}_{U, \xi'}/\mathfrak m_{X, \xi}\mathcal{O}_{U, \xi'} = \mathcal{O}_{U_\xi, \xi'} $$ is reduced, as required. $\square$Using this result, we may show that there exists a good notion of multiplicity by looking smooth locally.

Lemma 96.3.3. Let $U_1 \to \mathcal{X}$ and $U_2 \to \mathcal{X}$ be two smooth morphisms from schemes to a locally Noetherian algebraic stack $\mathcal{X}$. Let $T_1'$ and $T_2'$ be irreducible components of $|U_1|$ and $|U_2|$ respectively. Assume the closures of the images of $T_1'$ and $T_2'$ are the same irreducible component $T$ of $|\mathcal{X}|$. Then $m_{T_1', U_1} = m_{T_2', U_2}$.

Proof.Let $V_1$ and $V_2$ be dense subsets of $T_1'$ and $T'_2$, respectively, that are open in $U_1$ and $U_2$ respectively (see proof of Lemma 96.3.1). The images of $|V_1|$ and $|V_2|$ in $|\mathcal{X}|$ are non-empty open subsets of the irreducible subset $T$, and therefore have non-empty intersection. By Properties of Stacks, Lemma 89.4.3, the map $|V_1 \times_\mathcal{X} V_2| \to |V_1| \times_{|\mathcal{X}|} |V_2|$ is surjective. Consequently $V_1 \times_\mathcal{X} V_2$ is a non-empty algebraic space; we may therefore choose an étale surjection $V \to V_1 \times_\mathcal{X} V_2$ whose source is a (non-empty) scheme. If we let $T'$ be any irreducible component of $V$, then Lemma 96.3.1 shows that the closure of the image of $T'$ in $U_1$ (respectively $U_2$) is equal to $T'_1$ (respectively $T'_2$).Applying Lemma 96.3.2 twice we find that $$ m_{T_1', U_1} = m_{T', V} = m_{T_2', U_2}, $$ as required. $\square$

At this point we have done enough work to show the following definition makes sense.

Definition 96.3.4. Let $\mathcal{X}$ be a locally Noetherian algebraic stack. Let $T \subset |\mathcal{X}|$ be an irreducible component. The

multiplicityof $T$ in $\mathcal{X}$ is defined as $m_{T, \mathcal{X}} = m_{T', U}$ where $f : U \to \mathcal{X}$ is a smooth morphism from a scheme and $T' \subset |U|$ is an irreducible component with $f(T') \subset T$.This is independent of the choice of $f : U \to \mathcal{X}$ and the choice of the irreducible component $T'$ mapping to $T$ by Lemmas 96.3.1 and 96.3.3.

As a closing remark, we note that it is sometimes convenient to think of an irreducible component of $\mathcal{X}$ as a closed substack. To this end, if $\mathcal{T}$ is an irreducible component of $\mathcal{X}$, i.e., an irreducible component of $|\mathcal{X}|$, then we endow $\mathcal{T}$ with its induced reduced substack structure, see Properties of Stacks, Definition 89.10.4.

The code snippet corresponding to this tag is a part of the file `stacks-geometry.tex` and is located in lines 466–643 (see updates for more information).

```
\section{Multiplicities of components of algebraic stacks}
\label{section-multiplicities}
\noindent
If $X$ is a locally Noetherian scheme, then we may write $X$ (thought
of simply as a topological space) as a union of irreducible components,
say $X = \bigcup T_i$. Each irreducible component is the closure of a
unique generic point $\xi_i$, and the local ring $\mathcal O_{X,\xi_i}$
is a local Artin ring. We may define the {\it multiplicity of $X$ along $T_i$}
or the {\it multiplicity of $T_i$ in $X$} by
$$
m_{T_i, X} = \text{length}_{\mathcal O_{X, \xi_i}} \mathcal O_{X, \xi_i}
$$
In other words, it is the length of the local Artinian ring. Please
compare with
Chow Homology, Section \ref{chow-section-cycle-of-closed-subscheme}.
\medskip\noindent
Our goal here is to generalise this definition to locally
Noetherian algebraic stacks. If $\mathcal{X}$ is a stack,
then its topological space $|\mathcal{X}|$
(see Properties of Stacks, Definition
\ref{stacks-properties-definition-topological-space})
is locally Noetherian
(Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-Noetherian-topology}).
The irreducible components of $|\mathcal{X}|$ are sometimes
referred to as the irreducible components of $\mathcal{X}$.
If $\mathcal{X}$ is quasi-separated, then $|\mathcal{X}|$
is sober (Morphisms of Stacks, Lemma
\ref{stacks-morphisms-lemma-sober-qs}),
but it need not be in the
non-quasi-separated case. Consider for example the non-quasi-separated
algebraic space $X = \mathbf{A}^1_\mathbf{C}/\mathbf{Z}$.
Furthermore, there is no structure sheaf
on $|\mathcal{X}|$ whose stalks can be used to define multiplicities.
\begin{lemma}
\label{lemma-map-of-components}
Let $f : U \to \mathcal{X}$ be a smooth morphism from a scheme
to a locally Noetherian algebraic stack. The closure of the image of any
irreducible component of $|U|$ is an irreducible component of $|\mathcal{X}|$.
If $U \to \mathcal{X}$ is surjective, then all irreducible components of
$|\mathcal{X}|$ are obtained in this way.
\end{lemma}
\begin{proof}
The map $|U| \to |\mathcal{X}|$ is continuous and open by
Properties of Stacks, Lemma \ref{stacks-properties-lemma-topology-points}.
Let $T \subset |U|$ be an irreducible component. Since $U$ is locally
Noetherian, we can find a nonempty affine open $W \subset U$ contained in $T$.
Then $f(T) \subset |\mathcal{X}|$ is irreducible and contains the
nonempty open subset $f(W)$. Thus the closure of $f(T)$ is irreducible and
contains a nonempty open. It follows that this closure is an irreducible
component.
\medskip\noindent
Assume $U \to \mathcal{X}$ is surjective and let $Z \subset |\mathcal{X}|$
be an irreducible component. Choose a Noetherian open subset $V$
of $|\mathcal{X}|$ meeting $Z$. After removing the other irreducible
components from $V$ we may assume that $V \subset Z$.
Take an irreducible component of the nonempty
open $f^{-1}(V) \subset |U|$ and let $T \subset |U|$ be its closure.
This is an irreducible component of $|U|$ and the closure of $f(T)$
must agree with $Z$ by our choice of $T$.
\end{proof}
\noindent
The preceding lemma applies in particular in the case of smooth morphisms
between locally Noetherian schemes. This particular case is
implicitly invoked in the statement of the following lemma.
\begin{lemma}
\label{lemma-multiplicities}
Let $U \to X$ be a smooth morphism of locally Noetherian schemes.
Let $T'$ is an irreducible component of $U$. Let $T$ be the
irreducible component of $X$ obtained as the closure of the
image of $T'$. Then $m_{T', U} = m_{T, X}$.
\end{lemma}
\begin{proof}
Write $\xi'$ for the generic point of $T'$, and $\xi$ for the
generic point of $T$. Let $A = \mathcal{O}_{X, \xi}$ and
$B = \mathcal{O}_{U, \xi'}$. We need to show that
$\text{length}_A A = \text{length}_B B$. Since
$A \to B$ is a flat local homomorphism of rings
(since smooth morphisms are flat), we have
$$
\text{length}_A(A) \text{length}_B(B/\mathfrak m_A B) =
\text{length}_B(B)
$$
by Algebra, Lemma \ref{algebra-lemma-pullback-module}. Thus it suffices
to show $\mathfrak m_A B = \mathfrak m_B$, or equivalently, that
$B/\mathfrak m_A B$ is reduced. Since $U \to X$ is smooth,
so is its base change $U_{\xi} \to \Spec \kappa(\xi)$. As $U_{\xi}$ is a
smooth scheme over a field, it is reduced, and thus so its local ring
at any point
(Varieties, Lemma \ref{varieties-lemma-smooth-geometrically-normal}).
In particular,
$$
B/\mathfrak m_A B =
\mathcal{O}_{U, \xi'}/\mathfrak m_{X, \xi}\mathcal{O}_{U, \xi'} =
\mathcal{O}_{U_\xi, \xi'}
$$
is reduced, as required.
\end{proof}
\noindent
Using this result, we may show that there exists a good notion
of multiplicity by looking smooth locally.
\begin{lemma}
\label{lemma-multiplicity}
Let $U_1 \to \mathcal{X}$ and $U_2 \to \mathcal{X}$ be two smooth
morphisms from schemes to a locally Noetherian algebraic stack $\mathcal{X}$.
Let $T_1'$ and $T_2'$ be irreducible components of $|U_1|$
and $|U_2|$ respectively. Assume the closures of the images of
$T_1'$ and $T_2'$ are the same irreducible component $T$ of $|\mathcal{X}|$.
Then $m_{T_1', U_1} = m_{T_2', U_2}$.
\end{lemma}
\begin{proof}
Let $V_1$ and $V_2$ be dense subsets of $T_1'$ and $T'_2$, respectively,
that are open in $U_1$ and $U_2$ respectively (see proof of
Lemma \ref{lemma-map-of-components}).
The images of $|V_1|$ and $|V_2|$ in $|\mathcal{X}|$ are non-empty open
subsets of the irreducible subset $T$, and therefore have non-empty
intersection. By
Properties of Stacks, Lemma \ref{stacks-properties-lemma-points-cartesian},
the map $|V_1 \times_\mathcal{X} V_2| \to |V_1| \times_{|\mathcal{X}|} |V_2|$
is surjective. Consequently $V_1 \times_\mathcal{X} V_2$
is a non-empty algebraic space; we may therefore choose an
\'etale surjection $V \to V_1 \times_\mathcal{X} V_2$
whose source is a (non-empty) scheme.
If we let $T'$ be any irreducible component of $V$,
then Lemma \ref{lemma-map-of-components} shows that the closure of
the image of $T'$ in $U_1$ (respectively $U_2$) is equal to $T'_1$
(respectively $T'_2$).
\medskip\noindent
Applying Lemma \ref{lemma-multiplicities} twice we find
that
$$
m_{T_1', U_1} = m_{T', V} = m_{T_2', U_2},
$$
as required.
\end{proof}
\noindent
At this point we have done enough work to show the following
definition makes sense.
\begin{definition}
\label{definition-multiplicity}
Let $\mathcal{X}$ be a locally Noetherian algebraic stack. Let
$T \subset |\mathcal{X}|$ be an irreducible component.
The {\it multiplicity} of $T$ in $\mathcal{X}$ is defined as
$m_{T, \mathcal{X}} = m_{T', U}$ where $f : U \to \mathcal{X}$
is a smooth morphism from a scheme and $T' \subset |U|$
is an irreducible component with $f(T') \subset T$.
\end{definition}
\noindent
This is independent of the choice of $f : U \to \mathcal{X}$
and the choice of the irreducible component $T'$ mapping to $T$
by Lemmas \ref{lemma-map-of-components} and \ref{lemma-multiplicity}.
\medskip\noindent
As a closing remark, we note that it is sometimes convenient to think
of an irreducible component of $\mathcal{X}$ as a closed substack.
To this end, if $\mathcal{T}$ is an irreducible component of
$\mathcal{X}$, i.e., an irreducible component of $|\mathcal{X}|$,
then we endow $\mathcal{T}$ with its induced reduced substack structure, see
Properties of Stacks, Definition
\ref{stacks-properties-definition-reduced-induced-stack}.
```

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