Lemma 106.3.1. 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.

## 106.3 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 the *multiplicity of $T_ i$ in $X$* by

In other words, it is the length of the local Artinian ring. Please compare with Chow Homology, Section 42.9.

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 99.4.8) is locally Noetherian (Morphisms of Stacks, Lemma 100.8.3). 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 100.30.3), 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.

**Proof.**
The map $|U| \to |\mathcal{X}|$ is continuous and open by Properties of Stacks, Lemma 99.4.7. 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 106.3.2. 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

by Algebra, Lemma 10.52.13. 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 $U_{\xi } \to \mathop{\mathrm{Spec}}\kappa (\xi )$. As $U_{\xi }$ is a smooth scheme over a field, it is reduced, and thus so its local ring at any point (Varieties, Lemma 33.25.4). In particular,

is reduced, as required. $\square$

Using this result, we may show that there exists a good notion of multiplicity by looking smooth locally.

Lemma 106.3.3. 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 106.3.1). 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 99.4.3, 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 106.3.1 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 106.3.2 twice we find that

as required. $\square$

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

Definition 106.3.4. Let $\mathcal{X}$ be a locally Noetherian algebraic stack. Let $T \subset |\mathcal{X}|$ be an irreducible component. The *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 $f(T') \subset T$.

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 106.3.1 and 106.3.3.

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 $T \subset |\mathcal{X}|$ is an irreducible component, then we may consider the unique reduced closed substack $\mathcal{T} \subset \mathcal{X}$ with $|\mathcal{T}| = T$, see Properties of Stacks, Definition 99.10.4. If $\mathcal{X}$ is quasi-separated, then an irreducible component is an integral stack; see Morphisms of Stacks, Section 100.50 for further discussion.

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