The Stacks project

80.4 Multiplicities

A section with a few simple results on lengths and multiplicities.

Lemma 80.4.1. Let $S$ be a scheme and let $X$ be an algebraic space over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $x \in |X|$. Let $d \in \{ 0, 1, 2, \ldots , \infty \} $. The following are equivalent

  1. $\text{length}_{\mathcal{O}_{X, \overline{x}}} \mathcal{F}_{\overline{x}} = d$

  2. for some étale morphism $U \to X$ with $U$ a scheme and $u \in U$ mapping to $x$ we have $\text{length}_{\mathcal{O}_{U, u}} (\mathcal{F}|_ U)_ u = d$

  3. for any étale morphism $U \to X$ with $U$ a scheme and $u \in U$ mapping to $x$ we have $\text{length}_{\mathcal{O}_{U, u}} (\mathcal{F}|_ U)_ u = d$

Proof. Let $U \to X$ and $u \in U$ be as in (2) or (3). Then we know that $\mathcal{O}_{X, \overline{x}}$ is the strict henselization of $\mathcal{O}_{U, u}$ and that

\[ \mathcal{F}_{\overline{x}} = (\mathcal{F}|_ U)_ u \otimes _{\mathcal{O}_{U, u}} \mathcal{O}_{X, \overline{x}} \]

See Properties of Spaces, Lemmas 64.22.1 and 64.29.4. Thus the equality of the lengths follows from Algebra, Lemma 10.51.13 the fact that $\mathcal{O}_{U, u} \to \mathcal{O}_{X, \overline{x}}$ is flat and the fact that $\mathcal{O}_{X, \overline{x}}/\mathfrak m_ u\mathcal{O}_{X, \overline{x}}$ is equal to the residue field of $\mathcal{O}_{X, \overline{x}}$. These facts about strict henselizations can be found in More on Algebra, Lemma 15.44.1. $\square$

Definition 80.4.2. Let $S$ be a scheme and let $X$ be an algebraic space over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $x \in |X|$. Let $d \in \{ 0, 1, 2, \ldots , \infty \} $. We say $\mathcal{F}$ has length $d$ at $x$ if the equivalent conditions of Lemma 80.4.1 are satisfied.

Lemma 80.4.3. Let $S$ be a scheme. Let $i : Y \to X$ be a closed immersion of algebraic spaces over $S$. Let $\mathcal{G}$ be a quasi-coherent $\mathcal{O}_ Y$-module. Let $y \in |Y|$ with image $x \in |X|$. Let $d \in \{ 0, 1, 2, \ldots , \infty \} $. The following are equivalent

  1. $\mathcal{G}$ has length $d$ at $y$, and

  2. $i_*\mathcal{G}$ has length $d$ at $x$.

Proof. Choose an étale morphism $f : U \to X$ with $U$ a scheme and $u \in U$ mapping to $x$. Set $V = Y \times _ X U$. Denote $g : V \to Y$ and $j : V \to U$ the projections. Then $j : V \to U$ is a closed immersion and there is a unique point $v \in V$ mapping to $y \in |Y|$ and $u \in U$ (use Properties of Spaces, Lemma 64.4.3 and Spaces, Lemma 63.12.3). We have $j_*(\mathcal{G}|_ V) = (i_*\mathcal{G})|_ U$ as modules on the scheme $V$ and $j_*$ the “usual” pushforward of modules for the morphism of schemes $j$, see discussion surrounding Cohomology of Spaces, Equation (67.3.0.1). In this way we reduce to the case of schemes: if $i : Y \to X$ is a closed immersion of schemes, then

\[ (i_*\mathcal{G})_ x = \mathcal{G}_ y \]

as modules over $\mathcal{O}_{X, x}$ where the module structure on the right hand side is given by the surjection $i_ y^\sharp : \mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}$. Thus equality by Algebra, Lemma 10.51.5. $\square$

Lemma 80.4.4. Let $S$ be a scheme and let $X$ be a locally Noetherian algebraic space over $S$. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. Let $x \in |X|$. The following are equivalent

  1. for some étale morphism $U \to X$ with $U$ a scheme and $u \in U$ mapping to $x$ we have $u$ is a generic point of an irreducible component of $\text{Supp}(\mathcal{F}|_ U)$,

  2. for any étale morphism $U \to X$ with $U$ a scheme and $u \in U$ mapping to $x$ we have $u$ is a generic point of an irreducible component of $\text{Supp}(\mathcal{F}|_ U)$,

  3. the length of $\mathcal{F}$ at $x$ is finite and nonzero.

If $X$ is decent (equivalently quasi-separated) then these are also equivalent to

  1. $x$ is a generic point of an irreducible component of $\text{Supp}(\mathcal{F})$.

Proof. Assume $f : U \to X$ is an étale morphism with $U$ a scheme and $u \in U$ maps to $x$. Then $\mathcal{F}|_ U = f^*\mathcal{F}$ is a coherent $\mathcal{O}_ U$-module on the locally Noetherian scheme $U$ and in particular $(\mathcal{F}|_ U)_ u$ is a finite $\mathcal{O}_{U, u}$-module, see Cohomology of Spaces, Lemma 67.12.2 and Cohomology of Schemes, Lemma 30.9.1. Recall that the support of $\mathcal{F}|_ U$ is a closed subset of $U$ (Morphisms, Lemma 29.5.3) and that the support of $(\mathcal{F}|_ U)_ u$ is the pullback of the support of $\mathcal{F}|_ U$ by the morphism $\mathop{\mathrm{Spec}}(\mathcal{O}_{U, u}) \to U$. Thus $u$ is a generic point of an irreducible component of $\text{Supp}(\mathcal{F}|_ U)$ if and only if the support of $(\mathcal{F}|_ U)_ u$ is equal to the maximal ideal of $\mathcal{O}_{U, u}$. Now the equivalence of (1), (2), (3) follows from by Algebra, Lemma 10.61.3.

If $X$ is decent we choose an étale morphism $f : U \to X$ and a point $u \in U$ mapping to $x$. The support of $\mathcal{F}$ pulls back to the support of $\mathcal{F}|_ U$, see Morphisms of Spaces, Lemma 65.15.2. Also, specializations $x' \leadsto x$ in $|X|$ lift to specializations $u' \leadsto u$ in $U$ and any nontrivial specialization $u' \leadsto u$ in $U$ maps to a nontrivial specialization $f(u') \leadsto f(u)$ in $|X|$, see Decent Spaces, Lemmas 66.12.2 and 66.12.1. Using that $|X|$ and $U$ are sober topological spaces (Decent Spaces, Proposition 66.12.4 and Schemes, Lemma 26.11.1) we conclude $x$ is a generic point of the support of $\mathcal{F}$ if and only if $u$ is a generic point of the support of $\mathcal{F}|_ U$. We conclude (4) is equivalent to (1).

The parenthetical statement follows from Decent Spaces, Lemma 66.14.1. $\square$

Lemma 80.4.5. In Situation 80.2.1 let $X/B$ be good. Let $T \subset |X|$ be a closed subset and $t \in T$. If $\dim _\delta (T) \leq k$ and $\delta (t) = k$, then $t$ is a generic point of an irreducible component of $T$.

Proof. We know $t$ is contained in an irreducible component $T' \subset T$. Let $t' \in T'$ be the generic point. Then $k \geq \delta (t') \geq \delta (t)$. Since $\delta $ is a dimension function we see that $t = t'$. $\square$


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