## 31.4 Embedded points

Let $R$ be a ring and let $M$ be an $R$-module. Recall that a prime $\mathfrak p \subset R$ is an *embedded associated prime* of $M$ if it is an associated prime of $M$ which is not minimal among the associated primes of $M$. See Algebra, Definition 10.67.1. Here is the definition of embedded associated points for quasi-coherent sheaves on schemes as given in [IV Definition 3.1.1, EGA].

Definition 31.4.1. Let $X$ be a scheme. Let $\mathcal{F}$ be a quasi-coherent sheaf on $X$.

An *embedded associated point* of $\mathcal{F}$ is an associated point which is not maximal among the associated points of $\mathcal{F}$, i.e., it is the specialization of another associated point of $\mathcal{F}$.

A point $x$ of $X$ is called an *embedded point* if $x$ is an embedded associated point of $\mathcal{O}_ X$.

An *embedded component* of $X$ is an irreducible closed subset $Z = \overline{\{ x\} }$ where $x$ is an embedded point of $X$.

In the Noetherian case when $\mathcal{F}$ is coherent we have the following.

Lemma 31.4.2. Let $X$ be a locally Noetherian scheme. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. Then

the generic points of irreducible components of $\text{Supp}(\mathcal{F})$ are associated points of $\mathcal{F}$, and

an associated point of $\mathcal{F}$ is embedded if and only if it is not a generic point of an irreducible component of $\text{Supp}(\mathcal{F})$.

In particular an embedded point of $X$ is an associated point of $X$ which is not a generic point of an irreducible component of $X$.

**Proof.**
Recall that in this case $Z = \text{Supp}(\mathcal{F})$ is closed, see Morphisms, Lemma 29.5.3 and that the generic points of irreducible components of $Z$ are associated points of $\mathcal{F}$, see Lemma 31.2.9. Finally, we have $\text{Ass}(\mathcal{F}) \subset Z$, by Lemma 31.2.3. These results, combined with the fact that $Z$ is a sober topological space and hence every point of $Z$ is a specialization of a generic point of $Z$, imply (1) and (2).
$\square$

Lemma 31.4.3. Let $X$ be a locally Noetherian scheme. Let $\mathcal{F}$ be a coherent sheaf on $X$. Then the following are equivalent:

$\mathcal{F}$ has no embedded associated points, and

$\mathcal{F}$ has property $(S_1)$.

**Proof.**
This is Algebra, Lemma 10.157.2, combined with Lemma 31.2.2 above.
$\square$

Lemma 31.4.4. Let $X$ be a locally Noetherian scheme of dimension $\leq 1$. The following are equivalent

$X$ is Cohen-Macaulay, and

$X$ has no embedded points.

**Proof.**
Follows from Lemma 31.4.3 and the definitions.
$\square$

Lemma 31.4.5. Let $X$ be a locally Noetherian scheme. Let $U \subset X$ be an open subscheme. The following are equivalent

$U$ is scheme theoretically dense in $X$ (Morphisms, Definition 29.7.1),

$U$ is dense in $X$ and $U$ contains all embedded points of $X$.

**Proof.**
The question is local on $X$, hence we may assume that $X = \mathop{\mathrm{Spec}}(A)$ where $A$ is a Noetherian ring. Then $U$ is quasi-compact (Properties, Lemma 28.5.3) hence $U = D(f_1) \cup \ldots \cup D(f_ n)$ (Algebra, Lemma 10.29.1). In this situation $U$ is scheme theoretically dense in $X$ if and only if $A \to A_{f_1} \times \ldots \times A_{f_ n}$ is injective, see Morphisms, Example 29.7.4. Condition (2) translated into algebra means that for every associated prime $\mathfrak p$ of $A$ there exists an $i$ with $f_ i \not\in \mathfrak p$.

Assume (1), i.e., $A \to A_{f_1} \times \ldots \times A_{f_ n}$ is injective. If $x \in A$ has annihilator a prime $\mathfrak p$, then $x$ maps to a nonzero element of $A_{f_ i}$ for some $i$ and hence $f_ i \not\in \mathfrak p$. Thus (2) holds. Assume (2), i.e., every associated prime $\mathfrak p$ of $A$ corresponds to a prime of $A_{f_ i}$ for some $i$. Then $A \to A_{f_1} \times \ldots \times A_{f_ n}$ is injective because $A \to \prod _{\mathfrak p \in \text{Ass}(A)} A_\mathfrak p$ is injective by Algebra, Lemma 10.63.19.
$\square$

Lemma 31.4.6. Let $X$ be a locally Noetherian scheme. Let $\mathcal{F}$ be a coherent sheaf on $X$. The set of coherent subsheaves

\[ \{ \mathcal{K} \subset \mathcal{F} \mid \text{Supp}(\mathcal{K})\text{ is nowhere dense in }\text{Supp}(\mathcal{F}) \} \]

has a maximal element $\mathcal{K}$. Setting $\mathcal{F}' = \mathcal{F}/\mathcal{K}$ we have the following

$\text{Supp}(\mathcal{F}') = \text{Supp}(\mathcal{F})$,

$\mathcal{F}'$ has no embedded associated points, and

there exists a dense open $U \subset X$ such that $U \cap \text{Supp}(\mathcal{F})$ is dense in $\text{Supp}(\mathcal{F})$ and $\mathcal{F}'|_ U \cong \mathcal{F}|_ U$.

**Proof.**
This follows from Algebra, Lemmas 10.67.2 and 10.67.3. Note that $U$ can be taken as the complement of the closure of the set of embedded associated points of $\mathcal{F}$.
$\square$

Lemma 31.4.7. Let $X$ be a locally Noetherian scheme. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module without embedded associated points. Set

\[ \mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ X \longrightarrow \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{F})). \]

This is a coherent sheaf of ideals which defines a closed subscheme $Z \subset X$ without embedded points. Moreover there exists a coherent sheaf $\mathcal{G}$ on $Z$ such that (a) $\mathcal{F} = (Z \to X)_*\mathcal{G}$, (b) $\mathcal{G}$ has no associated embedded points, and (c) $\text{Supp}(\mathcal{G}) = Z$ (as sets).

**Proof.**
Some of the statements we have seen in the proof of Cohomology of Schemes, Lemma 30.9.7. The others follow from Algebra, Lemma 10.67.4.
$\square$

## Comments (2)

Comment #1239 by jojo on

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