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 $X$, and
$U$ contains all associated 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 (3) 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. Say $a \in A$ has annihilator a prime $\mathfrak p$. Since $a$ maps to a nonzero element of $A_{f_ i}$ for some $i$ we see that $f_ i \not\in \mathfrak p$. Thus (3) holds.
Note that $U$ is dense in $\mathop{\mathrm{Spec}}(A)$ if and only if it contains all minimal prime ideals. Since the set of associated primes of $A$ is equal to the union of the set of minimal primes and the set of embedded associated primes (see Algebra, Proposition 10.63.6), we see that (2) and (3) are equivalent.
Assume (3). This means that 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. In this way we see that (3) implies (1).
$\square$
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.
$\square$
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