## 31.2 Associated points

Let $R$ be a ring and let $M$ be an $R$-module. Recall that a prime $\mathfrak p \subset R$ is *associated* to $M$ if there exists an element of $M$ whose annihilator is $\mathfrak p$. See Algebra, Definition 10.63.1. Here is the definition of associated points for quasi-coherent sheaves on schemes as given in [IV Definition 3.1.1, EGA].

Definition 31.2.1. Let $X$ be a scheme. Let $\mathcal{F}$ be a quasi-coherent sheaf on $X$.

We say $x \in X$ is *associated* to $\mathcal{F}$ if the maximal ideal $\mathfrak m_ x$ is associated to the $\mathcal{O}_{X, x}$-module $\mathcal{F}_ x$.

We denote $\text{Ass}(\mathcal{F})$ or $\text{Ass}_ X(\mathcal{F})$ the set of associated points of $\mathcal{F}$.

The *associated points of $X$* are the associated points of $\mathcal{O}_ X$.

These definitions are most useful when $X$ is locally Noetherian and $\mathcal{F}$ of finite type. For example it may happen that a generic point of an irreducible component of $X$ is not associated to $X$, see Example 31.2.7. In the non-Noetherian case it may be more convenient to use weakly associated points, see Section 31.5. Let us link the scheme theoretic notion with the algebraic notion on affine opens; note that this correspondence works perfectly only for locally Noetherian schemes.

Lemma 31.2.2. Let $X$ be a scheme. Let $\mathcal{F}$ be a quasi-coherent sheaf on $X$. Let $\mathop{\mathrm{Spec}}(A) = U \subset X$ be an affine open, and set $M = \Gamma (U, \mathcal{F})$. Let $x \in U$, and let $\mathfrak p \subset A$ be the corresponding prime.

If $\mathfrak p$ is associated to $M$, then $x$ is associated to $\mathcal{F}$.

If $\mathfrak p$ is finitely generated, then the converse holds as well.

In particular, if $X$ is locally Noetherian, then the equivalence

\[ \mathfrak p \in \text{Ass}(M) \Leftrightarrow x \in \text{Ass}(\mathcal{F}) \]

holds for all pairs $(\mathfrak p, x)$ as above.

**Proof.**
This follows from Algebra, Lemma 10.63.15. But we can also argue directly as follows. Suppose $\mathfrak p$ is associated to $M$. Then there exists an $m \in M$ whose annihilator is $\mathfrak p$. Since localization is exact we see that $\mathfrak pA_{\mathfrak p}$ is the annihilator of $m/1 \in M_{\mathfrak p}$. Since $M_{\mathfrak p} = \mathcal{F}_ x$ (Schemes, Lemma 26.5.4) we conclude that $x$ is associated to $\mathcal{F}$.

Conversely, assume that $x$ is associated to $\mathcal{F}$, and $\mathfrak p$ is finitely generated. As $x$ is associated to $\mathcal{F}$ there exists an element $m' \in M_{\mathfrak p}$ whose annihilator is $\mathfrak pA_{\mathfrak p}$. Write $m' = m/f$ for some $f \in A$, $f \not\in \mathfrak p$. The annihilator $I$ of $m$ is an ideal of $A$ such that $IA_{\mathfrak p} = \mathfrak pA_{\mathfrak p}$. Hence $I \subset \mathfrak p$, and $(\mathfrak p/I)_{\mathfrak p} = 0$. Since $\mathfrak p$ is finitely generated, there exists a $g \in A$, $g \not\in \mathfrak p$ such that $g(\mathfrak p/I) = 0$. Hence the annihilator of $gm$ is $\mathfrak p$ and we win.

If $X$ is locally Noetherian, then $A$ is Noetherian (Properties, Lemma 28.5.2) and $\mathfrak p$ is always finitely generated.
$\square$

Lemma 31.2.3. Let $X$ be a scheme. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Then $\text{Ass}(\mathcal{F}) \subset \text{Supp}(\mathcal{F})$.

**Proof.**
This is immediate from the definitions.
$\square$

Lemma 31.2.4. Let $X$ be a scheme. Let $0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0$ be a short exact sequence of quasi-coherent sheaves on $X$. Then $\text{Ass}(\mathcal{F}_2) \subset \text{Ass}(\mathcal{F}_1) \cup \text{Ass}(\mathcal{F}_3)$ and $\text{Ass}(\mathcal{F}_1) \subset \text{Ass}(\mathcal{F}_2)$.

**Proof.**
For every point $x \in X$ the sequence of stalks $0 \to \mathcal{F}_{1, x} \to \mathcal{F}_{2, x} \to \mathcal{F}_{3, x} \to 0$ is a short exact sequence of $\mathcal{O}_{X, x}$-modules. Hence the lemma follows from Algebra, Lemma 10.63.3.
$\square$

Lemma 31.2.5. Let $X$ be a locally Noetherian scheme. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. Then $\text{Ass}(\mathcal{F}) \cap U$ is finite for every quasi-compact open $U \subset X$.

**Proof.**
This is true because the set of associated primes of a finite module over a Noetherian ring is finite, see Algebra, Lemma 10.63.5. To translate from schemes to algebra use that $U$ is a finite union of affine opens, each of these opens is the spectrum of a Noetherian ring (Properties, Lemma 28.5.2), $\mathcal{F}$ corresponds to a finite module over this ring (Cohomology of Schemes, Lemma 30.9.1), and finally use Lemma 31.2.2.
$\square$

Lemma 31.2.6. Let $X$ be a locally Noetherian scheme. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Then

\[ \mathcal{F} = 0 \Leftrightarrow \text{Ass}(\mathcal{F}) = \emptyset . \]

**Proof.**
If $\mathcal{F} = 0$, then $\text{Ass}(\mathcal{F}) = \emptyset $ by definition. Conversely, if $\text{Ass}(\mathcal{F}) = \emptyset $, then $\mathcal{F} = 0$ by Algebra, Lemma 10.63.7. To translate from schemes to algebra, restrict to any affine and use Lemma 31.2.2.
$\square$

Example 31.2.7. Let $k$ be a field. The ring $R = k[x_1, x_2, x_3, \ldots ]/(x_ i^2)$ is local with locally nilpotent maximal ideal $\mathfrak m$. There exists no element of $R$ which has annihilator $\mathfrak m$. Hence $\text{Ass}(R) = \emptyset $, and $X = \mathop{\mathrm{Spec}}(R)$ is an example of a scheme which has no associated points.

Lemma 31.2.8. Let $X$ be a locally Noetherian scheme. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. If $U \subset X$ is open and $\text{Ass}(\mathcal{F}) \subset U$, then $\Gamma (X, \mathcal{F}) \to \Gamma (U, \mathcal{F})$ is injective.

**Proof.**
Let $s \in \Gamma (X, \mathcal{F})$ be a section which restricts to zero on $U$. Let $\mathcal{F}' \subset \mathcal{F}$ be the image of the map $\mathcal{O}_ X \to \mathcal{F}$ defined by $s$. Then $\text{Supp}(\mathcal{F}') \cap U = \emptyset $. On the other hand, $\text{Ass}(\mathcal{F}') \subset \text{Ass}(\mathcal{F})$ by Lemma 31.2.4. Since also $\text{Ass}(\mathcal{F}') \subset \text{Supp}(\mathcal{F}')$ (Lemma 31.2.3) we conclude $\text{Ass}(\mathcal{F}') = \emptyset $. Hence $\mathcal{F}' = 0$ by Lemma 31.2.6.
$\square$

Lemma 31.2.9. Let $X$ be a locally Noetherian scheme. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $x \in \text{Supp}(\mathcal{F})$ be a point in the support of $\mathcal{F}$ which is not a specialization of another point of $\text{Supp}(\mathcal{F})$. Then $x \in \text{Ass}(\mathcal{F})$. In particular, any generic point of an irreducible component of $X$ is an associated point of $X$.

**Proof.**
Since $x \in \text{Supp}(\mathcal{F})$ the module $\mathcal{F}_ x$ is not zero. Hence $\text{Ass}(\mathcal{F}_ x) \subset \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$ is nonempty by Algebra, Lemma 10.63.7. On the other hand, by assumption $\text{Supp}(\mathcal{F}_ x) = \{ \mathfrak m_ x\} $. Since $\text{Ass}(\mathcal{F}_ x) \subset \text{Supp}(\mathcal{F}_ x)$ (Algebra, Lemma 10.63.2) we see that $\mathfrak m_ x$ is associated to $\mathcal{F}_ x$ and we win.
$\square$

The following lemma is the analogue of More on Algebra, Lemma 15.23.12.

Lemma 31.2.10. Let $X$ be a locally Noetherian scheme. Let $\varphi : \mathcal{F} \to \mathcal{G}$ be a map of quasi-coherent $\mathcal{O}_ X$-modules. Assume that for every $x \in X$ at least one of the following happens

$\mathcal{F}_ x \to \mathcal{G}_ x$ is injective, or

$x \not\in \text{Ass}(\mathcal{F})$.

Then $\varphi $ is injective.

**Proof.**
The assumptions imply that $\text{Ass}(\mathop{\mathrm{Ker}}(\varphi )) = \emptyset $ and hence $\mathop{\mathrm{Ker}}(\varphi ) = 0$ by Lemma 31.2.6.
$\square$

Lemma 31.2.11. Let $X$ be a locally Noetherian scheme. Let $\varphi : \mathcal{F} \to \mathcal{G}$ be a map of quasi-coherent $\mathcal{O}_ X$-modules. Assume $\mathcal{F}$ is coherent and that for every $x \in X$ one of the following happens

$\mathcal{F}_ x \to \mathcal{G}_ x$ is an isomorphism, or

$\text{depth}(\mathcal{F}_ x) \geq 2$ and $x \not\in \text{Ass}(\mathcal{G})$.

Then $\varphi $ is an isomorphism.

**Proof.**
This is a translation of More on Algebra, Lemma 15.23.13 into the language of schemes.
$\square$

## Comments (2)

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