The Stacks project

31.5 Weakly associated points

Let $R$ be a ring and let $M$ be an $R$-module. Recall that a prime $\mathfrak p \subset R$ is weakly associated to $M$ if there exists an element $m$ of $M$ such that $\mathfrak p$ is minimal among the primes containing the annihilator of $m$. See Algebra, Definition 10.66.1. If $R$ is a local ring with maximal ideal $\mathfrak m$, then $\mathfrak m$ is weakly associated to $M$ if and only if there exists an element $m \in M$ whose annihilator has radical $\mathfrak m$, see Algebra, Lemma 10.66.2.

Definition 31.5.1. Let $X$ be a scheme. Let $\mathcal{F}$ be a quasi-coherent sheaf on $X$.

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

  2. We denote $\text{WeakAss}(\mathcal{F})$ the set of weakly associated points of $\mathcal{F}$.

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

In this case, on any affine open, this corresponds exactly to the weakly associated primes as defined above. Here is the precise statement.

Lemma 31.5.2. 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. The following are equivalent

  1. $\mathfrak p$ is weakly associated to $M$, and

  2. $x$ is weakly associated to $\mathcal{F}$.

Proof. This follows from Algebra, Lemma 10.66.2. $\square$

Lemma 31.5.3. Let $X$ be a scheme. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Then

\[ \text{Ass}(\mathcal{F}) \subset \text{WeakAss}(\mathcal{F}) \subset \text{Supp}(\mathcal{F}). \]

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

Lemma 31.5.4. 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 $\text{WeakAss}(\mathcal{F}_2) \subset \text{WeakAss}(\mathcal{F}_1) \cup \text{WeakAss}(\mathcal{F}_3)$ and $\text{WeakAss}(\mathcal{F}_1) \subset \text{WeakAss}(\mathcal{F}_2)$.

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 10.66.4. $\square$

Lemma 31.5.5. Let $X$ be a scheme. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Then

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

Proof. Follows from Lemma 31.5.2 and Algebra, Lemma 10.66.5 $\square$

Lemma 31.5.6. Let $X$ be a scheme. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. If $\text{WeakAss}(\mathcal{F}) \subset U \subset X$ is open, 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, $\text{WeakAss}(\mathcal{F}') \subset \text{WeakAss}(\mathcal{F})$ by Lemma 31.5.4. Since also $\text{WeakAss}(\mathcal{F}') \subset \text{Supp}(\mathcal{F}')$ (Lemma 31.5.3) we conclude $\text{WeakAss}(\mathcal{F}') = \emptyset $. Hence $\mathcal{F}' = 0$ by Lemma 31.5.5. $\square$

Lemma 31.5.7. Let $X$ be a 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 $x \in \text{WeakAss}(\mathcal{F})$. In particular, any generic point of an irreducible component of $X$ is weakly associated to $\mathcal{O}_ X$.

Proof. Since $x \in \text{Supp}(\mathcal{F})$ the module $\mathcal{F}_ x$ is not zero. Hence $\text{WeakAss}(\mathcal{F}_ x) \subset \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$ is nonempty by Algebra, Lemma 10.66.5. On the other hand, by assumption $\text{Supp}(\mathcal{F}_ x) = \{ \mathfrak m_ x\} $. Since $\text{WeakAss}(\mathcal{F}_ x) \subset \text{Supp}(\mathcal{F}_ x)$ (Algebra, Lemma 10.66.6) we see that $\mathfrak m_ x$ is weakly associated to $\mathcal{F}_ x$ and we win. $\square$

Lemma 31.5.8. Let $X$ be a scheme. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. If $\mathfrak m_ x$ is a finitely generated ideal of $\mathcal{O}_{X, x}$, then

\[ x \in \text{Ass}(\mathcal{F}) \Leftrightarrow x \in \text{WeakAss}(\mathcal{F}). \]

In particular, if $X$ is locally Noetherian, then $\text{Ass}(\mathcal{F}) = \text{WeakAss}(\mathcal{F})$.

Proof. See Algebra, Lemma 10.66.9. $\square$

Lemma 31.5.9. Let $f : X \to S$ be a quasi-compact and quasi-separated morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $s \in S$ be a point which is not in the image of $f$. Then $s$ is not weakly associated to $f_*\mathcal{F}$.

Proof. Consider the base change $f' : X' \to \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ of $f$ by the morphism $g : \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}) \to S$ and denote $g' : X' \to X$ the other projection. Then

\[ (f_*\mathcal{F})_ s = (g^*f_*\mathcal{F})_ s = (f'_*(g')^*\mathcal{F})_ s \]

The first equality because $g$ induces an isomorphism on local rings at $s$ and the second by flat base change (Cohomology of Schemes, Lemma 30.5.2). Of course $s \in \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ is not in the image of $f'$. Thus we may assume $S$ is the spectrum of a local ring $(A, \mathfrak m)$ and $s$ corresponds to $\mathfrak m$. By Schemes, Lemma 26.24.1 the sheaf $f_*\mathcal{F}$ is quasi-coherent, say corresponding to the $A$-module $M$. As $s$ is not in the image of $f$ we see that $X = \bigcup _{a \in \mathfrak m} f^{-1}D(a)$ is an open covering. Since $X$ is quasi-compact we can find $a_1, \ldots , a_ n \in \mathfrak m$ such that $X = f^{-1}D(a_1) \cup \ldots \cup f^{-1}D(a_ n)$. It follows that

\[ M \to M_{a_1} \oplus \ldots \oplus M_{a_ r} \]

is injective. Hence for any nonzero element $m$ of the stalk $M_\mathfrak p$ there exists an $i$ such that $a_ i^ n m$ is nonzero for all $n \geq 0$. Thus $\mathfrak m$ is not weakly associated to $M$. $\square$

Lemma 31.5.10. Let $X$ be a 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

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

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

Then $\varphi $ is injective.

Proof. The assumptions imply that $\text{WeakAss}(\mathop{\mathrm{Ker}}(\varphi )) = \emptyset $ and hence $\mathop{\mathrm{Ker}}(\varphi ) = 0$ by Lemma 31.5.5. $\square$

Lemma 31.5.11. Let $X$ be a locally Noetherian scheme. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. Let $j : U \to X$ be an open subscheme such that for $x \in X \setminus U$ we have $\text{depth}(\mathcal{F}_ x) \geq 2$. Then

\[ \mathcal{F} \longrightarrow j_*(\mathcal{F}|_ U) \]

is an isomorphism and consequently $\Gamma (X, \mathcal{F}) \to \Gamma (U, \mathcal{F})$ is an isomorphism too.

Proof. We claim Lemma 31.2.11 applies to the map displayed in the lemma. Let $x \in X$. If $x \in U$, then the map is an isomorphism on stalks as $j_*(\mathcal{F}|_ U)|_ U = \mathcal{F}|_ U$. If $x \in X \setminus U$, then $x \not\in \text{Ass}(j_*(\mathcal{F}|_ U))$ (Lemmas 31.5.9 and 31.5.3). Since we've assumed $\text{depth}(\mathcal{F}_ x) \geq 2$ this finishes the proof. $\square$

Lemma 31.5.12. Let $X$ be a reduced scheme. Then the weakly associated points of $X$ are exactly the generic points of the irreducible components of $X$.

Proof. Follows from Algebra, Lemma 10.66.3. $\square$


Comments (4)

Comment #4072 by Matthieu Romagny on

Typo in introductory sentence of section 056K: "If R is a local ring with maximal ideal m, then m is weakly associated to M if and only if there exists an element m∈M whose annihilator has radical m, see Algebra, Lemma 0566".

Comment #4749 by Koito Yuu on

In the proof of Lemma 31.5.6, '' should be replaced by ''.


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