The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

10.62 Associated primes

Here is the standard definition. For non-Noetherian rings and non-finite modules it may be more appropriate to use the definition in Section 10.65.

Definition 10.62.1. Let $R$ be a ring. Let $M$ be an $R$-module. A prime $\mathfrak p$ of $R$ is associated to $M$ if there exists an element $m \in M$ whose annihilator is $\mathfrak p$. The set of all such primes is denoted $\text{Ass}_ R(M)$ or $\text{Ass}(M)$.

Lemma 10.62.2. Let $R$ be a ring. Let $M$ be an $R$-module. Then $\text{Ass}(M) \subset \text{Supp}(M)$.

Proof. If $m \in M$ has annihilator $\mathfrak p$, then in particular no element of $R \setminus \mathfrak p$ annihilates $m$. Hence $m$ is a nonzero element of $M_{\mathfrak p}$, i.e., $\mathfrak p \in \text{Supp}(M)$. $\square$

Lemma 10.62.3. Let $R$ be a ring. Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence of $R$-modules. Then $\text{Ass}(M') \subset \text{Ass}(M)$ and $\text{Ass}(M) \subset \text{Ass}(M') \cup \text{Ass}(M'')$.

Proof. Omitted. $\square$

Lemma 10.62.4. Let $R$ be a ring, and $M$ an $R$-module. Suppose there exists a filtration by $R$-submodules

\[ 0 = M_0 \subset M_1 \subset \ldots \subset M_ n = M \]

such that each quotient $M_ i/M_{i-1}$ is isomorphic to $R/\mathfrak p_ i$ for some prime ideal $\mathfrak p_ i$ of $R$. Then $\text{Ass}(M) \subset \{ \mathfrak p_1, \ldots , \mathfrak p_ n\} $.

Proof. By induction on the length $n$ of the filtration $\{ M_ i \} $. Pick $m \in M$ whose annihilator is a prime $\mathfrak p$. If $m \in M_{n-1}$ we are done by induction. If not, then $m$ maps to a nonzero element of $M/M_{n-1} \cong R/\mathfrak p_ n$. Hence we have $\mathfrak p \subset \mathfrak p_ n$. If equality does not hold, then we can find $f \in \mathfrak p_ n$, $f \not\in \mathfrak p$. In this case the annihilator of $fm$ is still $\mathfrak p$ and $fm \in M_{n-1}$. Thus we win by induction. $\square$

Lemma 10.62.5. Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. Then $\text{Ass}(M)$ is finite.

Proposition 10.62.6. Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. The following sets of primes are the same:

  1. The minimal primes in the support of $M$.

  2. The minimal primes in $\text{Ass}(M)$.

  3. For any filtration $0 = M_0 \subset M_1 \subset \ldots \subset M_{n-1} \subset M_ n = M$ with $M_ i/M_{i-1} \cong R/\mathfrak p_ i$ the minimal primes of the set $\{ \mathfrak p_ i\} $.

Proof. Choose a filtration as in (3). In Lemma 10.61.5 we have seen that the sets in (1) and (3) are equal.

Let $\mathfrak p$ be a minimal element of the set $\{ \mathfrak p_ i\} $. Let $i$ be minimal such that $\mathfrak p = \mathfrak p_ i$. Pick $m \in M_ i$, $m \not\in M_{i-1}$. The annihilator of $m$ is contained in $\mathfrak p_ i = \mathfrak p$ and contains $\mathfrak p_1 \mathfrak p_2 \ldots \mathfrak p_ i$. By our choice of $i$ and $\mathfrak p$ we have $\mathfrak p_ j \not\subset \mathfrak p$ for $j < i$ and hence we have $\mathfrak p_1 \mathfrak p_2 \ldots \mathfrak p_{i - 1} \not\subset \mathfrak p_ i$. Pick $f \in \mathfrak p_1 \mathfrak p_2 \ldots \mathfrak p_{i - 1}$, $f \not\in \mathfrak p$. Then $fm$ has annihilator $\mathfrak p$. In this way we see that $\mathfrak p$ is an associated prime of $M$. By Lemma 10.62.2 we have $\text{Ass}(M) \subset \text{Supp}(M)$ and hence $\mathfrak p$ is minimal in $\text{Ass}(M)$. Thus the set of primes in (1) is contained in the set of primes of (2).

Let $\mathfrak p$ be a minimal element of $\text{Ass}(M)$. Since $\text{Ass}(M) \subset \text{Supp}(M)$ there is a minimal element $\mathfrak q$ of $\text{Supp}(M)$ with $\mathfrak q \subset \mathfrak p$. We have just shown that $\mathfrak q \in \text{Ass}(M)$. Hence $\mathfrak q = \mathfrak p$ by minimality of $\mathfrak p$. Thus the set of primes in (2) is contained in the set of primes of (1). $\square$

slogan

Lemma 10.62.7. Let $R$ be a Noetherian ring. Let $M$ be an $R$-module. Then

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

Proof. If $M = (0)$, then $\text{Ass}(M) = \emptyset $ by definition. If $M \not= 0$, pick any nonzero finitely generated submodule $M' \subset M$, for example a submodule generated by a single nonzero element. By Lemma 10.39.2 we see that $\text{Supp}(M')$ is nonempty. By Proposition 10.62.6 this implies that $\text{Ass}(M')$ is nonempty. By Lemma 10.62.3 this implies $\text{Ass}(M) \not= \emptyset $. $\square$

Lemma 10.62.8. Let $R$ be a Noetherian ring. Let $M$ be an $R$-module. Any $\mathfrak p \in \text{Supp}(M)$ which is minimal among the elements of $\text{Supp}(M)$ is an element of $\text{Ass}(M)$.

Proof. If $M$ is a finite $R$-module, then this is a consequence of Proposition 10.62.6. In general write $M = \bigcup M_\lambda $ as the union of its finite submodules, and use that $\text{Supp}(M) = \bigcup \text{Supp}(M_\lambda )$ and $\text{Ass}(M) = \bigcup \text{Ass}(M_\lambda )$. $\square$

Lemma 10.62.9. Let $R$ be a Noetherian ring. Let $M$ be an $R$-module. The union $\bigcup _{\mathfrak q \in \text{Ass}(M)} \mathfrak q$ is the set of elements of $R$ which are zerodivisors on $M$.

Proof. Any element in any associated prime clearly is a zerodivisor on $M$. Conversely, suppose $x \in R$ is a zerodivisor on $M$. Consider the submodule $N = \{ m \in M \mid xm = 0\} $. Since $N$ is not zero it has an associated prime $\mathfrak q$ by Lemma 10.62.7. Then $x \in \mathfrak q$ and $\mathfrak q$ is an associated prime of $M$ by Lemma 10.62.3. $\square$

Lemma 10.62.10. Let $R$ is a Noetherian local ring, $M$ a finite $R$-module, and $f \in \mathfrak m$ an element of the maximal ideal of $R$. Then

\[ \dim (\text{Supp}(M/fM)) \leq \dim (\text{Supp}(M)) \leq \dim (\text{Supp}(M/fM)) + 1 \]

If $f$ is not in any of the minimal primes of the support of $M$ (for example if $f$ is a nonzerodivisor on $M$), then equality holds for the right inequality.

Proof. (The parenthetical statement follows from Lemma 10.62.9.) The first inequality follows from $\text{Supp}(M/fM) \subset \text{Supp}(M)$, see Lemma 10.39.9. For the second inequality, note that $\text{Supp}(M/fM) = \text{Supp}(M) \cap V(f)$, see Lemma 10.39.9. It follows, for example by Lemma 10.61.2 and elementary properties of dimension, that it suffices to show $\dim V(\mathfrak p) \leq \dim (V(\mathfrak p) \cap V(f)) + 1$ for primes $\mathfrak p$ of $R$. This is a consequence of Lemma 10.59.12. Finally, if $f$ is not contained in any minimal prime of the support of $M$, then the chains of primes in $\text{Supp}(M/fM)$ all give rise to chains in $\text{Supp}(M)$ which are at least one step away from being maximal. $\square$

Lemma 10.62.11. Let $\varphi : R \to S$ be a ring map. Let $M$ be an $S$-module. Then $\mathop{\mathrm{Spec}}(\varphi )(\text{Ass}_ S(M)) \subset \text{Ass}_ R(M)$.

Proof. If $\mathfrak q \in \text{Ass}_ S(M)$, then there exists an $m$ in $M$ such that the annihilator of $m$ in $S$ is $\mathfrak q$. Then the annihilator of $m$ in $R$ is $\mathfrak q \cap R$. $\square$

Remark 10.62.12. Let $\varphi : R \to S$ be a ring map. Let $M$ be an $S$-module. Then it is not always the case that $\mathop{\mathrm{Spec}}(\varphi )(\text{Ass}_ S(M)) \supset \text{Ass}_ R(M)$. For example, consider the ring map $R = k \to S = k[x_1, x_2, x_3, \ldots ]/(x_ i^2)$ and $M = S$. Then $\text{Ass}_ R(M)$ is not empty, but $\text{Ass}_ S(S)$ is empty.

Lemma 10.62.13. Let $\varphi : R \to S$ be a ring map. Let $M$ be an $S$-module. If $S$ is Noetherian, then $\mathop{\mathrm{Spec}}(\varphi )(\text{Ass}_ S(M)) = \text{Ass}_ R(M)$.

Proof. We have already seen in Lemma 10.62.11 that $\mathop{\mathrm{Spec}}(\varphi )(\text{Ass}_ S(M)) \subset \text{Ass}_ R(M)$. For the converse, choose a prime $\mathfrak p \in \text{Ass}_ R(M)$. Let $m \in M$ be an element such that the annihilator of $m$ in $R$ is $\mathfrak p$. Let $I = \{ g \in S \mid gm = 0\} $ be the annihilator of $m$ in $S$. Then $R/\mathfrak p \subset S/I$ is injective. Combining Lemmas 10.29.5 and 10.29.7 we see that there is a prime $\mathfrak q \subset S$ minimal over $I$ mapping to $\mathfrak p$. By Proposition 10.62.6 we see that $\mathfrak q$ is an associated prime of $S/I$, hence $\mathfrak q$ is an associated prime of $M$ by Lemma 10.62.3 and we win. $\square$

Lemma 10.62.14. Let $R$ be a ring. Let $I$ be an ideal. Let $M$ be an $R/I$-module. Via the canonical injection $\mathop{\mathrm{Spec}}(R/I) \to \mathop{\mathrm{Spec}}(R)$ we have $\text{Ass}_{R/I}(M) = \text{Ass}_ R(M)$.

Proof. Omitted. $\square$

Lemma 10.62.15. Let $R$ be a ring. Let $M$ be an $R$-module. Let $\mathfrak p \subset R$ be a prime.

  1. If $\mathfrak p \in \text{Ass}(M)$ then $\mathfrak pR_{\mathfrak p} \in \text{Ass}(M_{\mathfrak p})$.

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

Proof. If $\mathfrak p \in \text{Ass}(M)$ there exists an element $m \in M$ whose annihilator is $\mathfrak p$. As localization is exact (Proposition 10.9.12) we see that the annihilator of $m/1$ in $M_{\mathfrak p}$ is $\mathfrak pR_{\mathfrak p}$ hence (1) holds. Assume $\mathfrak pR_{\mathfrak p} \in \text{Ass}(M_{\mathfrak p})$ and $\mathfrak p = (f_1, \ldots , f_ n)$. Let $m/g$ be an element of $M_{\mathfrak p}$ whose annihilator is $\mathfrak pR_{\mathfrak p}$. This implies that the annihilator of $m$ is contained in $\mathfrak p$. As $f_ i m/g = 0$ in $M_{\mathfrak p}$ we see there exists a $g_ i \in R$, $g_ i \not\in \mathfrak p$ such that $g_ i f_ i m = 0$ in $M$. Combined we see the annihilator of $g_1\ldots g_ nm$ is $\mathfrak p$. Hence $\mathfrak p \in \text{Ass}(M)$. $\square$

Lemma 10.62.16. Let $R$ be a ring. Let $M$ be an $R$-module. Let $S \subset R$ be a multiplicative subset. Via the canonical injection $\mathop{\mathrm{Spec}}(S^{-1}R) \to \mathop{\mathrm{Spec}}(R)$ we have

  1. $\text{Ass}_ R(S^{-1}M) = \text{Ass}_{S^{-1}R}(S^{-1}M)$,

  2. $\text{Ass}_ R(M) \cap \mathop{\mathrm{Spec}}(S^{-1}R) \subset \text{Ass}_ R(S^{-1}M)$, and

  3. if $R$ is Noetherian this inclusion is an equality.

Proof. The first equality follows, since if $m \in S^{-1}M$, then the annihilator of $m$ in $R$ is the intersection of the annihilator of $m$ in $S^{-1}R$ with $R$. The displayed inclusion and equality in the Noetherian case follows from Lemma 10.62.15 since for $\mathfrak p \in R$, $S \cap \mathfrak p = \emptyset $ we have $M_{\mathfrak p} = (S^{-1}M)_{S^{-1}\mathfrak p}$. $\square$

Lemma 10.62.17. Let $R$ be a ring. Let $M$ be an $R$-module. Let $S \subset R$ be a multiplicative subset. Assume that every $s \in S$ is a nonzerodivisor on $M$. Then

\[ \text{Ass}_ R(M) = \text{Ass}_ R(S^{-1}M). \]

Proof. As $M \subset S^{-1}M$ by assumption we get the inclusion $\text{Ass}(M) = \text{Ass}(S^{-1}M)$ from Lemma 10.62.3. Conversely, suppose that $n/s \in S^{-1}M$ is an element whose annihilator is a prime ideal $\mathfrak p$. Then the annihilator of $n \in M$ is also $\mathfrak p$. $\square$

Lemma 10.62.18. Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak m$. Let $I \subset \mathfrak m$ be an ideal. Let $M$ be a finite $R$-module. The following are equivalent:

  1. There exists an $x \in I$ which is not a zerodivisor on $M$.

  2. We have $I \not\subset \mathfrak q$ for all $\mathfrak q \in \text{Ass}(M)$.

Proof. If there exists a nonzerodivisor $x$ in $I$, then $x$ clearly cannot be in any associated prime of $M$. Conversely, suppose $I \not\subset \mathfrak q$ for all $\mathfrak q \in \text{Ass}(M)$. In this case we can choose $x \in I$, $x \not\in \mathfrak q$ for all $\mathfrak q \in \text{Ass}(M)$ by Lemmas 10.62.5 and 10.14.2. By Lemma 10.62.9 the element $x$ is not a zerodivisor on $M$. $\square$

Lemma 10.62.19. Let $R$ be a ring. Let $M$ be an $R$-module. If $R$ is Noetherian the map

\[ M \longrightarrow \prod \nolimits _{\mathfrak p \in \text{Ass}(M)} M_{\mathfrak p} \]

is injective.

Proof. Let $x \in M$ be an element of the kernel of the map. Then if $\mathfrak p$ is an associated prime of $Rx \subset M$ we see on the one hand that $\mathfrak p \in \text{Ass}(M)$ (Lemma 10.62.3) and on the other hand that $(Rx)_{\mathfrak p} \subset M_{\mathfrak p}$ is not zero. This contradiction shows that $\text{Ass}(Rx) = \emptyset $. Hence $Rx = 0$ by Lemma 10.62.7. $\square$


Comments (4)

Comment #681 by Keenan Kidwell on

I have two comments about 02CE. First, and this is just notational, when it is stated that the product is not contained in , I take it that what is being used is that this is equivalent to none of the in the product being in , and this is supposed to hold because is minimal among primes showing up in the filtration quotients. But what if one of the for is ? I think what one needs to take is . Then the proof goes through. The second comment, which is maybe more serious, is that the argument shows that a minimal element of the is an associated prime, and since associated primes are among the , such a thing is necessarily a minimal associated. But if we start with a minimal associated prime, we know that it shows up in the filtration (and in the support), but how do we know it is minimal among these potentially larger sets? If with minimal among associated primes, is necessarily also associated?

Comment #1152 by Yu Zhao on

I think there is a typo in the proof of Lemma 10.62.12. In 3rd line of the proof, should " in is " be " in is "?


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