
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). $\square$

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. $\square$

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 $\mathfrak{p}_1\cdots\mathfrak{p}_{i-1}$ is not contained in $\mathfrak{p}_i$, I take it that what is being used is that this is equivalent to none of the $\mathfrak{p}_j$ in the product being in $\mathfrak{p}_i$, and this is supposed to hold because $\mathfrak{p}_i$ is minimal among primes showing up in the filtration quotients. But what if one of the $\mathfrak{p}_j$ for $1\leq j\leq i-1$ is $\mathfrak{p}_i$? I think what one needs to take is $\prod_{1\leq j\leq i-1,\mathfrak{p}_j\neq\mathfrak{p}_i}\mathfrak{p}_j$. Then the proof goes through. The second comment, which is maybe more serious, is that the argument shows that a minimal element of the $\mathfrak{p}_i$ is an associated prime, and since associated primes are among the $\mathfrak{p}-i$, 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 $\mathfrak{p}_j\subseteq\mathfrak{p}_i=\Ann_R(m)$ with $\mathfrak{p}_i$ minimal among associated primes, is $\mathfrak{p}_j$ 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 "$x$ in $R$ is $\mathfrak p$" be "$m$ in $R$ is $\mathfrak p$"?

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