Definition 10.65.1. Let $R$ be a ring. Let $M$ be an $R$-module. A prime $\mathfrak p$ of $R$ is *weakly associated* to $M$ if there exists an element $m \in M$ such that $\mathfrak p$ is minimal among the prime ideals containing the annihilator $\text{Ann}(m) = \{ f \in R \mid fm = 0\} $. The set of all such primes is denoted $\text{WeakAss}_ R(M)$ or $\text{WeakAss}(M)$.

## 10.65 Weakly associated primes

This is a variant on the notion of an associated prime that is useful for non-Noetherian ring and non-finite modules.

Thus an associated prime is a weakly associated prime. Here is a characterization in terms of the localization at the prime.

Lemma 10.65.2. Let $R$ be a ring. Let $M$ be an $R$-module. Let $\mathfrak p$ be a prime of $R$. The following are equivalent:

$\mathfrak p$ is weakly associated to $M$,

$\mathfrak pR_{\mathfrak p}$ is weakly associated to $M_{\mathfrak p}$, and

$M_{\mathfrak p}$ contains an element whose annihilator has radical equal to $\mathfrak pR_{\mathfrak p}$.

**Proof.**
Assume (1). Then there exists an element $m \in M$ such that $\mathfrak p$ is minimal among the primes containing the annihilator $I = \{ x \in R \mid xm = 0\} $ of $m$. As localization is exact, the annihilator of $m$ in $M_{\mathfrak p}$ is $I_{\mathfrak p}$. Hence $\mathfrak pR_{\mathfrak p}$ is a minimal prime of $R_{\mathfrak p}$ containing the annihilator $I_{\mathfrak p}$ of $m$ in $M_{\mathfrak p}$. This implies (2) holds, and also (3) as it implies that $\sqrt{I_{\mathfrak p}} = \mathfrak pR_{\mathfrak p}$.

Applying the implication (1) $\Rightarrow $ (3) to $M_{\mathfrak p}$ over $R_{\mathfrak p}$ we see that (2) $\Rightarrow $ (3).

Finally, assume (3). This means there exists an element $m/f \in M_{\mathfrak p}$ whose annihilator has radical equal to $\mathfrak pR_{\mathfrak p}$. Then the annihilator $I = \{ x \in R \mid xm = 0\} $ of $m$ in $M$ is such that $\sqrt{I_{\mathfrak p}} = \mathfrak pR_{\mathfrak p}$. Clearly this means that $\mathfrak p$ contains $I$ and is minimal among the primes containing $I$, i.e., (1) holds. $\square$

Lemma 10.65.3. For a reduced ring the weakly associated primes of the ring are the minimal primes.

**Proof.**
Let $(R, \mathfrak m)$ be a reduced local ring. Suppose $x \in R$ is an element whose annihilator has radical $\mathfrak m$. If $\mathfrak m \not= 0$, then $x$ cannot be a unit, so $x \in \mathfrak m$. Then in particular $x^{1 + n} = 0$ for some $n \geq 0$. Hence $x = 0$. Which contradicts the assumption that the annihilator of $\mathfrak m$ is contained in $\mathfrak m$. Thus we see that $\mathfrak m = 0$, i.e., $R$ is a field. By Lemma 10.65.2 this implies the statement of the lemma.
$\square$

Lemma 10.65.4. Let $R$ be a ring. Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence of $R$-modules. Then $\text{WeakAss}(M') \subset \text{WeakAss}(M)$ and $\text{WeakAss}(M) \subset \text{WeakAss}(M') \cup \text{WeakAss}(M'')$.

**Proof.**
We will use the characterization of weakly associated primes of Lemma 10.65.2. Let $\mathfrak p$ be a prime of $R$. As localization is exact we obtain the short exact sequence $0 \to M'_{\mathfrak p} \to M_{\mathfrak p} \to M''_{\mathfrak p} \to 0$. Suppose that $m \in M_{\mathfrak p}$ is an element whose annihilator has radical $\mathfrak pR_{\mathfrak p}$. Then either the image $\overline{m}$ of $m$ in $M''_{\mathfrak p}$ is zero and $m \in M'_{\mathfrak p}$, or the annihilator of $\overline{m}$ is $\mathfrak pR_{\mathfrak p}$. This proves that $\text{WeakAss}(M) \subset \text{WeakAss}(M') \cup \text{WeakAss}(M'')$. The inclusion $\text{WeakAss}(M') \subset \text{WeakAss}(M)$ is immediate from the definitions.
$\square$

Lemma 10.65.5. Let $R$ be a ring. Let $M$ be an $R$-module. Then

**Proof.**
If $M = (0)$ then $\text{WeakAss}(M) = \emptyset $ by definition. Conversely, suppose that $M \not= 0$. Pick a nonzero element $m \in M$. Write $I = \{ x \in R \mid xm = 0\} $ the annihilator of $m$. Then $R/I \subset M$. Hence $\text{WeakAss}(R/I) \subset \text{WeakAss}(M)$ by Lemma 10.65.4. But as $I \not= R$ we have $V(I) = \mathop{\mathrm{Spec}}(R/I)$ contains a minimal prime, see Lemmas 10.16.2 and 10.16.7, and we win.
$\square$

Lemma 10.65.6. Let $R$ be a ring. Let $M$ be an $R$-module. Then

**Proof.**
The first inclusion is immediate from the definitions. If $\mathfrak p \in \text{WeakAss}(M)$, then by Lemma 10.65.2 we have $M_{\mathfrak p} \not= 0$, hence $\mathfrak p \in \text{Supp}(M)$.
$\square$

Lemma 10.65.7. Let $R$ be a ring. Let $M$ be an $R$-module. The union $\bigcup _{\mathfrak q \in \text{WeakAss}(M)} \mathfrak q$ is the set elements of $R$ which are zerodivisors on $M$.

**Proof.**
Suppose $f \in \mathfrak q \in \text{WeakAss}(M)$. Then there exists an element $m \in M$ such that $\mathfrak q$ is minimal over $I = \{ x \in R \mid xm = 0\} $. Hence there exists a $g \in R$, $g \not\in \mathfrak q$ and $n > 0$ such that $f^ ngm = 0$. Note that $gm \not= 0$ as $g \not\in I$. If we take $n$ minimal as above, then $f (f^{n - 1}gm) = 0$ and $f^{n - 1}gm \not= 0$, so $f$ is a zerodivisor on $M$. Conversely, suppose $f \in R$ is a zerodivisor on $M$. Consider the submodule $N = \{ m \in M \mid fm = 0\} $. Since $N$ is not zero it has a weakly associated prime $\mathfrak q$ by Lemma 10.65.5. Clearly $f \in \mathfrak q$ and by Lemma 10.65.4 $\mathfrak q$ is a weakly associated prime of $M$.
$\square$

Lemma 10.65.8. Let $R$ be a 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 $\text{WeakAss}(M)$.

**Proof.**
Note that $\text{Supp}(M_{\mathfrak p}) = \{ \mathfrak pR_{\mathfrak p}\} $ in $\mathop{\mathrm{Spec}}(R_{\mathfrak p})$. In particular $M_{\mathfrak p}$ is nonzero, and hence $\text{WeakAss}(M_{\mathfrak p}) \not= \emptyset $ by Lemma 10.65.5. Since $\text{WeakAss}(M_{\mathfrak p}) \subset \text{Supp}(M_{\mathfrak p})$ by Lemma 10.65.6 we conclude that $\text{WeakAss}(M_{\mathfrak p}) = \{ \mathfrak pR_{\mathfrak p}\} $, whence $\mathfrak p \in \text{WeakAss}(M)$ by Lemma 10.65.2.
$\square$

Lemma 10.65.9. Let $R$ be a ring. Let $M$ be an $R$-module. Let $\mathfrak p$ be a prime ideal of $R$ which is finitely generated. Then

In particular, if $R$ is Noetherian, then $\text{Ass}(M) = \text{WeakAss}(M)$.

**Proof.**
Write $\mathfrak p = (g_1, \ldots , g_ n)$ for some $g_ i \in R$. It is enough the prove the implication “$\Leftarrow $” as the other implication holds in general, see Lemma 10.65.6. Assume $\mathfrak p \in \text{WeakAss}(M)$. By Lemma 10.65.2 there exists an element $m \in M_{\mathfrak p}$ such that $I = \{ x \in R_{\mathfrak p} \mid xm = 0\} $ has radical $\mathfrak pR_{\mathfrak p}$. Hence for each $i$ there exists a smallest $e_ i > 0$ such that $g_ i^{e_ i}m = 0$ in $M_{\mathfrak p}$. If $e_ i > 1$ for some $i$, then we can replace $m$ by $g_ i^{e_ i - 1} m \not= 0$ and decrease $\sum e_ i$. Hence we may assume that the annihilator of $m \in M_{\mathfrak p}$ is $(g_1, \ldots , g_ n)R_{\mathfrak p} = \mathfrak p R_{\mathfrak p}$. By Lemma 10.62.15 we see that $\mathfrak p \in \text{Ass}(M)$.
$\square$

Remark 10.65.10. Let $\varphi : R \to S$ be a ring map. Let $M$ be an $S$-module. Then it is not always the case that $\mathop{\mathrm{Spec}}(\varphi )(\text{WeakAss}_ S(M)) \subset \text{WeakAss}_ R(M)$ contrary to the case of associated primes (see Lemma 10.62.11). An example is to consider the ring map

and $M = S$. In this case $\mathfrak q = \sum x_ iS$ is a minimal prime of $S$, hence a weakly associated prime of $M = S$ (see Lemma 10.65.8). But on the other hand, for any nonzero element of $S$ the annihilator in $R$ is finitely generated, and hence does not have radical equal to $R \cap \mathfrak q = (x_1, x_2, x_3, \ldots )$ (details omitted).

Lemma 10.65.11. Let $\varphi : R \to S$ be a ring map. Let $M$ be an $S$-module. Then we have $\mathop{\mathrm{Spec}}(\varphi )(\text{WeakAss}_ S(M)) \supset \text{WeakAss}_ R(M)$.

**Proof.**
Let $\mathfrak p$ be an element of $\text{WeakAss}_ R(M)$. Then there exists an $m \in M_{\mathfrak p}$ whose annihilator $I = \{ x \in R_{\mathfrak p} \mid xm = 0\} $ has radical $\mathfrak pR_{\mathfrak p}$. Consider the annihilator $J = \{ x \in S_{\mathfrak p} \mid xm = 0 \} $ of $m$ in $S_{\mathfrak p}$. As $IS_{\mathfrak p} \subset J$ we see that any minimal prime $\mathfrak q \subset S_{\mathfrak p}$ over $J$ lies over $\mathfrak p$. Moreover such a $\mathfrak q$ corresponds to a weakly associated prime of $M$ for example by Lemma 10.65.2.
$\square$

Remark 10.65.12. Let $\varphi : R \to S$ be a ring map. Let $M$ be an $S$-module. Denote $f : \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ the associated map on spectra. Then we have

see Lemmas 10.62.11, 10.65.11, and 10.65.6. In general all of the inclusions may be strict, see Remarks 10.62.12 and 10.65.10. If $S$ is Noetherian, then all the inclusions are equalities as the outer two are equal by Lemma 10.65.9.

Lemma 10.65.13. Let $\varphi : R \to S$ be a ring map. Let $M$ be an $S$-module. Denote $f : \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ the associated map on spectra. If $\varphi $ is a finite ring map, then

**Proof.**
One of the inclusions has already been proved, see Remark 10.65.12. To prove the other assume $\mathfrak q \in \text{WeakAss}_ S(M)$ and let $\mathfrak p$ be the corresponding prime of $R$. Let $m \in M$ be an element such that $\mathfrak q$ is a minimal prime over $J = \{ g \in S \mid gm = 0\} $. Thus the radical of $JS_{\mathfrak q}$ is $\mathfrak qS_{\mathfrak q}$. As $R \to S$ is finite there are finitely many primes $\mathfrak q = \mathfrak q_1, \mathfrak q_2, \ldots , \mathfrak q_ l$ over $\mathfrak p$, see Lemma 10.35.21. Pick $x \in \mathfrak q$ with $x \not\in \mathfrak q_ i$ for $i > 1$, see Lemma 10.14.2. By the above there exists an element $y \in S$, $y \not\in \mathfrak q$ and an integer $t > 0$ such that $y x^ t m = 0$. Thus the element $ym \in M$ is annihilated by $x^ t$, hence $ym$ maps to zero in $M_{\mathfrak q_ i}$, $i = 2, \ldots , l$. To be sure, $ym$ does not map to zero in $S_{\mathfrak q}$.

The ring $S_{\mathfrak p}$ is semi-local with maximal ideals $\mathfrak q_ i S_{\mathfrak p}$ by going up for finite ring maps, see Lemma 10.35.22. If $f \in \mathfrak pR_{\mathfrak p}$ then some power of $f$ ends up in $JS_{\mathfrak q}$ hence for some $n > 0$ we see that $f^ t ym$ maps to zero in $M_{\mathfrak q}$. As $ym$ vanishes at the other maximal ideals of $S_{\mathfrak p}$ we conclude that $f^ t ym$ is zero in $M_{\mathfrak p}$, see Lemma 10.22.1. In this way we see that $\mathfrak p$ is a minimal prime over the annihilator of $ym$ in $R$ and we win. $\square$

Lemma 10.65.14. 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 $\text{WeakAss}_{R/I}(M) = \text{WeakAss}_ R(M)$.

**Proof.**
Omitted.
$\square$

Lemma 10.65.15. 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 $\text{WeakAss}_ R(S^{-1}M) = \text{WeakAss}_{S^{-1}R}(S^{-1}M)$ and

**Proof.**
Suppose that $m \in S^{-1}M$. Let $I = \{ x \in R \mid xm = 0\} $ and $I' = \{ x' \in S^{-1}R \mid x'm = 0\} $. Then $I' = S^{-1}I$ and $I \cap S = \emptyset $ unless $I = R$ (verifications omitted). Thus primes in $S^{-1}R$ minimal over $I'$ correspond bijectively to primes in $R$ minimal over $I$ and avoiding $S$. This proves the equality $\text{WeakAss}_ R(S^{-1}M) = \text{WeakAss}_{S^{-1}R}(S^{-1}M)$. The second equality 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 10.65.16. 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

**Proof.**
As $M \subset S^{-1}M$ by assumption we obtain $\text{WeakAss}(M) \subset \text{WeakAss}(S^{-1}M)$ from Lemma 10.65.4. Conversely, suppose that $n/s \in S^{-1}M$ is an element with annihilator $I$ and $\mathfrak p$ a prime which is minimal over $I$. Then the annihilator of $n \in M$ is $I$ and $\mathfrak p$ is a prime minimal over $I$.
$\square$

Lemma 10.65.17. Let $R$ be a ring. Let $M$ be an $R$-module. The map

is injective.

**Proof.**
Let $x \in M$ be an element of the kernel of the map. Set $N = Rx \subset M$. If $\mathfrak p$ is a weakly associated prime of $N$ we see on the one hand that $\mathfrak p \in \text{WeakAss}(M)$ (Lemma 10.65.4) and on the other hand that $N_{\mathfrak p} \subset M_{\mathfrak p}$ is not zero. This contradiction shows that $\text{WeakAss}(N) = \emptyset $. Hence $N = 0$, i.e., $x = 0$ by Lemma 10.65.5.
$\square$

Lemma 10.65.18. Let $R \to S$ be a ring map. Let $N$ be an $S$-module. Assume $N$ is flat as an $R$-module and $R$ is a domain with fraction field $K$. Then

via the canonical inclusion $\mathop{\mathrm{Spec}}(S \otimes _ R K) \subset \mathop{\mathrm{Spec}}(S)$.

**Proof.**
Note that $S \otimes _ R K = (R \setminus \{ 0\} )^{-1}S$ and $N \otimes _ R K = (R \setminus \{ 0\} )^{-1}N$. For any nonzero $x \in R$ multiplication by $x$ on $N$ is injective as $N$ is flat over $R$. Hence the lemma follows from Lemma 10.65.16.
$\square$

Lemma 10.65.19. Let $K/k$ be a field extension. Let $R$ be a $k$-algebra. Let $M$ be an $R$-module. Let $\mathfrak q \subset R \otimes _ k K$ be a prime lying over $\mathfrak p \subset R$. If $\mathfrak q$ is weakly associated to $M \otimes _ k K$, then $\mathfrak p$ is weakly associated to $M$.

**Proof.**
Let $z \in M \otimes _ k K$ be an element such that $\mathfrak q$ is minimal over the annihilator $J \subset R \otimes _ k K$ of $z$. Choose a finitely generated subextension $K/L/k$ such that $z \in M \otimes _ k L$. Since $R \otimes _ k L \to R \otimes _ k K$ is flat we see that $J = I(R \otimes _ k K)$ where $I \subset R \otimes _ k L$ is the annihilator of $z$ in the smaller ring (Lemma 10.39.4). Thus $\mathfrak q \cap (R \otimes _ k L)$ is minimal over $I$ by going down (Lemma 10.38.18). In this way we reduce to the case described in the next paragraph.

Assume $K/k$ is a finitely generated field extension. Let $x_1, \ldots , x_ r \in K$ be a transcendence basis of $K$ over $k$, see Fields, Section 9.26. Set $L = k(x_1, \ldots , x_ r)$. Say $[K : L] = n$. Then $R \otimes _ k L \to R \otimes _ k K$ is a finite ring map. Hence $\mathfrak q \cap (R \otimes _ k L)$ is a weakly associated prime of $M \otimes _ k K$ viewed as a $R \otimes _ k L$-module by Lemma 10.65.13. Since $M \otimes _ k K \cong (M \otimes _ k L)^{\oplus n}$ as a $R \otimes _ k L$-module, we see that $\mathfrak q \cap (R \otimes _ k L)$ is a weakly associated prime of $M \otimes _ k L$ (for example by using Lemma 10.65.4 and induction). In this way we reduce to the case discussed in the next paragraph.

Assume $K = k(x_1, \ldots , x_ r)$ is a purely transcendental field extension. We may replace $R$ by $R_\mathfrak p$, $M$ by $M_\mathfrak p$ and $\mathfrak q$ by $\mathfrak q(R_\mathfrak p \otimes _ k K)$. See Lemma 10.65.15. In this way we reduce to the case discussed in the next paragraph.

Assume $K = k(x_1, \ldots , x_ r)$ is a purely transcendental field extension and $R$ is local with maximal ideal $\mathfrak p$. We claim that any $f \in R \otimes _ k K$, $f \not\in \mathfrak p(R \otimes _ k K)$ is a nonzerodivisor on $M \otimes _ k K$. Namely, let $z \in M \otimes _ k K$ be an element. There is a finite $R$-submodule $M' \subset M$ such that $z \in M' \otimes _ k K$ and such that $M'$ is minimal with this property: choose a basis $\{ t_\alpha \} $ of $K$ as a $k$-vector space, write $z = \sum m_\alpha \otimes t_\alpha $ and let $M'$ be the $R$-submodule generated by the $m_\alpha $. If $z \in \mathfrak p(M' \otimes _ k K) = \mathfrak p M' \otimes _ k K$, then $\mathfrak pM' = M'$ and $M' = 0$ by Lemma 10.19.1 a contradiction. Thus $z$ has nonzero image $\overline{z}$ in $M'/\mathfrak p M' \otimes _ k K$ But $R/\mathfrak p \otimes _ k K$ is a domain as a localization of $\kappa (\mathfrak p)[x_1, \ldots , x_ n]$ and $M'/\mathfrak p M' \otimes _ k K$ is a free module, hence $f\overline{z} \not= 0$. This proves the claim.

Finally, pick $z \in M \otimes _ k K$ such that $\mathfrak q$ is minimal over the annihilator $J \subset R \otimes _ k K$ of $z$. For $f \in \mathfrak p$ there exists an $n \geq 1$ and a $g \in R \otimes _ k K$, $g \not\in \mathfrak q$ such that $g f^ n z \in J$, i.e., $g f^ n z = 0$. (This holds because $\mathfrak q$ lies over $\mathfrak p$ and $\mathfrak q$ is minimal over $J$.) Above we have seen that $g$ is a nonzerodivisor hence $f^ n z = 0$. This means that $\mathfrak p$ is a weakly associated prime of $M \otimes _ k K$ viewed as an $R$-module. Since $M \otimes _ k K$ is a direct sum of copies of $M$ we conclude that $\mathfrak p$ is a weakly associated prime of $M$ as before. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)