The Stacks project

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., $\mathfrak p \in \text{Supp}(M)$. $\square$

Proof. If $m' \in M'$, then the annihilator of $m'$ viewed as an element of $M'$ is the same as the annihilator of $m'$ viewed as an element of $M$. Hence the inclusion $\text{Ass}(M') \subset \text{Ass}(M)$. Let $m \in M$ be an element whose annihilator is a prime ideal $\mathfrak p$. If there exists a $g \in R$, $g \not\in \mathfrak p$ such that $m' = gm \in M'$ then the annihilator of $m'$ is $\mathfrak p$. If there does not exist a $g \in R$, $g \not\in \mathfrak p$ such that $gm \in M'$, then the annilator of the image $m'' \in M''$ of $m$ is $\mathfrak p$. This proves the inclusion $\text{Ass}(M) \subset \text{Ass}(M') \cup \text{Ass}(M'')$. We omit the proof of the final statement. $\square$

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. $\square$

Proof. Immediate from Lemma 10.63.4 and Lemma 10.62.1. $\square$

Proof. Choose a filtration as in (3). In Lemma 10.62.5 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 10.63.2 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). $\square$

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 10.40.2 we see that $\text{Supp}(M')$ is nonempty. By Proposition 10.63.6 this implies that $\text{Ass}(M')$ is nonempty. By Lemma 10.63.3 this implies $\text{Ass}(M) \not= \emptyset $. $\square$

Proof. If $M$ is a finite $R$-module, then this is a consequence of Proposition 10.63.6. 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 $\text{Ass}(M) = \bigcup \text{Ass}(M_\lambda )$. $\square$

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 10.63.7. Then $x \in \mathfrak q$ and $\mathfrak q$ is an associated prime of $M$ by Lemma 10.63.3. $\square$

Proof. (The parenthetical statement follows from Lemma 10.63.9.) The first inequality follows from $\text{Supp}(M/fM) \subset \text{Supp}(M)$, see Lemma 10.40.9. For the second inequality, note that $\text{Supp}(M/fM) = \text{Supp}(M) \cap V(f)$, see Lemma 10.40.9. It follows, for example by Lemma 10.62.2 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 10.60.13. 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. $\square$

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 $\mathfrak q \cap R$. $\square$

Proof. We have already seen in Lemma 10.63.11 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 10.30.5 and 10.30.7 we see that there is a prime $\mathfrak q \subset S$ minimal over $I$ mapping to $\mathfrak p$. By Proposition 10.63.6 we see that $\mathfrak q$ is an associated prime of $S/I$, hence $\mathfrak q$ is an associated prime of $M$ by Lemma 10.63.3 and we win. $\square$

Proof. Omitted. $\square$

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 $\mathfrak p \in \text{Ass}(M)$. $\square$

Proof. For $m \in S^{-1}M$, let $I \subset R$ and $J \subset S^{-1}R$ be the annihilators of $m$. Then $I$ is the inverse image of $J$ by the map $R \to S^{-1}R$ and $J = S^{-1}I$. The equality in (1) follows by the description of the map $\mathop{\mathrm{Spec}}(S^{-1}R) \to \mathop{\mathrm{Spec}}(R)$ in Lemma 10.17.5. For $m \in M$, let $I \subset R$ be the annihilator of $m$ in $R$ and let $J \subset S^{-1}R$ be the annihilator of $m/1 \in S^{-1}M$. We have $J = S^{-1}I$ which implies (2). The equality in the Noetherian case follows from Lemma 10.63.15 since for $\mathfrak p \in R$, $S \cap \mathfrak p = \emptyset $ we have $M_{\mathfrak p} = (S^{-1}M)_{S^{-1}\mathfrak p}$. $\square$

Proof. As $M \subset S^{-1}M$ by assumption we get the inclusion $\text{Ass}(M) \subset \text{Ass}(S^{-1}M)$ from Lemma 10.63.3. 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 $\mathfrak p$. $\square$

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 10.63.5 and 10.15.2. By Lemma 10.63.9 the element $x$ is not a zerodivisor on $M$. $\square$

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 10.63.3) 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 10.63.7. $\square$

Proof. By Lemma 10.63.5 the ring $S$ has finitely many associated prime ideals. By Lemma 10.61.3 the ring $S$ has infinitely many maximal ideals. Hence we can choose a maximal ideal $\mathfrak m \subset S$ which is not an associated prime of $S$. By prime avoidance (Lemma 10.15.2), we can choose a nonzero $f \in \mathfrak m$ which is not contained in any of the associated primes of $S$. By Lemma 10.63.9 the element $f$ is a nonzerodivisor and as $f \in \mathfrak m$ we see that $f$ is not a unit. $\square$