The Stacks project

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. $\square$

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 $x$ is contained in $\mathfrak m$. Thus we see that $\mathfrak m = 0$, i.e., $R$ is a field. By Lemma 10.66.2 this implies the statement of the lemma. $\square$

Proof. We will use the characterization of weakly associated primes of Lemma 10.66.2. 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 radical of 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. $\square$

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 10.66.4. But as $I \not= R$ we have $V(I) = \mathop{\mathrm{Spec}}(R/I)$ contains a minimal prime, see Lemmas 10.17.2 and 10.17.7, and we win. $\square$

Proof. The first inclusion is immediate from the definitions. If $\mathfrak p \in \text{WeakAss}(M)$, then by Lemma 10.66.2 we have $M_{\mathfrak p} \not= 0$, hence $\mathfrak p \in \text{Supp}(M)$. $\square$

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 10.66.5. Clearly $f \in \mathfrak q$ and by Lemma 10.66.4 $\mathfrak q$ is a weakly associated prime of $M$. $\square$

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 10.66.5. Since $\text{WeakAss}(M_{\mathfrak p}) \subset \text{Supp}(M_{\mathfrak p})$ by Lemma 10.66.6 we conclude that $\text{WeakAss}(M_{\mathfrak p}) = \{ \mathfrak pR_{\mathfrak p}\} $, whence $\mathfrak p \in \text{WeakAss}(M)$ by Lemma 10.66.2. $\square$

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 10.66.6. Assume $\mathfrak p \in \text{WeakAss}(M)$. By Lemma 10.66.2 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 10.63.15 we see that $\mathfrak p \in \text{Ass}(M)$. $\square$

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 10.66.2. $\square$

Proof. One of the inclusions has already been proved, see Remark 10.66.12. 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 10.36.21. Pick $x \in \mathfrak q$ with $x \not\in \mathfrak q_ i$ for $i > 1$, see Lemma 10.15.2. 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 10.36.22. If $f \in \mathfrak pR_{\mathfrak p}$ then some power of $f$ ends up in $JS_{\mathfrak q}$ hence for some $t > 0$ 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 10.23.1. In this way we see that $\mathfrak p$ is a minimal prime over the annihilator of $ym$ in $R$ and we win. $\square$

Proof. Special case of Lemma 10.66.13. $\square$

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 10.66.2 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 obtain $\text{WeakAss}(M) \subset \text{WeakAss}(S^{-1}M)$ from Lemma 10.66.4. 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 $I$. $\square$

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 10.66.4) 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 10.66.5. $\square$

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 10.66.16. $\square$

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 10.40.4). Thus $\mathfrak q \cap (R \otimes _ k L)$ is minimal over $I$ by going down (Lemma 10.39.19). 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 10.66.13. 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 10.66.4 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 10.66.15. 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 10.20.1 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. $\square$