The Stacks project

Proof. Some of the arguments in the proof will be repeated in the proofs of later lemmas which are more precise than this one (because they deal with a given module $M$ or a given prime $\mathfrak p$ and not with the collection of all of them).

Proof of (1). Let $\mathfrak p$ be a prime of $R$. Then we have

the first equality by Lemma 10.63.14 and the second by Lemma 10.63.16 part (1). This prove that $A = A'$. The inclusion $A_{fin} \subset A'_{fin}$ is clear.

Proof of (2). Each of the inclusions is immediate from the definitions except perhaps $A_{fin} \subset A$ which follows from Lemma 10.63.16 and the fact that we require $\mathfrak p = R \cap \mathfrak q$ in the formulation of $A_{fin}$.

Proof of (3). The equality $A = A_{fin}$ follows from Lemma 10.63.16 part (3) if $S$ is Noetherian. Let $\mathfrak q = (g_1, \ldots , g_ m)$ be a finitely generated prime ideal of $S$. Say $z \in N \otimes _ R M$ is an element whose annihilator is $\mathfrak q$. We may pick a finite submodule $M' \subset M$ such that $z$ is the image of $z' \in N \otimes _ R M'$. Then $\text{Ann}_ S(z') \subset \mathfrak q = \text{Ann}_ S(z)$. Since $N \otimes _ R -$ commutes with colimits and since $M$ is the directed colimit of finite $R$-modules we can find $M' \subset M'' \subset M$ such that the image $z'' \in N \otimes _ R M''$ is annihilated by $g_1, \ldots , g_ m$. Hence $\text{Ann}_ S(z'') = \mathfrak q$. This proves that $B = B_{fin}$ if $S$ is Noetherian.

Proof of (4). If $N$ is flat, then the functor $N \otimes _ R -$ is exact. In particular, if $M' \subset M$, then $N \otimes _ R M' \subset N \otimes _ R M$. Hence if $z \in N \otimes _ R M$ is an element whose annihilator $\mathfrak q = \text{Ann}_ S(z)$ is a prime, then we can pick any finite $R$-submodule $M' \subset M$ such that $z \in N \otimes _ R M'$ and we see that the annihilator of $z$ as an element of $N \otimes _ R M'$ is equal to $\mathfrak q$. Hence $B = B_{fin}$. Let $\mathfrak p'$ be a prime of $R$ and let $\mathfrak q$ be a prime of $S$ which is an associated prime of $N/\mathfrak p'N$. This implies that $\mathfrak p'S \subset \mathfrak q$. As $N$ is flat over $R$ we see that $N/\mathfrak p'N$ is flat over the integral domain $R/\mathfrak p'$. Hence every nonzero element of $R/\mathfrak p'$ is a nonzerodivisor on $N/\mathfrak p'$. Hence none of these elements can map to an element of $\mathfrak q$ and we conclude that $\mathfrak p' = R \cap \mathfrak q$. Hence $A_{fin} = A'_{fin}$. Finally, by Lemma 10.63.17 we see that $\text{Ass}_ S(N/\mathfrak p'N) = \text{Ass}_ S(N \otimes _ R \kappa (\mathfrak p'))$, i.e., $A'_{fin} = A$.

Proof of (5). We only need to prove $A'_{fin} = B_{fin}$ as the other equalities have been proved in (4). To see this let $M$ be a finite $R$-module. By Lemma 10.62.1 there exists a filtration by $R$-submodules

such that each quotient $M_ i/M_{i-1}$ is isomorphic to $R/\mathfrak p_ i$ for some prime ideal $\mathfrak p_ i$ of $R$. Since $N$ is flat we obtain a filtration by $S$-submodules

such that each subquotient is isomorphic to $N/\mathfrak p_ iN$. By Lemma 10.63.3 we conclude that $\text{Ass}_ S(N \otimes _ R M) \subset \bigcup \text{Ass}_ S(N/\mathfrak p_ iN)$. Hence we see that $B_{fin} \subset A'_{fin}$. Since the other inclusion is part of (2) we win. $\square$

Proof. If $\mathfrak p \in \text{Ass}_ R(M)$ then there exists an injection $R/\mathfrak p \to M$. As $N$ is flat over $R$ we obtain an injection $R/\mathfrak p \otimes _ R N \to M \otimes _ R N$. Since $R/\mathfrak p \otimes _ R N = N/\mathfrak pN$ we conclude that $\text{Ass}_ S(N/\mathfrak pN) \subset \text{Ass}_ S(M \otimes _ R N)$, see Lemma 10.63.3. Hence the right hand side is contained in the left hand side.

Write $M = \bigcup M_\lambda$ as the union of its finitely generated $R$-submodules. Then also $N \otimes _ R M = \bigcup N \otimes _ R M_\lambda$ (as $N$ is $R$-flat). By definition of associated primes we see that $\text{Ass}_ S(N \otimes _ R M) = \bigcup \text{Ass}_ S(N \otimes _ R M_\lambda )$ and $\text{Ass}_ R(M) = \bigcup \text{Ass}(M_\lambda )$. Hence we may assume $M$ is finitely generated.

Let $\mathfrak q \in \text{Ass}_ S(M \otimes _ R N)$, and assume $R$ is Noetherian and $M$ is a finite $R$-module. To finish the proof we have to show that $\mathfrak q$ is an element of the right hand side. First we observe that $\mathfrak qS_{\mathfrak q} \in \text{Ass}_{S_{\mathfrak q}}((M \otimes _ R N)_{\mathfrak q})$, see Lemma 10.63.15. Let $\mathfrak p$ be the corresponding prime of $R$. Note that

If $\mathfrak pR_{\mathfrak p} \not\in \text{Ass}_{R_{\mathfrak p}}(M_{\mathfrak p})$ then there exists an element $x \in \mathfrak pR_{\mathfrak p}$ which is a nonzerodivisor in $M_{\mathfrak p}$ (see Lemma 10.63.18). Since $N_{\mathfrak q}$ is flat over $R_{\mathfrak p}$ we see that the image of $x$ in $\mathfrak qS_{\mathfrak q}$ is a nonzerodivisor on $(M \otimes _ R N)_{\mathfrak q}$. This is a contradiction with the assumption that $\mathfrak qS_{\mathfrak q} \in \text{Ass}_ S((M \otimes _ R N)_{\mathfrak q})$. Hence we conclude that $\mathfrak p$ is one of the associated primes of $M$.

Continuing the argument we choose a filtration

such that each quotient $M_ i/M_{i-1}$ is isomorphic to $R/\mathfrak p_ i$ for some prime ideal $\mathfrak p_ i$ of $R$, see Lemma 10.62.1. (By Lemma 10.63.4 we have $\mathfrak p_ i = \mathfrak p$ for at least one $i$.) This gives a filtration

with subquotients isomorphic to $N/\mathfrak p_ iN$. If $\mathfrak p_ i \not= \mathfrak p$ then $\mathfrak q$ cannot be associated to the module $N/\mathfrak p_ iN$ by the result of the preceding paragraph (as $\text{Ass}_ R(R/\mathfrak p_ i) = \{ \mathfrak p_ i\}$). Hence we conclude that $\mathfrak q$ is associated to $N/\mathfrak pN$ as desired. $\square$

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 10.63.17 combined with Lemma 10.63.16 part (1). $\square$

Proof. This is equivalent to Lemma 10.65.3 by Lemmas 10.63.14, 10.39.7, and 10.65.4. $\square$