Lemma 10.65.1. Let $R \to S$ be a ring map. Let $N$ be an $S$-module. Let $A$, $A'$, $A_{fin}$, $B$, and $B_{fin}$ be the subsets of $\mathop{\mathrm{Spec}}(S)$ introduced above.

1. We always have $A = A'$.

2. We always have $A_{fin} \subset A$, $B_{fin} \subset B$, $A_{fin} \subset A'_{fin} \subset B_{fin}$ and $A \subset B$.

3. If $S$ is Noetherian, then $A = A_{fin}$ and $B = B_{fin}$.

4. If $N$ is flat over $R$, then $A = A_{fin} = A'_{fin}$ and $B = B_{fin}$.

5. If $R$ is Noetherian and $N$ is flat over $R$, then all of the sets are equal, i.e., $A = A' = A_{fin} = A'_{fin} = B = B_{fin}$.

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

$\text{Ass}_ S(N \otimes _ R \kappa (\mathfrak p)) = \text{Ass}_{S/\mathfrak pS}(N \otimes _ R \kappa (\mathfrak p)) = \text{Ass}_{S \otimes _ R \kappa (\mathfrak p)}(N \otimes _ R \kappa (\mathfrak p))$

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

$0 = M_0 \subset M_1 \subset \ldots \subset M_ n = M$

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

$0 = N \otimes _ R M_0 \subset N \otimes _ R M_1 \subset \ldots \subset N \otimes _ R M_ n = N \otimes _ R M$

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. $\square$

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