Lemma 10.65.3. Let $R \to S$ be a ring map. Let $M$ be an $R$-module, and let $N$ be an $S$-module. If $N$ is flat as $R$-module, then

and if $R$ is Noetherian then we have equality.

Lemma 10.65.3. Let $R \to S$ be a ring map. Let $M$ be an $R$-module, and let $N$ be an $S$-module. If $N$ is flat as $R$-module, then

\[ \text{Ass}_ S(M \otimes _ R N) \supset \bigcup \nolimits _{\mathfrak p \in \text{Ass}_ R(M)} \text{Ass}_ S(N/\mathfrak pN) \]

and if $R$ is Noetherian then we have equality.

**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

\[ (M \otimes _ R N)_{\mathfrak q} = M \otimes _ R N_{\mathfrak q} = M_{\mathfrak p} \otimes _{R_{\mathfrak p}} N_{\mathfrak q} \]

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

\[ 0 = M_0 \subset M_1 \subset \ldots \subset M_ n = M \]

such that each quotient $M_ i/M_{i-1}$ is isomorphic to $R/\mathfrak p_ i$ for some prime ideal $\mathfrak p_ i$ of $R$, 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

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

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. $\square$

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