## 10.65 Relative assassin

Discussion of relative assassins. Let $R \to S$ be a ring map. Let $N$ be an $S$-module. In this situation we can introduce the following sets of primes $\mathfrak q$ of $S$:

$A$: with $\mathfrak p = R \cap \mathfrak q$ we have that $\mathfrak q \in \text{Ass}_ S(N \otimes _ R \kappa (\mathfrak p))$,

$A'$: with $\mathfrak p = R \cap \mathfrak q$ we have that $\mathfrak q$ is in the image of $\text{Ass}_{S \otimes \kappa (\mathfrak p)}(N \otimes _ R \kappa (\mathfrak p))$ under the canonical map $\mathop{\mathrm{Spec}}(S \otimes _ R \kappa (\mathfrak p)) \to \mathop{\mathrm{Spec}}(S)$,

$A_{fin}$: with $\mathfrak p = R \cap \mathfrak q$ we have that $\mathfrak q \in \text{Ass}_ S(N/\mathfrak pN)$,

$A'_{fin}$: for some prime $\mathfrak p' \subset R$ we have $\mathfrak q \in \text{Ass}_ S(N/\mathfrak p'N)$,

$B$: for some $R$-module $M$ we have $\mathfrak q \in \text{Ass}_ S(N \otimes _ R M)$, and

$B_{fin}$: for some finite $R$-module $M$ we have $\mathfrak q \in \text{Ass}_ S(N \otimes _ R M)$.

Let us determine some of the relations between theses sets.

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.

We always have $A = A'$.

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

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

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

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.
$\square$

We define the relative assassin of $N$ over $S/R$ to be the set $A = A'$ above. As a motivation we point out that it depends only on the fibre modules $N \otimes _ R \kappa (\mathfrak p)$ over the fibre rings. As in the case of the assassin of a module we warn the reader that this notion makes most sense when the fibre rings $S \otimes _ R \kappa (\mathfrak p)$ are Noetherian, for example if $R \to S$ is of finite type.

Definition 10.65.2. Let $R \to S$ be a ring map. Let $N$ be an $S$-module. The *relative assassin of $N$ over $S/R$* is the set

\[ \text{Ass}_{S/R}(N) = \{ \mathfrak q \subset S \mid \mathfrak q \in \text{Ass}_ S(N \otimes _ R \kappa (\mathfrak p)) \text{ with }\mathfrak p = R \cap \mathfrak q\} . \]

This is the set named $A$ in Lemma 10.65.1.

The spirit of the next few results is that they are about the relative assassin, even though this may not be apparent.

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.
$\square$

Lemma 10.65.4. Let $R \to S$ be a ring map. Let $N$ be an $S$-module. Assume $N$ is flat as an $R$-module and $R$ is a domain with fraction field $K$. Then

\[ \text{Ass}_ S(N) = \text{Ass}_ S(N \otimes _ R K) = \text{Ass}_{S \otimes _ R K}(N \otimes _ R K) \]

via the canonical inclusion $\mathop{\mathrm{Spec}}(S \otimes _ R K) \subset \mathop{\mathrm{Spec}}(S)$.

**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).
$\square$

Lemma 10.65.5. Let $R \to S$ be a ring map. Let $M$ be an $R$-module, and let $N$ be an $S$-module. Assume $N$ is flat as $R$-module. Then

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

where we use Remark 10.17.8 to think of the spectra of fibre rings as subsets of $\mathop{\mathrm{Spec}}(S)$. If $R$ is Noetherian then this inclusion is an equality.

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

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