6
$\begingroup$

Let $R$ be a Noetherian ring, and let $M$ and $N$ be finitely generated $R$ module. Do we know any formulas for $\operatorname{Ass}(M\otimes_R N)$ in terms of $\operatorname{Ass}(M)$, $\operatorname{Ass}(N)$ or in terms of $\operatorname{Supp}(M)$ or $\operatorname{Supp}(N)$?

Recall that we have such a formula for the support, i.e., $\operatorname{Supp}(M\otimes_R N)=\operatorname{Supp}M\cap \operatorname{Supp}N$. We also have a formula for $\operatorname{Ass}(\operatorname{Hom}(M,N))$.

I have not seen any formula for Ass of tensor products, it would be nice to have such a formula in at least a few special cases.

$\endgroup$
1
  • 1
    $\begingroup$ I think that in general almost nothing can be said. See for example $M =A/I$ and $N = A/J$. $\endgroup$
    – Andrea
    Apr 10, 2013 at 12:42

1 Answer 1

4
$\begingroup$

The following is THM 23.2 in Matsumura's "Commutative Ring Theory":

Let $\phi:A \rightarrow B$ be a homomorphism of Noetherian rings, and let $E$ be an $A$-module and $G$ be a $B$-module. Suppose $G$ is flat over $A$; then:

(i) If $\mathfrak{p} \in \text{Spec}(A)$ and $ G/\mathfrak{p}G \ne 0$, then, letting $\phi^a: \text{Spec}(B) \rightarrow \text{Spec}(A)$ be the induced map (this is Matsumura's notation) we have $\phi^a(\text{Ass}_B(G/\mathfrak{p}G))=\text{Ass}_A(G/\mathfrak{p}G)=\{\mathfrak{p}$}

(ii) $\text{Ass}_B(E \otimes_A G) = \bigcup_{\mathfrak{p} \in \text{Ass}_A(E)}\text{Ass}_B(G/\mathfrak{p}G)$

So, this answers the question in the case where $M$ is a flat $R$-module; take $A=R=B$, $E=N$, and $G=M$, and apply (ii).

So this isn't very interesting when $R$ is local, for instance...but it is the best I can do.

$\endgroup$
1
  • $\begingroup$ I know this result, but as you mention, it is not as useful once the ring is local. $\endgroup$
    – messi
    Apr 11, 2013 at 10:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .