Does $\operatorname{Ann}_B(B \otimes_A M) = (\operatorname{Ann}_A(M))^e$ when $M$ is finitely generated?

Question

I am working on solving part (viii) of exercise 19 in chapter 3 of Atiyah-MacDonald. In the problem, we have rings $A, B$, a ring homomorphism $f : A\to B$, and a finitely generated $A$-module $M$; we must show that $\operatorname{supp}_B(B\otimes_A M) = (f^*)^{-1}(\operatorname{supp}_A(M))$ where $f^* : \operatorname{Spec}(B) \to \operatorname{Spec}(A)$ is the pullback.

My approach boils down to showing $V(\operatorname{Ann}_B(B\otimes_A M)) = V((\operatorname{Ann}_A(M))^e)$ in $\operatorname{Spec}(B)$, and I believe this can be accomplished by showing $\operatorname{Ann}_B(B\otimes_A M) = (\operatorname{Ann}_A(M))^e$ in $B$ (or at least the radicals are equal, but I think my given relation is true). Does this latter equality hold?

I can at least show the inclusion $(\operatorname{Ann}_A(M))^e \subseteq \operatorname{Ann}_B(B\otimes_A M)$, which is pretty simple. However, given a $t \in \operatorname{Ann}_B(B\otimes_A M)$, I am having trouble proceeding and and finding elements of $\operatorname{Ann}_A(M)$ which generate $t$ over $B$. Hints are appreciated if there is something simple I'm not seeing!

Answer 1

Counterexample. Let $A$ be a domain.

Suppose that there exist ideals $I, J$ of $A$ such that $(JI : I) \not= J$. Then, taking $M = I$ and $B = A/J$ you get a counterexample. Indeed, $B \otimes M \cong I/IJ$ has annihilator $(IJ: I)$, which by construction has nonzero image in $B$. On the other hand, since $A$ is a domain and $M$ is an ideal of $A$, $\operatorname{Ann}_A(M) = 0$.

This might seem a little bit contrived, but here are two concrete ways in which the above situation commonly arises:

$(1)$ Suppose that $A$ is any domain that is not integrally closed in its fraction field $K$. Then there exists $k \in K \setminus A$ such that $A[k]$ is a finite $A$-module. Choose any $a \in A$ that clears denominators for the finite set of generators of $A[k]$. Then the set $I = aA[k]$ is a finitely generated ideal of $A$. Moreover $k I \subseteq I$. Write $k = b/c$ with $b, c \in A$. Then $b I \subseteq cI$ but, by assumption that $k \notin A$, $b \notin cA$. Thus, $b \in (cI : I) \setminus cI$.

$(2)$ Call an ideal $I$ a cancellation ideal if for any ideals $J, K$, $IJ = IK$ implies $J = K$. It is elementary to show that, for a fixed ideal $I$, $I$ is a cancellation ideal iff $(IJ : J) = J$ for all $J$. It is is also not hard to show that $I$ is a cancellation ideal iff it is locally principal (i.e. $I_P$ is principally generated for each prime $P$). On the other hand, a finitely generated ideal $I$ is invertible (meaning $(I(A :_K I )= A$) iff it is locally principal. Thus, a finitely generated ideal $I$ is invertible iff $(IJ : I) = J$ for all ideals $J$.

Any domain that is not Prüfer (in particular, any Noetherian domain that is not Dedekind) contains a finitely generated ideal $I$ that is not invertible, and thus contains a pair of ideals $J, I$ such that $(JI : I) \not= J$.

Note. If you assume additionally $A \rightarrow B$ is flat, then the property you're asking about does hold.