Quotient isomorphic to tensor Given a non zero element $x \in M$ and the nonzero submodule $M' = Ax$ of a commutative ring with unity $A$, and B a flat $A$-module, I'm trying to see that as I can show that
$$
(\operatorname{Ann}(x))^e \subset m^e \subsetneq B
$$
where $m$ is a maximal ideal in $A$, then
$$
M'_B \simeq A/\operatorname{Ann}(x) \otimes_A B \simeq B/ (\operatorname{Ann}(x))^e \neq 0
$$
But although I can intuitively see why this would be true I can't show the last isomorphism, is it something trivial? I'm studying through  Atiyah and Macdonald but I don't recall seeing such a property.
 A: Throughout this answer I am assuming that by $\operatorname{Ann}(x)^e$ you mean the product of $\operatorname{Ann}(x)$ and $B$. If so, then this is true for an arbitrary ideal and module. See Atiyah-Macdonald, Chapter 2, Exercise 2 if you would like to prove it for yourself. I will construct an alternative proof from the usual one utilizing a nice fact about flat modules.
To show this, we can take the exact sequence
$$0 \to \operatorname{Ann}(x) \to A \to A/\operatorname{Ann}(x) \to 0$$
and tensor it with $B$ to get an exact sequence (by flatness)
$$0 \to \operatorname{Ann}(x) \otimes_A B \to B \to A/\operatorname{Ann}(x) \otimes_A B \to 0$$
Thus $B/(\operatorname{Ann}(x) \otimes_A B) \cong A/\operatorname{Ann}(x)\otimes_A B$, and it suffices to construct an isomorphism of $\operatorname{Ann}(x) \otimes_A B$ and $\operatorname{Ann}(x)B$. We can do this by taking the inclusion map of our ideal into $A$ and tensoring by $B$, to get a map $\operatorname{Ann}(x) \otimes_A B \to B$ sending $a \otimes_A b$ to $ab$. It is injective because $B$ is flat, and the image of this map is precisely $\operatorname{Ann}(x) B$, giving the desired isomorphism.
