Tensor product of faithful modules In commutative algebra, is it true that the tensor product of two faithful modules is a faithful module?
I have written for myself a proof for the case of finitely generated modules over reduced rings: Suppose $M,N$ are finite $A$-modules, with $A$ reduced. Then
$\mathrm{V(Ann}(M\otimes_A N))=\mathrm{Supp}(M\otimes_A N)=\mathrm{Supp(M)}\cap\mathrm{Supp(N)}=\mathrm{V}(\mathrm{Ann}(M))\cap\mathrm{V(Ann}(N))=\mathrm{V}(0)\cap\mathrm{V}(0)=\mathrm{Spec}(A),$
and hence $\sqrt{\mathrm{Ann}(M\otimes_A N)}=\sqrt{(0)}=(0),$ hence $\mathrm{Ann}(M\otimes_A N)=0$.
 A: It's not true in general.  E.g. if $M = \bigoplus_{n =1 }^{\infty} \mathbb Z/n\mathbb Z,$ and $N = \mathbb Q$, then both $M$ and $N$ are faithful $\mathbb Z$-modules, but $M\otimes N = 0.$
[Both this example and that of Manny Reyes illustrate the general principle
that a torsion ab. gp. tensored with a divisible ab. gp. gives zero, and it's possible to find faithful torsion modules and faithful divisible modules (or even
faithful modules that are both divisible and torsion, as in Manny Reyes's answer); of course these won't be finitely generated.]
A: Consider the $\mathbb{Z}$-module $M = \mathbb{Q}/\mathbb{Z}$.  I'll be lazy and denote the class of an element $x \in \mathbb{Q}$ by $x \in \mathbb{Q}/\mathbb{Z}$.
$M$ is faithful: if $0 \neq n \in \mathbb{Z}$ then (for instance) $1/2n \in M$ satisfies $n \cdot (1/2n) = 1/2 \neq 0$ in $M$.
But $M \otimes_{\mathbb{Z}} M = 0$: each element of $M$ is a multiple of an element of the form $1/m$, and $1/m \otimes 1/n = n/mn \otimes 1/n = 1/mn \otimes n/n = 1/mn \otimes 0 = 0$ (as $n/n = 1 = 0$ in $M$).
