Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathrm{Ann}(M\otimes_A N)=0$.

share|improve this question
add comment

2 Answers

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.]

share|improve this answer
    
Nice principle! –  Manny Reyes Aug 23 '12 at 1:58
    
Any words on the situation if we exclude the product being zero? –  rschwieb Aug 23 '12 at 12:27
    
@rschwieb: Dear rschweib, You could always just add something like $\mathbb Z/n\mathbb Z$, for some $n$, to each of $M$ and $N$, so that the tensor product is non-zero (it contains $\mathbb Z/n\mathbb Z$ as a direct summand) but not faithful (it has exponent $n$). Regards, –  Matt E Aug 23 '12 at 14:36
add comment

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 $M$).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.