$\mathrm{Tor}(M/\mathrm{Tor}(M))=0$ for an $A$-module $M$?

Question

Definition. Let $M$ be an $A$-module. Define $\mathrm{Tor}(M)=\{m \in M : \exists a \in A-\{0\}, am = 0 \}$).

Show that the quotient module $M/\mathrm{Tor}(M)$ doesn't have a non-zero torsion element.

I think we have to show that $\mathrm{Tor}(M/\mathrm{Tor}(M))=0$. So, $r \bar{m} = r(m + \mathrm{Tor}(M))= rm + \mathrm{Tor}(M)=0$.

Here I know that $rm=0$, but how could I show that $m=0$ ($r \not= 0$)?

Answer 1

You should assume that $a · \bar{m} = 0$ and conclude $\bar{m} = 0$. So you have $a · m + \mathrm{Tor}(M) = 0$, i.e. $a · m ∈ \mathrm{Tor}(M)$, and you want $m ∈ \mathrm{Tor}(M)$. Is it clear now how to proceed?

Answer 2

Unless $A$ is an integral domain, this is false!

Set $A=M=\mathbb Z/4\mathbb Z$. Then $t(M)=\{0,2\}$, and $M/t(M)\simeq\mathbb Z/2\mathbb Z$. Now notice that $2\cdot 1=0$ in $\mathbb Z/2\mathbb Z$.