The following is from Davis' Lecture Notes in Algebraic Topology. I'm very confused about the definition of ${}_{a}M=\{m\in M\mid am=0\}$. If $a$ is a non zero divisor then $am=0$ implies $m=0$, but then ${}_{a}M=\{0\}$.
Is ${}_{a}M$ not always trivial? If it is always trivial is there a pedagogical reason for writing it in this way?
I understand that if $a$ was not a zero divisor then ${}_{a}M$ need not be trivial.
Here is the proposition for context.
You have to be careful to distinguish the ring structure on $R$ from the $R$-module structure on $M$. The example given in Captain Lama's comment generalizes: if $a\in R$ then $R/a$ is an $R$-module via $r\cdot [b] = [rb]$, and $a\cdot m = 0$ for all $m\in R/a$ wether or not $a$ is a zero-divisor in $R$.
To work out a concrete example, consider $R = \mathbb{Z}$ and $M = \mathbb{Z}/k$ where $k = p q$ is a product of two distinct primes. Then $${}_pM = \{ [m] \in \mathbb{Z}/k\ |\ [pm] = 0\} = \{ [m] \in \mathbb{Z}/k\ |\ k\text{ divides } pm\} = q\cdot \mathbb{Z}/k \cong \mathbb{Z}/p.$$
This is consistent with the proposition, since $Tor(\mathbb{Z}/p, \mathbb{Z}/k) \cong Hom(\mathbb{Z}/p,\mathbb{Z}/k) \cong \mathbb{Z}/p$.