What does $\overline{r}m:=rm$ mean? On this Wikipedia article, it says that you can define an $R$-module $M$ as an $R/Ann_R(M)$-module using the action $\overline{r}m:=rm.$ 
What does that action actually mean? What is $\overline{r}$?
 A: When a projection map $R\to S$ is in play, the notation $\bar{r}$ often denotes the image of $r$ in $S$ under the projection map. E.g. the elements of $\Bbb Z/n\Bbb Z$ are the residues $\overline{0},\overline{1},\overline{2},\cdots,\overline{n-1}$.
A priori one would have to wonder if $\overline{r}m:=rm$ is well-defined, since $\overline{r}=\overline{s}$ is possible in the quotient ring $R/{\rm Ann}_R(M)$ even if $r\ne s$ in $R$, and a priori different elements of $R$ act on the elements $m\in M$ in different ways. But if $r-s\in{\rm Ann}_R(M)$ then $(r-s)m=0$ implies that the elements $rm=sm$ are equal, and so $\overline{r}m$ is independent of choice of representative $r\in R$ for the residue $\overline{r}\in R/{\rm Ann}_R(M)$.
A: The symbol $\bar r$ denotes the equivalence class of $r$ in $\text{Ann}_R(M)$. The point here is that you want to define the action of the quotient on the module. This is well-defined because if $\bar{r_1}=\bar{r_2}$, this means that $r_1-r_2\in\text{Ann}_R(M)$, so for any $m$ we have $(r_1-r_2)m=0$, i.e. $r_1m=r_2m$. 
A: Note that an $R$-module is the same as an abelian group $M$ together with a ring homomorphism $R\longrightarrow\operatorname{End}_\mathbb{Z}(M)$. The kernel of  this homomorphism is precisely $\operatorname{Ann}_R(M)$ (actually this is how one should define the annihilator, so one does not have to check it is an ideal). Now the homomorphism theorem yields a uniquely determined ring homomorphism $R/\!\operatorname{Ann}_R(M)\hookrightarrow\operatorname{End}_\mathbb{Z}(M)$, realizing $M$ as a faithful $R/\!\operatorname{Ann}_R(M)$-module. One does not have to check well-definedness or the module axioms at all. By looking at the concrete description of this homomorphism, one sees that this is indeed the action you describe. As the others already said, $\bar{r}$ denots the image of $r\in R$ in $R/\!\operatorname{Ann}_R(M)$.
