If M a R-module then M is naturally a R/Ann(M)-Module

Question

I am trying to show this is true by taking $(r+Ann(M)).m=r.m$. But I never use the fact that if $r\in Ann(M)$ and $x\in M$ then $rx=0$.

$(r+Ann(M)).(m_0+m_1)=r.(m_0+m_1)=r.m_0+r.m_1= (r+Ann(M)).m_0+(r+Ann(M))m_1$

$((r_0+$Ann$(M))+(r_1+$Ann$(M)))m=(r_0+r_1+$Ann$(M)).m=(r_0+r_1).m=(r_0+$Ann$(M)).m+(r_1+$Ann$(M)).m$

$(r_0+$Ann$(M))[(r_1+$Ann$(M)).m]=(r_0+$Ann$(M)).[r_1.m]=r_0r_1m=[(r_0+$Ann$(M)).(r_1+$Ann$(M))].m$

What am I missing here?

Answer 1

You need this fact to prove that multiplication is well-defined.

Write $A=\text{Ann}(M)$. You define $(r+A)\cdot m=rm$. For this to make sense, you must have that if $r+A=r'+A$ then $rm=r'm$. Equivalently, if $r-r'\in A$ then $(r-r')m=0$. This is where you use the fact that $A$ annihilates $M$.

Answer 2

There is a more natural way to think about it. Recall the definition of a $R$-module. By saying an abelian group $M$ is a $R$-module, we mean that there is a ring map $\phi:R \rightarrow End_{\mathbb{Z}}(M)$. Now $Ker\phi = Ann(M)$. So $M$ is naturally a $R/Ann(M)$-module.