Natural $(R/I)$-module structure for an $R$-module $M$ annihilated by $I$

Question

Suppose that $I$ is a two-sided ideal in the ring $R$, and that $M$ is a module over the quotient ring $R/I$. Why can we naturally regard $M$ as a $R$-module that is annihilated by $I$? Conversely, suppose that $M$ is a $R$-module annihilated by $I$. Why is $M$ naturally a module over $R/I$?

Answer 1

The obvious module action is $rm:=(r+ I)m$.

Conversely, given an ideal $I$ and an $R$ module $M$ we would like to use $(r+ I)m:= rm$, but this is not well-defined unless $I$ is contained in the annihilator of $M$.

Explicitly, given $r+I=r'+I$ and $m=m'$ it needs to be established that $(r+I)m=(r'+I)m'$, and proving this makes use of the condition that $I$ annihilates $M$.