Show if M is free of rank n as R-module, then $\frac{M}{IM}$ is free of rank n as $\frac{R}{I}$ module:

Let R be a ring and$I\subset R$ a two-sided ideal and M an R-module with

$IM=\{\sum {r}_{i}{x}_{i}\mid {r}_{i}\in I,{x}_{i}\in M\}.$

Let R be a ring and