# Show if M is free of rank n as R-module,

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
$IM=\left\{\sum {r}_{i}{x}_{i}\mid {r}_{i}\in I,{x}_{i}\in M\right\}.$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Larry Hogan
Step 1
Your proof that $\frac{M}{IM}$ is a submodule of M is surely wrong. Take $R=M=\mathbb{Z}$ and $I=2\mathbb{Z}$. Then $\frac{M}{IM}=\frac{\mathbb{Z}}{2\mathbb{Z}}$ that doesn't even embed in M.
Saying that M is a free module of rank n is the same as saying that $M\stackrel{\sim }{=}{R}^{n\mid }$ and it's not restrictive to take $M={R}^{n}$. Now $I{R}^{n}={I}^{n}$ and
$\frac{M}{IM}=\frac{{R}^{n}}{{I}^{n}}\stackrel{\sim }{=}{\left(\frac{R}{I}\right)}^{n}$
as R-modules via the map
$\left({x}_{1},{x}_{2},\cdots ,{x}_{n}\right)+{I}^{n}↦\left({x}_{1}+I,{x}_{2}+I,\cdots ,{x}_{n}+I\right)$
This is also an isomorphism of $\frac{R}{I}$ -modules, as you can verify.