burubukuamaw

2022-04-02

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

sorrisi7yny

Beginner2022-04-03Added 9 answers

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

$({x}_{1},{x}_{2},\cdots ,{x}_{n})+{I}^{n}\mapsto ({x}_{1}+I,{x}_{2}+I,\cdots ,{x}_{n}+I)$

This is also an isomorphism of$\frac{R}{I}$ -modules, as you can verify.

Your proof that

Saying that M is a free module of rank n is the same as saying that

as R-modules via the map

This is also an isomorphism of

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