Show if M is free of rank n



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Show if M is free of rank n as R-module, then MIM is free of rank n as RI module:
Let R be a ring and IR a two-sided ideal and M an R-module with

Answer & Explanation



Beginner2022-04-03Added 9 answers

Step 1
Your proof that MIM is a submodule of M is surely wrong. Take R=M=Z and I=2Z. Then MIM=Z2Z that doesn't even embed in M.
Saying that M is a free module of rank n is the same as saying that M=Rn and it's not restrictive to take M=Rn. Now IRn=In and
as R-modules via the map
This is also an isomorphism of RI -modules, as you can verify.

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