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

Araceli Soto 2022-04-30 Answered
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
IM={rixiriI,xiM}.
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Answers (1)

Larry Hogan
Answered 2022-05-01 Author has 20 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
MIM=RnIn=(RI)n
as R-modules via the map
(x1,x2,,xn)+In(x1+I,x2+I,,xn+I)
This is also an isomorphism of RI -modules, as you can verify.
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