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 I⊂R a two-sided ideal and M an R-module withIM={∑rixi∣ri∈I,xi∈M}.

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sorrisi7yny · Accepted Answer

Step 1Your proof that MIM is a submodule of M is surely wrong. Take R=M=Z and I=2Z. Then MIM=Z2Z that doesn&#039;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&#039;s not restrictive to take M=Rn. Now IRn=In andMIM=RnIn=∼(RI)nas 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.

Show if M is free of rank n

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