Maximal Ideal Theorem: Proving M/IM=O for R-Module M and Ideal I

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Homework Statement



let M be an R-module and I an ideal of R
suppose Mm=0 for all maximal ideals m
then, M=IM




The Attempt at a Solution



O\rightarrow IM \rightarrow M \rightarrow M/IM \rightarrow O is exact seq
To show that M/IM=O (then M=IM)
 
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Is M a left R-module or a right R-module? You have R operating both on the right (Mm = 0) and the left (M = IM) of M. Also, assuming that all R operations are on the left (say), then what if I is a maximal ideal? We'd then have that M = MI = 0, which doesn't seem correct.

Petek
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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