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Torsion-free modules

  1. Jan 25, 2009 #1
    Hi all, I came across this problem in a book and I can`t seem to crack it.
    It says that if we have an integral domain R and M is any non-principal ideal of R,
    M is torsion-free of rank 1 and is NOT a free R-module.

    Why is this true?

  2. jcsd
  3. Jan 25, 2009 #2
    ah so i see why it needs to be torsion free. (M lies in R, so if there is r such that rm=0, R cant be an integral domain...)

    but what about the rank? Why must it be 1? I am also surprised that it is not free..
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