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Q (the rationals) not free abelian yet iso to Z^2?

  1. Nov 25, 2011 #1
    So I just proved that Q (the rationals) is not free abelian since it must have an empty basis. But Q is isomorphic to Z^2 = ƩZ which is equivalent to Q having a nonempty basis. I must be wrong about the isomorphism but I don't see why.
    Last edited: Nov 25, 2011
  2. jcsd
  3. Nov 25, 2011 #2


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    there is a map ZxZ*-->Q but it isn't injective.
  4. Nov 25, 2011 #3
    D'oh! Thanks, mathwonk.
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