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

  • Thread starter TopCat
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  • #1
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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.
 
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  • #2
mathwonk
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there is a map ZxZ*-->Q but it isn't injective.
 
  • #3
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D'oh! Thanks, mathwonk.
 

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