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

[TopCat]

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.

 

[mathwonk]

there is a map ZxZ*-->Q but it isn't injective.

 

[TopCat]

D'oh! Thanks, mathwonk.