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

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SUMMARY

The discussion confirms that the set of rational numbers, Q, is not free abelian due to its requirement for an empty basis. However, it is established that Q is isomorphic to Z^2, which implies the existence of a nonempty basis. The confusion arises from the mapping Z x Z* to Q, which is not injective, indicating a flaw in the initial assumption of isomorphism. The participant acknowledges the need for clarification on the isomorphism issue.

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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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there is a map ZxZ*-->Q but it isn't injective.
 
D'oh! Thanks, mathwonk.
 

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