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41

Any two nonzero rationals are linearly dependent: if a,b∈Qa,b∈Q, a≠0≠ba≠0≠b, then there exist nonzero integers nn and mm such that na+mb=0na+mb=0.

So if QQ were free, it would be free of rank 11, and hence cyclic. But QQ is not a cyclic ZZ module (it is divisible, so it is not isomorphic to ZZ, the only infinite cyclic ZZ-module.

So QQ cannot be free.

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answered Apr 8 '12 at 17:11

Arturo Magidin

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  I don't understand this argument. ZZ is not a field, so if I'm not mistaken, having two linear dependent elements of a ZZ-module doesn't mean you can express one in terms of the other. – James Well May 17 '19 at 21:24 

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  @JamesWell: You are mistaken. A free ZZ-module must have a basis. A basis is a set of elements {mi}i∈I{mi}i∈I such that (i) every element of the module is a (finite) ZZ-linear combination of the mimi; and (ii) the only (finite) ZZ-linear combinations of the mimi equal to 00 are trivial. So the first part shows that if it has a basis, it has at most one element; and the second part shows that no one element set can span. – Arturo Magidin May 17 '19 at 21:42 

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  @JamesWell It’s not about being able to express one element in terms of the other; being linearly dependent in modules is not equivalent to having one element by in the span of the rest. That’s not the definition of linear dependence, that’s a consequence in the case of vector spaces. – Arturo Magidin May 17 '19 at 21:44

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16

Suppose a/ba/b and c/dc/d are two members of a set of free generators and both fractions are in lowest terms. Find e=lcm(b,d)e=lcm⁡(b,d) and write both fractions as (something/e(something/e). Then

ab=1e+⋯+1e and cd=1e+⋯+1e,ab=1e+⋯+1e and cd=1e+⋯+1e,

where in general the numbers of terms in the two sums will be different.

Then a/ba/b and c/dc/d are not two independent members of a set of generators, since both are in the set generated by 1/e1/e. So QQ must be generated by just one generator, so Q={0,±f,±2f,±3f,…}Q={0,±f,±2f,±3f,…}. But that fails to include the average of ff and 2f2f, which is rational.

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edited May 2 '13 at 5:09

answered Apr 8 '12 at 22:13

Michael Hardy

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  Having written this, I see that it's not really so different from what Arturo Magidin wrote, except in style. So each reader can choose his or her preferred style. – Michael Hardy Apr 8 '12 at 22:15

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It follows from the definition of free modules.

Let us suppose to the contradictory that QQ is a free ZZ module, so by definition of free modules, for a given injective map α:X→Qα:X→Q and for any map f:X→Zf:X→Z, there exist a unique ZZ-homomorphism g:Q→Zg:Q→Z such that f=gαf=gα. Every ZZ module homomophism is a group homomorphism and we know that there is only trivial group homomorphism from QQ to ZZ. Since we can define a lot of distinct maps from XX to ZZ and we don't have any homomorphism from QQ to ZZ corresponding to non-zero maps f:X→Zf:X→Z, thus QQ is not a free module over ZZ.

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