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I am really confused from the definition.

How do we know that $\mathbb Q$ is not a free $\mathbb Z$-module?

In class people use it as a trivial fact, but I don't seem to understand.

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    $\begingroup$ If $\Bbb{Q}$ was free, it must have a basis (I assume you mean free abelian (free as a $\Bbb{Z}$-module)). Show the rationals aren't cyclic (so not of rank 1), and that any two rational numbers are not LI over $\Bbb{Z}$ (so not of rank > 1). $\endgroup$ Commented Apr 8, 2012 at 16:09
  • $\begingroup$ What is relationship between cyclicness of Q and it's freeness over Z? $\endgroup$
    – Pritam Roy
    Commented Sep 12, 2022 at 17:47

3 Answers 3

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Any two nonzero rationals are linearly dependent: if $a,b\in\mathbb{Q}$, $a\neq 0 \neq b$, then there exist nonzero integers $n$ and $m$ such that $na + mb = 0$.

So if $\mathbb{Q}$ were free, it would be free of rank $1$, and hence cyclic. But $\mathbb{Q}$ is not a cyclic $\mathbb{Z}$ module (it is divisible, so it is not isomorphic to $\mathbb{Z}$, the only infinite cyclic $\mathbb{Z}$-module.

So $\mathbb{Q}$ cannot be free.

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    $\begingroup$ I don't understand this argument. $\mathbb Z$ is not a field, so if I'm not mistaken, having two linear dependent elements of a $\mathbb Z$-module doesn't mean you can express one in terms of the other. $\endgroup$
    – James Well
    Commented May 17, 2019 at 21:24
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    $\begingroup$ @JamesWell: You are mistaken. A free $\mathbb{Z}$-module must have a basis. A basis is a set of elements $\{m_i\}_{i\in I}$ such that (i) every element of the module is a (finite) $\mathbb{Z}$-linear combination of the $m_i$; and (ii) the only (finite) $\mathbb{Z}$-linear combinations of the $m_i$ equal to $0$ 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. $\endgroup$ Commented May 17, 2019 at 21:42
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    $\begingroup$ @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. $\endgroup$ Commented May 17, 2019 at 21:44
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    $\begingroup$ You began with basics but concluded the answer in an advanced language which wasn't necessary. I understood it but a beginner would trip. $\endgroup$ Commented Sep 17, 2021 at 13:23
  • $\begingroup$ @permutation_matrix: Please point the "advanced language" in my answer that does not appear in the question, and show me that it "wasn't necessary" by stating exactly how to avoid it. A beginner who is tripped by this answer is tripped by the question itself. $\endgroup$ Commented Sep 17, 2021 at 14:29
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Suppose $a/b$ and $c/d$ are two members of a set of free generators and both fractions are in lowest terms. Find $e=\operatorname{lcm}(b,d)$ and write both fractions as $(\text{something}/e$). Then $$ \frac a b = \frac 1 e + \cdots + \frac 1 e\text{ and }\frac c d = \frac 1 e + \cdots + \frac 1 e, $$ where in general the numbers of terms in the two sums will be different.

Then $a/b$ and $c/d$ are not two independent members of a set of generators, since both are in the set generated by $1/e$. So $\mathbb{Q}$ must be generated by just one generator, so $\mathbb{Q} = \{ 0, \pm f, \pm 2f, \pm 3f, \ldots \}$. But that fails to include the average of $f$ and $2f$, which is rational.

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    $\begingroup$ 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. $\endgroup$ Commented Apr 8, 2012 at 22:15
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It follows from the definition of free modules.

Let us suppose to the contradictory that $\mathbb{Q}$ is a free $\mathbb{Z}$ module, so by definition of free modules, for a given injective map $\alpha: X \rightarrow \mathbb{Q}$ and for any map $f : X \rightarrow \mathbb{Z}$, there exist a unique $\mathbb{Z}$-homomorphism $g: \mathbb{Q} \rightarrow \mathbb{Z}$ such that $f=g\alpha$. Every $\mathbb{Z}$ module homomophism is a group homomorphism and we know that there is only trivial group homomorphism from $\mathbb{Q}$ to $\mathbb{Z}$. Since we can define a lot of distinct maps from $X$ to $\mathbb{Z}$ and we don't have any homomorphism from $\mathbb{Q}$ to $\mathbb{Z}$ corresponding to non-zero maps $f:X \rightarrow \mathbb{Z}$, thus $\mathbb{Q}$ is not a free module over $\mathbb{Z}$.

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  • $\begingroup$ If the only homomorphism $\mathbb{Q}\to \mathbb{Z}$ is the zero morphism, then you do have uniqueness. The problem is with the existence. $\endgroup$
    – Arnaud D.
    Commented Jul 26, 2018 at 9:31
  • $\begingroup$ @Arnaud: I did not get your point of uniqueness. According the definition of free modules, the maps $f$ and $g$ are in one to one correspondence. In our case, we can define a lot of distinct constant maps $f: X \rightarrow \mathbb{Z}$, so if $\mathbb{Q}$ is free $\mathbb{Z}$ module, then there must be more than one homomorphism from $\mathbb{Q}$ to $\mathbb{Z}$, which is not the case. $\endgroup$
    – eyp
    Commented Jul 27, 2018 at 5:08
  • $\begingroup$ That's what I was saying : the problem is that there are too few homomorphisms $\mathbb{Q}\to \mathbb{Z}$ in comparison with maps $X\to \mathbb{Z}$ (assuming $X$ non-empty), so in the universal property, you can't guarantee the existence of a homomorphism $\mathbb{Q}\to \mathbb{Z}$ for any map; on the other hand, in the case where there is a homomorphism, it is unique. $\endgroup$
    – Arnaud D.
    Commented Jul 27, 2018 at 9:46
  • $\begingroup$ @ Arnaud: I got your point, thank you for correcting me. I have edit my answer, please tell, if it is still wrong. $\endgroup$
    – eyp
    Commented Jul 27, 2018 at 11:27

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