# How can we show that $\mathbb Q$ is not a free $\mathbb Z$-module?

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.

• 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). Commented Apr 8, 2012 at 16:09
• What is relationship between cyclicness of Q and it's freeness over Z? Commented Sep 12, 2022 at 17:47

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.

• 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. Commented May 17, 2019 at 21:24
• @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. Commented May 17, 2019 at 21:42
• @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. Commented May 17, 2019 at 21:44
• You began with basics but concluded the answer in an advanced language which wasn't necessary. I understood it but a beginner would trip. Commented Sep 17, 2021 at 13:23
• @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. Commented Sep 17, 2021 at 14:29

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.

• 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. Commented Apr 8, 2012 at 22:15

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 $\mathbb{Z}$.

• If the only homomorphism $\mathbb{Q}\to \mathbb{Z}$ is the zero morphism, then you do have uniqueness. The problem is with the existence. Commented Jul 26, 2018 at 9:31
• @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.
– eyp
Commented Jul 27, 2018 at 5:08
• 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. Commented Jul 27, 2018 at 9:46
• @ Arnaud: I got your point, thank you for correcting me. I have edit my answer, please tell, if it is still wrong.
– eyp
Commented Jul 27, 2018 at 11:27