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

Solution 1:

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.

Solution 2:

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.

Solution 3:

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 $\mathbb{Z}$.