# How can we show that $\mathbb Q$ is not free?

I am really confused from the definition.

How do we know that $\mathbb Q$ is not free?

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

Edit : yes I meant free being free $\mathbb Z$ module, thanks!

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"Free" in what sense? – JeffE Apr 8 '12 at 16:05
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). – David Wheeler Apr 8 '12 at 16:09
Thank you, that gives me an answer! – Emily Apr 8 '12 at 16:13

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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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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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