Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
"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
up vote 7 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.