I'm trying to decide whether the multiplicative group of nonzero rational numbers is a free module over the integers or not. It is the second part of a question from Musili's Introduction to rings and modules. The first one, which I could solve, is:

Show that the multiplicative group of positive rational numbers, considered as a module over $\mathbb Z$, is a free module with the set of all positive prime numbers as a basis.

My attempt was unimportant. Since $\mathbb Z$ is PID, I've looked for a nonfree submodule, my bet being the negative. On another hand, I've tried to find a basis. My guess is that this basis doesn't exist, but I could not prove it. I'd be glad if someone could help me. Thanks in advance!

  • $\begingroup$ Don't you just have to show a product of powers of primes cannot equal one unless all powers are zero? Or something like that? $\endgroup$ Jun 9, 2017 at 20:47
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    $\begingroup$ The question body is about positive rationals, the title about non-zero rationals - that's a great differenece $\endgroup$ Jun 9, 2017 at 20:53
  • $\begingroup$ See math.stackexchange.com/questions/2233946/…. $\endgroup$
    – lhf
    Jun 9, 2017 at 21:13

1 Answer 1


The statement that $\mathbb Q_{>0}^\times$ considered as a $\mathbb Z$-module is free follows from the fundamental theorem of arithmetic: every integer greater than 1 either is prime itself or is the product of prime numbers, and this product is unique, up to the order of the factors.

You need to apply it twice: once to the numerator, once to the denominator, assuming they are coprime.

As Hagen von Eitzen and others have noted, the multiplicative group $\mathbb Q^\times$ isn't free, because $-1$ has finite order: $(-1)^2 = 1$.

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    $\begingroup$ But $\Bbb Q^\times$ is certainly not free (in contrast to $\Bbb Q_{>0}$) $\endgroup$ Jun 9, 2017 at 20:52
  • $\begingroup$ My problem is to find the basis to generate the negative rationals... The function $\psi: \mathbb{Z}\times \mathbb{Q^*}\rightarrow \mathbb{Q^*}$ is $(z,q)\mapsto q^z$, right? $\endgroup$
    – rgm
    Jun 9, 2017 at 20:55
  • $\begingroup$ @rgm $\Bbb Q^*$ has torsion, so it is not free. $\endgroup$ Jun 9, 2017 at 21:01
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    $\begingroup$ It can't be that $\;\Bbb Q^*\;$ is a free abelian group = free $\;\Bbb Z\,-$ module, since it has a finite order non-trivial element: $\;-1\;$ . $\endgroup$
    – DonAntonio
    Jun 9, 2017 at 21:02
  • $\begingroup$ Excellent observation, @DonAntonio! Thank you! :) $\endgroup$
    – rgm
    Jun 9, 2017 at 21:08

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