Is the multiplicative group $\mathbb Q\backslash\{0\}$ a free module over $\mathbb Z$?

Question

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!

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? — Gregory Grant, Jun 09 '17 at 20:47
The question body is about positive rationals, the title about non-zero rationals - that's a great differenece — Hagen von Eitzen, Jun 09 '17 at 20:53
See https://math.stackexchange.com/questions/2233946/what-is-a-set-of-generators-for-the-multiplicative-group-of-rationals. — lhf, Jun 09 '17 at 21:13

Santiago · Accepted Answer · 2017-06-09T21:52:59.980

4

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: $(-1)^2 = 1$.

edited Jun 09 '17 at 21:52

answered Jun 09 '17 at 20:48

Santiago

1,586

3

But $\Bbb Q^\times$ is certainly not free (in contrast to $\Bbb Q_{>0}$) – Hagen von Eitzen Jun 09 '17 at 20:52
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? – rgm Jun 09 '17 at 20:55
@rgm $\Bbb Q^*$ has torsion, so it is not free. – Adam Hughes Jun 09 '17 at 21:01
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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: $;-1;$ . – DonAntonio Jun 09 '17 at 21:02
Excellent observation, @DonAntonio! Thank you! :) – rgm Jun 09 '17 at 21:08
The body of the question said "positive integers," so I changed the title. – Thomas Andrews Jun 09 '17 at 21:10
@ThomasAndrews The body of the question also mentions simply "the group of nonzero rational numbers" – DonAntonio Jun 09 '17 at 21:25
@DonAntonio Whoops, yep, quite right. I took the inset text to be the actual question, when that was the question OP had solved. – Thomas Andrews Jun 09 '17 at 21:35
@ThomasAndrews Actually, Musili's book's exercise (34 of chapter 5, I think) first talks about the poisitive rationals, and then it asks about the non-zero ones. – DonAntonio Jun 09 '17 at 21:43
I'm sorry for the confusion, an edit was necessary - the statement is true for positive rationals. – Santiago Jun 09 '17 at 21:54

Is the multiplicative group $\mathbb Q\backslash\{0\}$ a free module over $\mathbb Z$?

1 Answers1