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I'm trying to think of a set of generators for the group $(\mathbb{Q^\times, *})$, which I think is countably generated (as opposed to finitely generated). (I wonder if such a set could be the primes of $\mathbb{Z}$ (together with $1$ and $-1),$ meaning the set $\{1, -1, \frac{1}{2}, -\frac{1}{2}, ...\}$ generates all nonzero rationals?)

Harsh Kumar
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étale-cohomology
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2 Answers2

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You're quite right: $(\mathbb{Q^\times, *})$ is generated by the (positive) primes and $-1$: $$ \mathbb{Q}^\times \cong C_2 \times \bigoplus_p \mathbb Z $$

This is essentially the fundamental theorem of arithmetic.

lhf
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    And the uniqueness asserted by the F.T.A. essentially says that $(\Bbb Q^{\times}, \ast)$ is freely generated by the primes, giving a canonical identification of that group with (since there are countably many primes) $\left(\bigoplus_{\textrm{$p$ prime}} \Bbb Z, +\right)$. – Travis Willse Apr 14 '17 at 14:16
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    @Travis: Up to the roots of unity you're forgetting about. –  Apr 14 '17 at 14:21
  • @Travis I was about to ask if this was the minimal generating set. I guess it is, then? – étale-cohomology Apr 14 '17 at 14:23
  • @Hurkyl Oops, you're right, of course: That statement applies to $\Bbb Q^+$, not $\Bbb Q^{\times}$. To get all of $\Bbb Q^{\times}$ you need to throw in some negative number. Then, the element $-1$ generates a torsion subgroup, and the canonical identification is with $\left((\Bbb Z / 2 \Bbb Z) \oplus \bigoplus_{\textrm{$p$ prime}} \Bbb Z, +\right)$. – Travis Willse Apr 14 '17 at 14:30
  • @Travis Also, what does $\bigoplus_p\mathbb{Z}$ mean? What group is that? Or could you mean $\bigoplus_p \mathbb{Z}/p\mathbb{Z}$? – étale-cohomology Apr 14 '17 at 14:30
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    @étale-cohomology It just means a countable sum of copies of $\Bbb Z$ indexed by primes. One can write any positive rational $q$ uniquely as $2^{n_2} 3^{n_3} 5^{n_5} \cdots$ for finitely many nonzero $n_a$. Then, this identification maps $q$ to $(n_2, n_3, n_5, \ldots) \in \Bbb Z \oplus \Bbb Z \oplus \Bbb Z \oplus \cdots)$, and it's straightforward to check that multiplication of rational numbers corresponds to the usual addition of sequences. – Travis Willse Apr 14 '17 at 14:33
  • @étale-cohomology it means a direct sum over all primes where each "summand" is the integers. This is correct. – quid Apr 14 '17 at 14:33
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    And yes, this generating set (of $\Bbb Q^+)$ is minimal. No prime can be expressed as a product and quotient of other primes, so no generator can be omitted. – Travis Willse Apr 14 '17 at 14:34
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I assume your list continues on $\{\frac{1}{3},-\frac{1}{3},\frac{1}{5},-\frac{1}{5}, \ldots\}$.

You do indeed give a list of generators; but your list is redundant:

  • $1$ is contained in every group; it need not be listed
  • $-\frac{1}{2} = (-1) \cdot \frac{1}{2}$, so you only need to include one of each associate; it's traditional to take the positive one in your list and remove the negative one.

Also, it's traditional to list the primes as the generators, rather than their inverses. That is, the typical set of generators one uses for $\mathbb{Q}^\times$ is the set

$$ \{ -1, 2, 3, 5, 7, 11, \ldots \} $$

This is a minimal set of generators; the only relationship between its elements is

$$(-1) = (-1)^{-1}$$