Group $\mathbb Q^*$ as direct product/sum

Solution 1:

Yes, you're right.

Your statement can be generalized to the multiplicative group $K^*$ of the fraction field $K$ of a unique factorization domain $R$. Can you see how?

In fact, if I'm not mistaken it follows from this that for any number field $K$, the group $K^*$ is the product of a finite cyclic group (the group of roots of unity in $K$) with a free abelian group of countable rank, so of the form

$K^* \cong \newcommand{\Z}{\mathbb{Z}}$ $\Z/n\Z \oplus \bigoplus_{i=1}^{\infty} \Z.$

Here it is not enough to take the most obvious choice of $R$, namely the full ring of integers in $K$, because this might not be a UFD. But one can always choose an $S$-integer ring (obtained from $R$ by inverting finitely many prime ideals) with this property and then apply Dirichlet's S-Unit Theorem.