A multiplicative subgroup of rational numbers Let $\Bbb Q^+$ be the set of positive rational numbers and $K=\{\sqrt {t^2 +s^2}:t, s\in \Bbb Q^+\}$. It is easy to see that $K$ forms a multiplicative group since 
$$
\sqrt{(p^2+q^2)(r^2+s^2)}=\sqrt{(pr+qs)^2+(ps-qr)^2}.
$$
Is $K$ with multiplication a free abelian group? If true, what is a generating set?
 A: Taking the square root is an isomorphism of $\mathbb{R}_{>0}$ so we can assume that $K=\{t^2+s^2 : t,s\in \mathbb{Q}\}$ which is just the norms of elements in $\mathbb{Q}[i]$. Note that we also have $K=\{ \frac{a^2+b^2}{c^2+d^2} : a,b,c,d\in \mathbb{Z} \}$ (Since $\mathbb{Q}[i]=Frac(\mathbb{Z}[i])$).
The set $K_0 = \{a^2 +b^2 : a,b\in \mathbb{Z} \}$ is a free monoid generated by 2, the primes $p$ where $p\equiv_4 1$ and the square primes $q^2$ where $q\equiv_4 3$ (these are the norms of the prime elements in $\mathbb{Z}[i]$ which is a UFD). It follows that $K$ is a free group over these elements.
This can be generalized - if $R$ is any UFD, then any nonzero element in $R$ has the form $u \prod p_i ^{k_i}$ where the $p_i$ are primes, $k_i\geq 0$ and $u$ is invertible. Thus, the nonzero elements of its fraction field $Q$ have the form $u \prod p_i ^{k_i}$ with $p_i$ primes and $k_i \in \mathbb{Z}$ and $u$ invertible in $R$. Hence, taking modulo the subgroup $R^\times$ of $Q^\times$ we get a free group. In the case above, $R^\times$ is exactly the kernel of the norm function, so $Im(norm)$ is a free group.
