Quotient by squares of finite rank

Question

Let $F$ be a number field and $\mathcal{O}$ its ring of integers.

Why is $\mathcal{O}^\times/\mathcal{O}^{\times 2}$ of finite rank?

In particular, what do we get for $K=\mathbf{Q}$? And for $K=\mathbf{Q}_p$?

Answer 1

First of all, note that $\Bbb Z^\times=\{\pm1\}$ so that the reason is very obvious in this case.

More generally, if $\cal O$ is the ring of integers in a number field $F$ a theorem of Dirichlet says that $$ {\cal O}^\times\simeq T\times\Bbb Z^r $$ where $T$ is a finite group (actually, the roots of $1$ in $F$) and $r=r_1+r_2-1$ where $r_1$ is the number of real embeddings and $r_2$ the number of complex embedding of $F$.

Thus, the result is pretty clear because $$ ({\cal O}^\times)^2\simeq (T)^2\times2\Bbb Z^r. $$