Free abelian group of finite rank with subgroup of equal rank

Question

A theorem says that if $G$ is a free abelian group of finite rank $n$ and $H\leq$$G$ then $H$ is free of finite rank and $rank(H)\leq$$rank(G)$. But is it possible to build a free abelian group $G$ of finite rank $n$ sucht that $G$ has a proper subgroup $H$ of rank $n$? I don't see clear why this should be possible.

Answer 1

A concrete example from number theory is $\mathbb{Z}[\sqrt{m}] \hookrightarrow \mathbb{Z}_{\mathbb{Q}(\sqrt{m})}$ where $m \neq 0,1$ is a square-free integer and $\mathbb{Z}_{\mathbb{Q}(\sqrt{m})}$ is the ring of algebraic integers of the quadratic field $\mathbb{Q}(\sqrt{m})$. If $m=1 \pmod 4$ then $\mathbb{Z}_{\mathbb{Q}(\sqrt{m})}\cong \mathbb{Z}[\frac{1+\sqrt{m}}{2} ]$ so $\mathbb{Z}[\sqrt{m}]$ is a proper rank 2 subgroup of a rank 2 group.