If domain $R$ is a free $\mathbb{Z}$-module, and $I$ is an integral ideal of $R$, have $R$ and $I$ the same rank as $\mathbb{Z}$-modules? If an integral domain $R$ is a free $\mathbb{Z}$-module, and $I$ is an ideal of $R$, do $R$ and $I$ have the same rank as $\mathbb{Z}$-modules? Assume $I$ is non-zero.

If $R$ is of finite rank, then yes. If $a$ is a nonzero element of $I$, then $Ra$ is a free abelian group inside $I$ of rank equal to that of $R$, so that $I$ has rank at least equal to that of $R$. As it has rank at most equal to that of $R$, we have equality.
