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.

I believe so. Let $\{b_1,\cdots,b_n\}$ be a $\Bbb Z$-basis for $R$, and let $w\in I$, nonzero element. Then $\{wb_i\}$ are still $\Bbb Z$-linearly independent, so the rank of $I$ is at least $n$. And at most $n$ as well, certainly.
