Proving quotient of $\mathcal{O}_K $ with ideal $I$ is finite.

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I need help showing the following : Suppose $K$ is a number field of degree $n$ and $\mathfrak{a} $ is a non-zero ideal of $\mathcal{O}_K $. Then $\mathfrak{a} $ as an additive finitely generated abelian subgroup of $\mathcal{O}_K^+ $ has rank $n$.

Here is what I know

$\mathcal{O}_K$ Is a finitely generated abelian group of rank $n$. I know that $\mathcal{O}_K /I $ is a finite group. Now I have some feeling that if $I$ has rank $m<n $ then $(I,+) \cong \mathbb{Z}^m $ and this somehow means that the quotient is infinite as $\mathcal{O}_K^+ \cong \mathbb{Z}^n $ but don’t know how to precisely pin this down.

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Question: "Here is what I know: $O_K$ is a finitely generated abelian group of rank n. I know that $O_K/I$ is a finite group. Now I have some feeling that if $I$ has rank $m<n$ then $(I,+)≅Z^m$ and this somehow means that the quotient is infinite as $O^+_K≅Z^n$ but don’t know how to precisely pin this down."

Answer: You find a proof in Neukirch "Algebraic number theory", Proposition 2.10 (page 12).