Question about norm of ideals of some ring of integers $\mathcal{O}$ I want to understand the proof of the theorem below and everything written in italic is my comments about some statements that are not clear for me. So, if you clarify  the parts that I wrote in italic and if its possible give another proof, I would appreciate it.

Theorem. Let $I$ be a non-zero ideal of $\mathcal{O}$. Then, 
i.We have that $N(\alpha\mathcal{O})=|N_{K/ \mathbb{Q}}(\alpha)|, \alpha\in\mathcal{O}$.
ii.The norm of $I$ is finite.
Proof.
i.$\mathcal{O}$ is a free Abelian group of rank $n= [K:\mathbb{Q}]$, thus, there exist a $\mathbb{Z}-$basis $\alpha_1,\dots,\alpha_n$ of $\mathcal{O}$ such that $\mathcal{O}=\alpha_1\mathbb{Z} \oplus\dots\ \oplus\alpha_n\mathbb{Z}$.
-Now, how can we write $\mathcal{O}=\alpha_1\mathbb{Z} \oplus\dots\ \oplus\alpha_n\mathbb{Z}$? I was thinking that we can write $\mathcal{O}= \bigoplus_{i=1}^{n}R$ where $R$ is some $\mathbb{Z}-$module so the summands would be the same.
Then, it is a general result(how?) on free Abelian groups that if $H$ is a subgroup of $G$, both of same rank with $\mathbb{Z}$-bases $x_1, \dots,x_n$ and $y_1,\dots,y_n$ respectively with $y_i=\displaystyle\sum_{j}a_{ij}x_j$, then $|G/H|=|det(a_{ij})|$.
-I have no idea about how to prove this 'general' result. I have found some questions related to this result but it would be great if you can clarify it.
We apply this theorem in our case where $G=\mathcal{O}$ and $H=\alpha\mathcal{O}$. Since one basis is obtained from the other by multiplication by $\alpha$, we have that $|\mathcal{O}/ \alpha\mathcal{O}|=det(\mu_\alpha)=|N_{K/ \mathbb{Q}}(\alpha)|.$
Is $\mu_\alpha$ the transformation matrix of the linear operator-field homomorphism- $\varphi : K \to \mathbb{C}$ where $\varphi(x) = \alpha x$?
ii. Let $0 \neq \alpha \in I$. Since $I$ is an ideal of $\mathcal{O}$ and $\alpha \mathcal{O} \subset I$, we have a surjective map $\mathcal{O}/ \alpha \mathcal{O} \to \mathcal{O} / I$. So, the result follows from part one.
Do we define the map as $\varphi(x + \alpha\mathcal{O})=x+I$, and how did we conclude the result? 

 A: I think this should clarify what you need here


*

*$\mathcal{O}_K=\{ \sum_{j=1}^n c_j \alpha_j , c_j \in \mathbb{Z}\}$ is a rank-$n$ free $\mathbb{Z}$-module (or free abelian group), which means we have the isomorphism of $\mathbb{Z}$-modules
$$\varphi : \mathbb{Z}^n \to \mathcal{O}_K,\qquad \varphi(c) =\sum_{j=1}^n c_j \alpha_j$$

*We didn't define the multiplication in $\mathcal{O}_K$ (ie. the result of $\alpha_i\alpha_j$). All we need to know is that the multiplication by any $\beta \in \mathcal{O}_K$ is given by some matrix $B \in \mathbb{Z}^{n \times n}$ :
$$\beta \sum_{j=1}^n c_j \alpha_j = \beta\varphi(c) = \varphi(B c)=\sum_{j=1}^n (Bc)_j \alpha_j$$

*And from the $\mathbb{Z}^n$ side we have the quotient of free $\mathbb{Z}$-modules formula
$$|\mathcal{O}_K/\beta\mathcal{O}_K|=|\varphi^{-1}(\mathcal{O}_K)/\varphi^{-1}(\beta\mathcal{O}_K)| = |\mathbb{Z}^n/B\mathbb{Z}^n| = |\det(B)| = |N_{K/\mathbb{Q}}(\beta)|$$
Note how changing the basis $\alpha_j$ by another one means replacing $B$ by $M B M^{-1}$ with $M,M^{-1} \in \mathbb{Z}^{n\times n}$ and hence $|\det(M)| = 1$.
