if $P$ is a prime ideal of $O_K$, then $O_K/P$ is finite let $P$ be a non-zero prime ideal of $O_K$, where $K$ is a number field (i.e. the degree $[K:\mathbb{Q}]$ is finite) then $O_K/P$ is finite. I'm working through a proof for this claim, however there is some group theory used in the proof which I don't understand.
Choose $\alpha\in P$, such that $\alpha\neq 0$, then $N=|Nm(\alpha)|=\alpha\Pi_{i=2}^d \phi_i(\alpha)$ where $\phi_i$ are the embeddings for $\alpha$, letting $\phi_1$ be the identity map. So $N=\alpha\beta$, where both $\alpha,\beta\in O_k$. Therefore $N\in P$, by definition of an ideal. So $\langle N\rangle\subseteq P$ and $(O_K/P)\subseteq (O_K/\langle N\rangle)$.
I understand everything up until this point, but now $O_K\cong \mathbb{Z}^d$, where $d$ is the degree of the minimal polynomial of $\alpha$. But I don't understand where this result comes from. Next the proof says that $(O_K/\langle N\rangle)\cong(\mathbb{Z}/N\mathbb{Z})^d$, which is finite, hence $(O_K/P)\subseteq (O_K/\langle N\rangle)$ must also be finite. Since $\langle N\rangle = NO_K$, does this mean that $\langle N\rangle\cong N\mathbb{Z}^d$, and then can we jump to the conclusion that $(O_K/\langle N\rangle)\cong(\mathbb{Z}/N\mathbb{Z})^d$.

By the theorem for finitely generated modules over a PID we know that there exists an integral basis $x_1,\ldots ,x_n$ of $\mathcal{O}_K$ such that $a_1x_1,\ldots ,a_nx_n$ is an integral basis of the ideal $P$, for some integers $a_i$. This gives a group homomorphism
$$
\mathcal{O}_K/P\simeq \mathbb{Z}/a_1\mathbb{Z}\times \cdots \times \mathbb{Z}/a_n\mathbb{Z},
$$
so that $\mathcal{O}_K/P$ has $\mid a_1\cdots a_n\mid$ elements, and thus is finite.
