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 $(O_K/\langle N\rangle)\cong(\mathbb{Z}/N\mathbb{Z})^d$.

Let $(p)=P\cap \mathbf Z$. As $P$ is prime, $p$ is a prime number, and we have a commutative diagram:
$$\begin{matrix}
\mathbf Z &\mkern-12mu\hookrightarrow\mkern-12mu&\mathcal O_K\\
\downarrow&&\downarrow\\
\mathbf Z/p\mathbf Z&\mkern-12mu\hookrightarrow\mkern-12mu &\mathcal O_K/P
\end{matrix}$$
$\mathcal O_K$  is a free $\mathbf Z$-module of rank $d=[K:\mathbf Q]$, so $\mathcal O_K/p\mkern1.5mu \mathcal O_K$ is a $\mathbf Z/p\mathbf Z$-vector space of dimension $d$, and its quotient $\mathcal O_K/P$ is a field extension of degree $\le d$. This proves  $\mathcal O_K/P$ is finite and
$$\lvert\mathcal O_K/P\rvert\le p^d.$$
