Finitely many ideals of bounded norm

Let $\mathfrak{a}$ be a nonzero integral ideal with $N\mathfrak{a}=m$; by [F1] the additive group ${\mathcal{O}}_K\mathrm{/}\mathfrak{a}$ has exactly $m$ elements, so by [F3] the order of each of its elements divides $m$, and for $x\in {\mathcal{O}}_K$ this gives $m(x+\mathfrak{a})=\mathfrak{a}$, that is $mx\in \mathfrak{a}$, so $m{\mathcal{O}}_K\subseteq \mathfrak{a}$.

Fix a $\mathbb{Z}$-basis $e_1,\dots ,e_n$ of ${\mathcal{O}}_K$, which exists by [F2]; every class of ${\mathcal{O}}_K\mathrm{/}m{\mathcal{O}}_K$ has exactly one representative ${\sum }_i{a}_i{e}_i$ with $0\le a_i<m$, because subtracting suitable multiples of $m$ from the coordinates gives existence, while ${\sum }_i{b}_i{e}_i\in m{\mathcal{O}}_K$ with $\mathrm{\mid }{b}_i\mathrm{\mid }<m$ for all $i$ forces each $b_i$ to be an integer multiple of $m$, hence $b_i=0$, giving uniqueness; thus $\mathrm{\mid }{\mathcal{O}}_K\mathrm{/}m{\mathcal{O}}_K\mathrm{\mid }=m^n$.

Conversely every nonzero integral ideal $\mathfrak{a}$ with $m{\mathcal{O}}_K\subseteq \mathfrak{a}$ has image $\mathfrak{a}\mathrm{/}m{\mathcal{O}}_K$ an ideal of the quotient ring ${\mathcal{O}}_K\mathrm{/}m{\mathcal{O}}_K$, and distinct ideals $\mathfrak{a},\mathfrak{b}$ containing $m{\mathcal{O}}_K$ have distinct images: if $\mathfrak{a}\mathrm{/}m{\mathcal{O}}_K=\mathfrak{b}\mathrm{/}m{\mathcal{O}}_K$ and $x\in \mathfrak{a}$, then $x+m{\mathcal{O}}_K\in \mathfrak{b}\mathrm{/}m{\mathcal{O}}_K$, so $x\in \mathfrak{b}+m{\mathcal{O}}_K\subseteq \mathfrak{b}$, and interchanging $\mathfrak{a}$ and $\mathfrak{b}$ gives equality.

A finite ring has only finitely many ideals, since its underlying set has only finitely many subsets, so for each fixed $m$ there are finitely many nonzero integral ideals with $N\mathfrak{a}=m$ by steps 2.1 and 1.2.

A nonzero integral ideal of norm $\le B$ has norm equal to one of the positive integers $m\le B$, of which there are only the finitely many values $1\le m\le \lfloor B\rfloor$, and a finite union of finite sets is finite, so only finitely many nonzero integral ideals satisfy $N\mathfrak{a}\le B$.