$\{aZ[\sqrt{d}]:N(a)=n\} $ is finite Prove that for any integer $ n$ the set of ideals of the form $ aZ[\sqrt{d}] $, where $ a\in Z[\sqrt{d}] $ has norm $ n $, is a finite set.
I was thinking that since $Z[\sqrt{d}]$ is Notherian,  $ (\sum_{i=1}^{n}a_iZ[\sqrt{d}])_{n\in N} $ is an increasing sequence of ideals will lead to a contradiction.  But then realized that if any  two of the ideal are comaximal/coprime then the sequence of sum  will be $Z[\sqrt(d)]$ after some point, and the contradiction doesn't exist.
Any hints?
 A: Let $K$ be a number field and $N_{K/\mathbb{Q}}$ be the field norm. Also let $\mathcal{O}_K$ be the ring of integers, and let $\mathcal{O}$ be an order in $\mathcal{O}_K$ (submodule of the same rank as $\mathcal{O}_K$, which also means it has finite index).
Define a group operation on the set $G = \{u\mathcal{O}:u \in \mathcal{O}_K^*\}$ by $(u_1\mathcal{O})(u_2\mathcal{O}) = (u_1u_2)\mathcal{O}$.
This is well-defined because if $u_1\mathcal{O} = u_1'\mathcal{O}$ then $u_1'/u_1$ is a unit in $\mathcal{O}^*$, and similarly for $u_2$ and $u_2'$, so $(u_1'u_2')/(u_1u_2)$ is also a unit in $\mathcal{O}^*$ and then $(u_1'u_2')\mathcal{O} = (u_1u_2)\mathcal{O}$. Associativity, identity and invertibility are trivial to show.
Consider the map $\varphi: \mathcal{O}_K^* \rightarrow G$ defined by $\varphi(u) = u\mathcal{O}$.
This is a surjective group homomorphism and its kernel is $\mathcal{O}^*$, thus $G \cong \mathcal{O}_K^*/\mathcal{O}^*$.
Now, the set of ideals of the form $a\mathcal{O}$ where $N_{K/\mathbb{Q}}(a) = \pm n$ is in bijection with $G$ by choosing a fixed element $a_0$ of norm $\pm n$ and mapping $a\mathcal{O}$ to $\frac{a}{a_0}\mathcal{O}$ (note that $N_{K/\mathbb{Q}}(a/a_0) = \pm 1$ and so $a/a_0$ is a unit in $\mathcal{O}_K^*$, and conversely every unit $u \in \mathcal{O}_K^*$ has norm $\pm 1$ so $a = ua_0$ has norm $\pm n$).
So the number of such ideals is equal to $|G| = |\mathcal{O}_K^*/\mathcal{O}^*|$.
For any order $\mathcal{O} \subset \mathcal{O}_K$, the group $\mathcal{O}_K^*/\mathcal{O}^*$ is finite (see Neukirch Theorem 12.12 in Chapter I).

You could also prove this as follows.
Let $m = [\mathcal{O}_K:\mathcal{O}]$, then $m\mathcal{O}_K \subset \mathcal{O}$ and $\mathcal{O}_K / (m\mathcal{O}_K)$ is a finite ring. Hence $(\mathcal{O}_K/m\mathcal{O}_K)^*$ is a finite group. Let $u \in \mathcal{O}_K^*$, then $u + m\mathcal{O}_K \in (\mathcal{O}_K/m\mathcal{O}_K)^*$.
By the Dirichlet unit theorem, there are finitely many fundamental units $u_1, ..., u_r \in \mathcal{O}_K^*$ such that any unit can be written as $\mu u_1^{e_1}...u_r^{e_r}$ where $\mu$ is a root of unity.
Hence from the previous remark, $u_i + m\mathcal{O}_K$ has finite order $k_i$ in $(\mathcal{O}_K/\mathcal{O})^*$ for each $i$. Pick some $k$ which is a common multiple of all $k_i$ and also of the order of the group $\mu(K)$ of roots of unity in $K$. Then for any unit $u \in \mathcal{O}_K^*$, we have $$u^k + m\mathcal{O}_K = (\mu + m\mathcal{O}_K)^k(u_1 + m\mathcal{O}_K)^k...(u_r + m\mathcal{O}_K)^k = 1 + m\mathcal{O}_K$$ which is the identity in $(\mathcal{O}_K/m\mathcal{O}_K)^*$. Hence, $u^k \in 1 + m\mathcal{O}_K \subset \mathcal{O}$.
Thus there is a finite $k$ such that $u^k \in \mathcal{O}^*$ for all $u \in \mathcal{O}_K^*$. Hence $\mathcal{O}_K^*/\mathcal{O}^*$ is a finitely-generated abelian group of finite exponent, which means it must be finite (by the structure theorem for finitely-generated abelian groups).

So we finally showed that the number of ideals in $\mathcal{O}$ of the form $a\mathcal{O}$ such that $N_{K/\mathbb{Q}}(a) = \pm n$ is finite (and clearly also the subset of those ideals with $N_{K/\mathbb{Q}}(a) = n$).
Note that the question here is the particular case where $K = \mathbb{Q}(\sqrt{d})$ and $\mathcal{O} = \mathbb{Z}[\sqrt{d}]$.
Also note that if $\mathcal{O} = \mathcal{O}_K$ then this is trivial because then $\mathcal{O}_K^*/\mathcal{O}^*$ is the trivial group, which in our specific example would occur if $d$ is squarefree and $d \equiv 2, 3 \pmod 4$. In fact, there will be only one distinct ideal $a\mathbb{Z}[\sqrt{d}]$ for any $n$.
