How do I prove that $Z[w]/(a)\cong Z_{N(a)}$? Define $S=\{\sqrt{D}:D\equiv 2,3 \pmod 4\}$
Define $T=\{\frac{1-\sqrt{D}}{2}:D\equiv 1 \pmod 4 \}$.
Let $w\in S\cup T$.
Let $N:\mathbb{Z}[w] \rightarrow \mathbb{Z}$ be the norm.
Then how do I prove that $\mathbb{Z}[w]/(a)\cong \mathbb{Z}_{N(a)}$?
 A: The assertion is false: $\mathbb Z[\omega]/(a)$ may not be cyclic, for example $\mathbb Z[i]/(3)$ is not the cyclic group with $9$ elements, since it is the field with $9$ elements, whose additive group is isomorphic to $\mathbb Z/3\mathbb Z \times \mathbb Z/3\mathbb Z$.
Maybe you only want to show that $\mathbb Z[\omega]/(a)$ has $N(a)$ elements, which is true for any $a \in \mathbb Z[\omega]$.
Let me show this:
By the proof of the structure theorem of finitely generated abelian groups, we get a commutative diagram of exact sequences:
$\require{AMScd}$
\begin{CD}
0@>>>\mathbb Z[\omega] @>\cdot a>> \mathbb Z[\omega] @>>> \mathbb Z[\omega]/(a) @>>> 0\\
& @VSVV @VTVV @VVV\\
0 @>>> \mathbb Z^2 @>M>> \mathbb Z^2 @>>> A @>>> 0 \\
\end{CD}
where $S$ and $T$ are matrices in $GL_2(\mathbb Z)$ and the isomorphism $\mathbb Z[\omega]/(a) \cong A$ is induced by abstract nonsense. $M$ can be chosen to be a diagonal matrix.
We conclude $|Z[\omega]/(a)|=|A| = |\mathbb Z^2/im(M)|$. We have $\mathbb Z^2/im(M)=\mathbb Z/k\mathbb Z \times \mathbb Z/l\mathbb Z$, where $k,l$ are the diagonal entries of $M$. So $|Z[\omega]/(a)|=|kl|=|\det M| = |\det (\cdot a)|$. The least equality holds because $S,T \in GL_2(\mathbb Z)$.
Showing $\det (\cdot a) = N(a)$ is an easy exercise, I gave you some hints in the comments.
