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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 $\mathbb{Z}[w]/(a)\cong \mathbb{Z}_{N(a)}$?

Bill Dubuque
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Rubertos
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1 Answers1

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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.

MooS
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  • How do I show that it has $N(a)$ elements? – Rubertos Apr 13 '15 at 08:39
  • By the proof (with presentations and smith normal form) of the structure theorem of finitely generated abelian groups, we deduce that the number of elements in $\mathbb Z[\omega]/(a)$ is precisely the absolute value of the determinant of the multiplication with $a$ on $\mathbb Z[\omega]$. This determinant can easily be checked to be the norm of $a$. This is trivial if we have $a \in \mathbb Z$. If we have $a \notin \mathbb Z$, note that we can compute the determinant after switchting to the quotient field. There you can use the basis ${1,a}$ to compute the determinant. – MooS Apr 13 '15 at 09:11
  • @Rubertos If this presentation is too abstract then see here for a more concrete way to proceed (see esp. the link to Franz Lemmermeyer's notes). – Bill Dubuque Apr 13 '15 at 15:15