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I have a few questions about ideals the ring of integers $\mathbb{Z}[\zeta_{n}]$ in a cyclotomic number field. Specifically, I'm trying to classify the ideals of norm 2.

I know that the Gaussian integers are a PID and for any ideal $(a)$ where $a=x+iy$ we have that $|\mathbb{Z}[i]/(a)|=N(a)=N(x+iy)=x^{2}+y^{2}$. Therefore the only ideal of norm 2 is $(1+i)$.

With a little work I was able to prove that this also holds for the Eisenstein integers with $N(x+\omega y)=x^{2}+y^{2}-xy$. This time $x^{2}+y^{2}-xy=2$ has no integer solutions, so conjecturally $\mathbb{Z}[i]$ has no ideals of norm 2.

In $\mathbb{Z}[\zeta_{5}]$ the norm can be written $N(\alpha)=\frac{1}{4}(A^{2}-5B^{2})$ for integers $A$ and $B$. $N(\alpha)=2 \Rightarrow A^{2}-5B^{2}=8 \Rightarrow A^{2} \equiv 3 \pmod 5$. But 3 is not a quadratic residue modulo 5, so there are no ideals of norm 2.

It seems like the $n$ for which $\mathbb{Z}[\zeta_{n}]$ is a Euclicean domain was a tough question and that there are 46 such $n$. For such $n$, my questions are these:

1) Is it true that $|\mathbb{Z}[\zeta_{n}]/(\alpha)|=N(\alpha)$?

2) For which $n$ does $\mathbb{Z}[\zeta_{n}]$ have ideals of norm 2? How many?

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    There are more general norms on extensions of number fields which when applied to these few cases gets you the ones you know (however these more general norms dont turn the ring of integers into$a$Euclidean domain. http://en.wikipedia.org/wiki/Field_norm2012-12-12

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Theorem: There exists a prime ideal $I$ of $\mathbf{Z}[\zeta_n]$ of norm 2 if and only if $n$ is a power of 2, in which case there is exactly one such ideal.

Proof: Suppose $n$ is not a power of 2. Then there is a prime factor $\ell > 2$ of $n$. By the transitivity of the norm map, the norm of $I$ from $\mathbf{Z}[\zeta_n]$ to $\mathbf{Z}[\zeta_\ell]$ is an ideal of $\mathbf{Z}[\zeta_\ell]$ whose norm down to $\mathbf{Z}$ is 2; so we may assume $n = \ell$ is an odd prime. As in fretty's answer, $I$ must be one of the prime ideals of $\mathbf{Z}[\zeta_\ell]$ dividing $(2)$; these are all Galois-conjugate and there are $(\ell - 1) / f$ of them where $f$ is the order of $2$ modulo $\ell$, and the product of their norms is $2^{\ell-1}$, so each must have norm $2^f$. So none can have norm 2, as $\ell > 1$ so the order of 2 modulo $\ell$ cannot be 1.

Conversely, suppose $n$ is a power of 2. Then the ideal generated by $\zeta_n - 1$ has norm 2, and 2 is totally ramified in the field, so this ideal is the unique ideal of norm 2. QED.

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    Great answer, thanks!2012-12-12
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Notice that if an ideal $A = P_1^{e_1}P_2^{e_2} ... P_g^{e_g}$ has norm $2$ then $g=1$ and $e_i=1$ for all $i$ (by unique factorisation in $\mathbb{Z}$ along with the fact that the norm of a proper ideal cannot be $1$). The inertia degree is then $f = \phi(n)$.

Thus $A$ is a prime ideal. But since $2 = N(A)\in A$, it must be that $A|2\mathbb{Z}[\zeta_n]$. So $A$ must be a prime ideal in the factorisation of $2\mathbb{Z}[\zeta_n]$.

But by basic theorems in algebraic number theory, in cyclotomic fields $2\mathbb{Z}[\zeta_n]$ will factorise into $\phi(n)/f$ prime ideals, where $f$ is the order of $2$ mod $n$ (when $n$ is odd). For even $n$ you have ramification to contend with but similar results hold.

Now for $2\mathbb{Z}[\zeta_n]$ to have a prime ideal divisor $A$ satisfying the above we must have $A = 2\mathbb{Z}[\zeta_n]$, thus $2\mathbb{Z}[\zeta_n]$ is prime. In other words (when $n$ is odd) we must have that $2$ generates $\left(\mathbb{Z}/n\mathbb{Z}\right)^{\times}$. But there are only certain $n$ for which this can happen...

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    Yes, this is what I realised in the comments. I was being silly I think. So yes, the only way you can construct ideals of norm $2$ is to look at $n$ that are divisible by $p$ (so that $p$ is ramified).2012-12-15