Ideals of norm 2 in $\mathbb{Z}[\zeta_{n}]$ 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?
 A: 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.
A: 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...
