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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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Do you have a definition of the index of an ideal on hand? I've not heard that terminology with rings. –  Mario Carneiro Dec 12 '12 at 10:11
    
Neither have I TBH. If $I$ is an ideal and $R$ a commutative ring let the index of $I$ be $|R/I|$ if it is finite and $\infty$ otherwise. –  Jackson Walters Dec 12 '12 at 10:13
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1) holds in every number field if you replace $\mathbb Z[\zeta_n]$ with the ring of integers and $N(\alpha)$ with $|N(\alpha)|$. What you call index is the well known concept of the (absolute) ideal norm. Thus you are looking for ideals of norm 2. Using the unique prime ideal factorization, one can answer 2) –  Hans Giebenrath Dec 12 '12 at 10:14
    
Are you sure the Eisenstein integers have no ideals of index 2? What about $(1-\omega)$? (Edit: NVM, this is index 3.) –  Mario Carneiro Dec 12 '12 at 10:16
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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. en.wikipedia.org/wiki/Field_norm –  fretty Dec 12 '12 at 10:59

2 Answers 2

up vote 4 down vote accepted

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! –  Jackson Walters Dec 12 '12 at 21:08

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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I should maybe have justified more why $A = 2\mathbb{Z}[\zeta_n]$. Basically the $f$ as being the order of $2$ mod $n$ is the same as the inertia degree of the underlying prime ideals in the factorisation of $2\mathbb{Z}[\zeta_n]$. So for there to be such a prime ideal with $f = \phi(n)$, we must have $\phi(n)/\phi(n) = 1$ prime ideal in the factorisation. When $n$ is odd $2$ cannot ramify, so we really do get that $A = 2\mathbb{Z}[\zeta_n]$, for even $n$ you might get a non-trivial power of a prime ideal but I am convinced this case doesn't work. –  fretty Dec 12 '12 at 11:16
    
This is a bit over my head as I haven't had a course in ANT yet, but I can more or less follow along. Does all of this imply that 2 being a generator for $(\mathbb{Z}/n\mathbb{Z})^{\times}$ is a necessary condition for $A$ to be an ideal of norm 2 in $\mathbb{Z}[\zeta_{n}]$? –  Jackson Walters Dec 12 '12 at 11:23
    
Actually no, I have made a mistake...clearly this result does not agree with what you did for $n=3,5$. I think there is no such ideal for $n$ odd since you would require $2$ to split completely, and this will never happen (since then $2$ would have to have order $1$ which is false). –  fretty Dec 12 '12 at 11:44
    
For even $n$ you do get solutions since now $2$ ramifies and it is possible to get $f=1$. Programatically speaking for a given even $n$ the number of such ideals of norm $2$ will be the number of linear factors in the factorisation of the $n$th cyclotomic polynomial when reduced mod $2$ (not counting repetitions). –  fretty Dec 12 '12 at 11:49
    
I can't find my error in my answer yet but hopefully someone else on here will point it out soon! –  fretty Dec 12 '12 at 11:50

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