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Does anyone know an answer to this question? Question: In an cyclotomic field which algebraic integers have integer absolute value?

Revision 1: -1

I like to add this to the above question, Let's take w to be a primitive n-th root of unity, for which set of exponents A of {0,1,...n-1} we have the absolute value of the sum_{i \in A} w^i is an integer. this might not be any help to make it solvable but at least avoid some repetitions

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To begin with, everything of the form $n\zeta^r$, where $n$ and $r$ are integers, and $\zeta$ is the root of unity generating the field. –  Gerry Myerson Sep 3 '12 at 22:45
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My guess would be (1) roots of unity (2) elements in imaginary quadratic subfields and (1+2) things which can be written as products of the above. Can anyone prove it? –  David Speyer Sep 3 '12 at 22:56
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Possible counterexample: $z^2$ where $z = 1 + \zeta^2 + \zeta^5 + \zeta^6$ and $\zeta$ is a $13$th root of unity. There is no imaginary quadratic subfield of ${\bf Q}(\zeta)$, but $|z^2| = 3$ because $\lbrace 0, 2, 5, 6 \rbrace$ is a perfect difference set $\bmod 13$. –  Noam D. Elkies Sep 4 '12 at 1:09
    
Nice! Well, that explains why I couldn't prove it. I guess it could still be a product of quadratic imaginaries coming from, for example $\mathbb{Q}(\sqrt{-3})$ and $\mathbb{Q}(\sqrt{-39})$, but even if that is true the methods of proof I was trying wouldn't have found it. –  David Speyer Sep 4 '12 at 1:39
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If $a$ and $b$ are positive integers for which $a+b$ is the square of an integer, then $\sqrt{a}+i\sqrt{b}$ is an algebraic integer in a cyclotomic field whose absolute value is an integer. More generally, the problem reduces to solving $x^2+y^2=z^2$ in real algebraic integers $x,y$ belong to a cyclotomic field, and a rational integer $z$. –  Richard Stanley Sep 4 '12 at 13:53

3 Answers 3

What follows is not mine. It was posted by Scaroth, and then deleted by Scaroth, with the explanation, "Given the subsequent comments of the OP, the indication was that this answer was not even read, so I will delete it." However, the answer was read by at least 9 people who gave it upvotes. Whether or not it helped OP, several people found it helpful. I have no way to contact Scaroth to ask him/her to reconsider, so I will simply repost it.

$\newcommand\p{\mathfrak{p}}$ $\newcommand\Q{\mathbf{Q}}$

Denote the cylotomic field by $F$. Since complex conjugation commutes with every automorphism of $G = \mathrm{Gal}(F/\mathbf{Q})$, it follows that complex conjugation preserves absolute values in $F$. Hence you are asking exactly for a classification of so-called $n^2$-Weil numbers. (By definition, $m$-Weil numbers are algebraic integers which have absolute value $\sqrt{m}$ for every complex embedding.) It is generally thought that if you fix an integer $m$, then there are only finitely many $m$-Weil numbers up to roots of unity in all cyclotomic fields simultaneously, but this is not yet known. On the other hand, if you fix $F$ and increase $m$, such numbers are easy to come by. Indeed, it's easy to find such $\beta$ so that the Galois group $G$ of $F$ acts faithfully on the conjugates of not only the element $\beta$, but the ideal $(\beta)$ as well (which is not the case for $(\zeta \alpha)$ for a root of unity $\zeta$ whenever $\mathbf{Q}(\alpha) \ne F$.) The answer to your question, therefore, is "there are a lot of them." There's no easy classification of Weil numbers, however, since otherwise the conjecture mentioned above would be known. If there's a more precise property of the set of numbers that you are interested in, please ask a follow up question.

Recall that, in a cylcotomic field (more generally, a CM field), the real units have finite index inside $\mathcal{O}^{\times}_F$. It follows that the group of units of the form $\epsilon \cdot \overline{\epsilon}$ has finite index, and hence that there exists a integer $N$ such that:

For every (totally) real unit $\eta$, the unit $\eta^N$ is of the form $\epsilon \cdot \overline{\epsilon}$ for some unit $\epsilon$.

Let us now construct some Weil numbers. Suppose that the prime $p$ splits completely in $F$. There are $[F:\mathbf{Q}] = 2d$ such primes, and thus $2^d$ ways to choose a set $S$ of $d$ of primes which includes exactly one from each pair $\{\p,\overline{\p}}$. Let $$I = \prod_{S} \p.$$ Then $I \overline{I} = (p)$. Let $h$ denote the class number of $F$, and write $I^h = (\alpha)$. Then $$\alpha \overline{\alpha} = p^h \eta$$ for some totally real unit $\eta$. It follows that $$\alpha^{N} \overline{\alpha}^N = p^{hN} \epsilon \overline{\epsilon},$$ and hence that $\beta:=\alpha^N/\epsilon$ has absolute value $p^{hN}$. Moreover, from the factorization of $\beta$, we see that the ideal $(\beta)$ is only fixed by $H \subset \mathrm{Gal}(F/\mathbf{Q})$ which fix the set $S$. In particular, if $|G| > 4$, one can choose $S$ such that the stabilizer of $\beta$ is trivial and thus $F = \Q(\beta)$.

There are many variations on the above argument. For example, one can probably find $p$-Weil numbers for prime $p$ by finding primes $p$ which split completely in some ray class field of conductor $M \infty$ (designed so that totally positive units which are $1 \mod M$ are of the form $\epsilon \cdot \overline{\epsilon}$).

(Given the subsequent comments of the OP, the indication was that this answer was not even read, so I will delete it.)

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Thank you, Gerry. I also thought it was unfortunate that scaroth deleted his answer, so that only those with more than 10k points could see it. The OP is not the only one who can benefit. –  Todd Trimble Sep 8 '12 at 23:10
    
Good heavens, who voted this down?! Gerry was being very helpful. –  Todd Trimble Sep 8 '12 at 23:43
    
@Todd By the way, I think scaroth is female. I could be wrong, but might as well not assume. And I agree that scaroth's answer was very helpful. –  David Speyer Sep 9 '12 at 12:48
    
@David: yes, of course you're right. An unfortunate slip on my part. –  Todd Trimble Sep 9 '12 at 14:51

Katie asks further about the case where we have a subset $A$ of $\mathbb{Z}/n$ and asks for $\sum_{k \in A} \zeta^k$ to have integer absolute value. She writes that, for $n$ prime, the only solutions should be the trivial ones $|A| =0$, $1$, $p-1$ or $p$. The point of this note is that she is correct, and that there probably isn't a good description like this for $n$ not prime. This answer is built on Noam Elkies's very helpful comment above.

Let $p$ be prime and let $A \subseteq \mathbb{Z}/p$. Let $a=|A|$. Let $z=\sum_{k \in A} \zeta^k$. Let $b_k = \# \{ (i,j) \in A^2 : i-j \equiv k \bmod p \}$ So $$z \bar{z} = a + \sum_{k=1}^{p-1} b_k \zeta^k$$ Since the minimal polynomial of $\zeta$ is $1+\zeta+\zeta^2+\cdots+\zeta^{p-1}$, the only way that $z \bar{z}$ can be an integer is if all the $b_k$ are equal to some common value, say $b$. In this case, $z \bar{z} = a-b$. Furthermore, we want $\sqrt{z \bar{z}}$ to be an integer, say $n$. So $a=n^2+b$ for some nonnegative integer $n$.

Now, we must have $a(a-1) = b(p-1)$ since $\sum_{k=1}^{p-1} b_k$ is clearly $a(a-1)$. So $(n^2+b)(n^2+b-1) = b(p-1)$. If $b=0$ then $|A|=0$ or $1$. If not, we can divide by $b$ to write: $$p=\frac{(n^2+b)(n^2+b-1)}{b}+1 = \frac{(n^2+n+b)(n^2-n+b)}{b}.$$ (That factorization came out of nowhere, as far as I'm concerned.) Since $p$ is prime, that means that at least one of $n^2+n+b$ and $n^2-n+b$ is $\leq b$. But $n^2 \pm n \geq 0$ for nonnegative integer $n$. So $n=0$ or $1$. This gives $p=b$ and $p=b+2$, and then $a=p$ and $a=p-1$ respectively.


Now, if $n$ is not prime, then there are lots of sets $A$ such that all the $b_k$ are equal. The special case $b=1$ is called a perfect difference set, and they are pretty plentiful. For example, $(0,5,6,9,19)$ is a perfect difference set modulo $21$, giving $$|1+\zeta_{21}^5+\zeta_{21}^6+\zeta_{21}^9+\zeta_{21}^{19}| = \sqrt{5-1} = 2.$$ According to the mathworld link above, there are perfect difference sets modulo $q^2+q+1$ for every perfect power $q$; taking $q=p^2$ we deduce that we can always find $A$ in $\mathbb{Z}/(p^4+p^2+1)$ with absolute value $p$.

Based on skimming the results of a quick google search, my impression is that there are lots of methods known for constructing perfect difference sets, but no classification. Probably someone has studied the case $b>1$, but I didn't see it.

But even if there were a classification, that wouldn't be a complete answer. Because $1+\zeta+\cdots + \zeta^{n-1}$ is not the minimal polynomial of $\zeta$ for $n$ composite. And I don't see how to control the other solutions that might come up because of this.

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There is a primitive way to do it, given $n$.
You can of course an element of such a ring of integers can be written (non-uniquely, but this does not matter) as $\sum a_i\zeta_n^i$, and the absolute value then is $\sum_{i,j} a_ia_jcos(2\pi(i-j)/n)$. So just figure the relations in these cosine values over $\mathbb{Z}$ (for a specific value of $n$), and then write down some relations you need between the $a_i$'s and $a_j$'s to make sure you get an integer in the end.

Of course, using a basis instead of a spanning-set simplifies these relations, but it doesn't matter: you can just write the relations and then decompose any element this way and just check the relations.

Hope this helps.

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As a slight modification, which avoids handling the cosines, I suggest the following: Set $a=\sum a_i\zeta^i$. Then $b=a\bar a$ is an algebraic integer, and it is an integer if and only if $b^\sigma=b$ for each $\sigma$ in the Galois group of $\mathbb Q(\zeta)/\mathbb Q$. But this Galois group acts by replacing $\zeta$ with $\zeta^k$ for $k$ prime to $n$, so $\sum a_i\zeta^i\sum a_i\zeta^{-i}=\sum a_i\zeta^{ik}\sum a_i\zeta^{-ik}$ for each such $k$. Introducing a basis then gives a system of quadratic equations of the $a_i$. (Isn't the question actually whether $\sqrt{b}$ is an integer?) –  Peter Mueller Sep 3 '12 at 21:20
    
For $n=3$ all elements are integers because the abolute value is just the algebraic norm, which is always an integer. The thing you are making sure of is that $(a_0a_1+a_0a_2+a_1a_2)+(a_0a_1+a_0a_2+a_1a_2)$ is even, which of course it is. –  Will Sawin Sep 3 '12 at 21:21
    
Also your formulas for the cosines are wrong. For $n=3$ the cosine is either $-1/2$ or $1$, and for $n=5$ it is $1$ or $(1\pm \sqrt{5})/4$, not the other way around. –  Will Sawin Sep 3 '12 at 21:22
    
Actually as Peter points out for $n=3$ you need to know that the algebraic norm is a perfect square, which ends up being something about the ideal generated by the number that you can check locally at each prime. –  Will Sawin Sep 3 '12 at 21:24
    
Oops, I screwed up, let me fix it. –  Joseph Victor Sep 4 '12 at 0:22

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