which algebraic integers in a cyclotomic field give you integer absolute value? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T03:59:12Z http://mathoverflow.net/feeds/question/106269 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106269/which-algebraic-integers-in-a-cyclotomic-field-give-you-integer-absolute-value which algebraic integers in a cyclotomic field give you integer absolute value? katie 2012-09-03T19:44:54Z 2012-09-09T12:15:32Z <p>Does anyone know an answer to this question? Question: In an cyclotomic field which algebraic integers have integer absolute value?</p> <p>Revision 1: -1</p> <p>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</p> http://mathoverflow.net/questions/106269/which-algebraic-integers-in-a-cyclotomic-field-give-you-integer-absolute-value/106275#106275 Answer by Joseph Victor for which algebraic integers in a cyclotomic field give you integer absolute value? Joseph Victor 2012-09-03T20:55:20Z 2012-09-04T00:24:36Z <p>There is a primitive way to do it, given $n$.<br> 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.</p> <p>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. </p> <p>Hope this helps. </p> http://mathoverflow.net/questions/106269/which-algebraic-integers-in-a-cyclotomic-field-give-you-integer-absolute-value/106693#106693 Answer by Gerry Myerson for which algebraic integers in a cyclotomic field give you integer absolute value? Gerry Myerson 2012-09-08T22:56:38Z 2012-09-08T22:56:38Z <p>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. </p> <p>$\newcommand\p{\mathfrak{p}}$ $\newcommand\Q{\mathbf{Q}}$</p> <p>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 <i> $n^2$-Weil numbers</i>. (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 <i> fix </i> an integer $m$, then there are only finitely many $m$-Weil numbers up to roots of unity in <i> all</i> 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 <i> ideal </i> $(\beta)$ 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.</p> <p>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:</p> <p>For every (totally) real unit $\eta$, the unit $\eta^N$ is of the form $\epsilon \cdot \overline{\epsilon}$ for some unit $\epsilon$.</p> <p>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 $F = \Q(\beta)$.</p> <p>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 $\epsilon \cdot \overline{\epsilon}$). </p> <p>(Given the subsequent comments of the OP, the indication was that this answer was not even read, so I will delete it.)</p> http://mathoverflow.net/questions/106269/which-algebraic-integers-in-a-cyclotomic-field-give-you-integer-absolute-value/106725#106725 Answer by David Speyer for which algebraic integers in a cyclotomic field give you integer absolute value? David Speyer 2012-09-09T12:15:32Z 2012-09-09T12:15:32Z <p>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.</p> <p>Let $p$ be prime and let $A \subseteq \mathbb{Z}/p$. Let $a=|A|$. Let $z=\sum_{k \in A} \zeta^k$. Let <code>$b_k = \# \{ (i,j) \in A^2 : i-j \equiv k \bmod p \}$</code> So <code>$$z \bar{z} = a + \sum_{k=1}^{p-1} b_k \zeta^k$$</code> 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 $n$.</p> <p>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: <code>$$p=\frac{(n^2+b)(n^2+b-1)}{b}+1 = \frac{(n^2+n+b)(n^2-n+b)}{b}.$$</code> (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.</p> <hr> <p>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 <a href="http://mathworld.wolfram.com/PerfectDifferenceSet.html" rel="nofollow">perfect difference set</a>, 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 $p$.</p> <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.</p> <p>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.</p>