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A cyclotomic field is a number field obtained by adjoining a complex primitive root of unity to $\mathbb Q$, the field of rational numbers.

3 votes

which algebraic integers in a cyclotomic field give you integer absolute value?

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 …
4 votes

Ring of algebraic integers in a quadratic extension of a cyclotomic field

UPDATE: Previous argument was flawed. Here is what can salvage. I can show there is no solution with $n$ an odd prime, or with $n$ odd and $\omega$ cyclotomic. Let $\sigma$ denote complex conjuga …
David E Speyer's user avatar
5 votes

Realizability of a real representation using real cyclotomic coefficients

$\def\CC{\mathbb{C}}\def\RR{\mathbb{R}}\def\QQ{\mathbb{Q}}\def\ZZ{\mathbb{Z}}$The main result of this answer will be that, if $V$ is a representation defined over $\RR$ and over $\QQ(\zeta_M)$, then t …
David E Speyer's user avatar
3 votes
Accepted

How can I prove that a sequence of squares of graph norms is never cyclotomic?

This is a vague thought: is there some simple recurrence for the characteristic polynomials of the charctertistic polynomials of the corresponding matrices. For example, if you look at the A_n chains, …
David E Speyer's user avatar
6 votes
Accepted

Trace of n-th root of unity in cyclotomic extension of p-adic rationals

Certainly not in general: If $p \equiv 1 \bmod n$ (but $n \neq \pm 1$) then $\mathbb{Q}_p(\zeta_n) = \mathbb{Q}_p$) so $Tr(\zeta_n) = \zeta_n$ which is not in $\mathbb{Q}$. To give another example, ta …
David E Speyer's user avatar
4 votes
Accepted

How to find a cyclotomic polynomial of degree d that decompose into d irreducible polynomial...

Sorry, you are out of luck. Let $\phi_n$ denote the $n$-th cyclotomic polynomial. I will show that (1) $\phi_n$ factors completely into linear factors modulo $p$ (a prime) if and only if $n$ is of th …
David E Speyer's user avatar
9 votes
Accepted

Diophantine equation $\cos(2\pi x)\cos(2\pi y) = \cos(2\pi z)$

The solution to the Dipohantine equation: You have, indeed, found all the solutions. Put $\alpha = e^{2 \pi i x}$, $\beta = e^{2 \pi i y}$, $\gamma = e^{2 \pi i z}$, so you want $$(\alpha+\alpha^{-1}) …
David E Speyer's user avatar