What sums of roots of unity can be expressed "simply"?

Question

Given $\zeta_n = e^{\frac{2i\pi}{n}}$ for natural $n$,

When is $S = \zeta_{n_0}^{m_0} + \zeta_{n_1}^{m_1} + \dots = M\sqrt[r]{R}^k\zeta_{n}^m$, for rational $M, R, m$ and natural $r, k$?

So far I have found the following nontrivial results (and the respective reflections from raising summands to powers when appropriate)

$\zeta_n^0 + \zeta_n^1 + \dots + \zeta_n^{n-1} = 0$

$\zeta_{1} + \zeta_{3} = \zeta_{6}$

$\zeta_{1} + \zeta_{4} = \sqrt{2}\zeta_{8}$

$\zeta_{3} + \zeta_{6} = \sqrt{3}\zeta_{4}$

Are there any nontrivial $S$ involving $\zeta_5$ or $\zeta_7$? Is there a general characterization that prevents some terms from being part of $S$? Is there an area of number theory studying this?

Answer 1

The comment by Daniel Schepler led me to a solution, I think.

I'll write something up to have an answer posted, but there's bound to be lots of mistakes -- I think nothing fundamentally wrong though.

Reducing the Statement

First, the statement

When is $S = \zeta_{n_0}^{m_0} + \zeta_{n_1}^{m_1} + \dots = M\sqrt[r]{R}^k\zeta_{n}^m$, for rational $M, R, m$ and natural $r, k$?

can be greatly simplified.

1. Multiply the left-hand side by the inverse of $\zeta_{n}^m$. This is another root of unity, allowing us to describe all of $S$ with a subset:

$\zeta_{n_0}^{m_0} + \zeta_{n_1}^{m_1} + \dots = M\sqrt[r]{R}^k$

2. Let $M = \frac{A}{B}$. Multiply the left-hand side by $B$. $B$ is an integer, so again we focus on a subset:

$\zeta_{n_0}^{m_0} + \zeta_{n_1}^{m_1} + \dots = A\sqrt[r]{R}^k$

3. $A\sqrt[r]{R}^k$ = $A\sqrt[r]{R^k}$, so we absorb $A$ into the radical, $\sqrt[r]{A^rR^k}$, and the inside of the radical is another rational number $R'$ because $r, k \in \mathbb{Z}$ per the statement.

$\zeta_{n_0}^{m_0} + \zeta_{n_1}^{m_1} + \dots = \sqrt[r]{R'}$

4. We raise the left-hand side to the power of $r$, producing another sum of roots of unity, allowing us again to reduce to a subset of $S$. While we're at it, replace $R'$ by an integer $N$ obtainable by multiplying our sum by the denominator of $R$.

$\zeta_{n_0}^{m_0} + \zeta_{n_1}^{m_1} + \dots = N$

5. Finally, $N$ factors, so we shall consider sums that sum up to a prime, and we can combine those to obtain sums that sum up to a natural.

Thus our question is reduced to the $S'$ satisfying the following, and we can reconstruct $S$ by multiplication by integers (2)(4), exponentiation by naturals (4), or multiplication by roots of unity (1). Crucially, no addition is needed to reconstruct all $S$ from all $S'$, other than that resulting from exponentiation.

When is $S' = \zeta_{n_0}^{m_0} + \zeta_{n_1}^{m_1} + \dots = p$, for odd prime $p$?

Classifying in Terms of Gauss Sums

EDIT: The following was written via analyzing Sage computations while I didn't understand this well. I don't understand it well enough to correct it yet, but I warn that it's far more confusing than it really is.

The theory of Gauss sums allows us to assign to every primitive Dirichlet character a sum of roots of unity modulo the modulus of the character. The Wikipedia page goes into excellent detail about how to do so.

Every Gauss sum of a primitive Dirichlet character $\chi_p$ has magnitude $\sqrt{p}$. For every odd prime $p$ there is a unique Dirichlet character (sending $2$ to $-1$) $\chi_p(p-1, ⋅)$ with purely real or purely imaginary Gauss sum:

$G(\chi_p(p-1, ⋅)) = \zeta_p^{c_0} + \zeta_p^{c_1} + ... = \sqrt{p}\zeta_4^k$ for some $c_j$ and $k$. (The $c_j$ and $k$ are known, but no need to specify them here.)

Multiply the sum by the inverse of $\zeta_4^k$ and square it. This leads to unique $S'(p)$ for each odd prime $p$:

$S'(p) = \zeta_4^{-k}G(\chi_p(p-1, ⋅))^2 = \zeta_p^{c_0} + \zeta_p^{c_1} + \dots$

We can reconstruct any $S$ from $S'$ via the operations described earlier.

This allows us to assign to every $p$ a corresponding set $\{a, b, c, \dots\}$ specifying the powers of the roots of unity present in the square of its sum, together with a modulus $M$ corresponding to the "LCM" of the summands. For convenience we'll specify the sum that adds to the prime's square root and add the power of $\zeta_4$ as a multiple. It also turns out that every root of unity in the sum is doubled except for a singular $1$, so we'll skip it.

$\sqrt{p}$	$M$	$\zeta_4^k$	$1 + 2\zeta_M^a + 2\zeta_M^b + \dots$
$\sqrt{3}$	$6$	$3$	$4$
$\sqrt{5}$	$5$	$0$	$1, 4$
$\sqrt{7}$	$42$	$3$	$1, 11, 25, 29, 30$
$\sqrt{11}$	$110$	$3$	$1, 20, 23, 60, 67, 70, 80, 89$
$\sqrt{13}$	$156$	$2$	$\dots$
$\sqrt{17}$	$272$	$2$	$\dots$
$\sqrt{19}$	$342$	$3$	$\dots$
$\sqrt{23}$	$506$	$1$	$\dots$