## Number Theory #1 Sum of Roots of Unity

This entry was written by Lawrence Sun.

Problem: Let $n,m$ be positive integers. Prove that the sum of $m$ $n^{th}$ roots of unity, whose sum is not $0$, has modulus at least $m^{-\varphi(n)}$.

Inspired by a result of Gerald Myerson.

Solution:
Consider the field $\mathbb{Q}[\omega_n]$ where $\omega_n$ denotes a primitive $n^{th}$ root of unity. Now, for each $\gcd(k,n) = 1$ denote $f_k$ the automorphism which maps $\omega_n$ to $\omega_n^k$. If we let $U_n$ be the set of residues modulo $n$ relatively prime to $n$ we have

$\text{Aut}(\mathbb{Q}[\omega_n]) = \{f_i\}_{i \in U_n}$

$k = \zeta_1 + \zeta_2 + ... + \zeta_m$

be our sum of roots of unity. Clearly $k \in \mathbb{Q}[\omega_n]$ as $\zeta_i \in \mathbb{Q}[\omega_n]$ for each $i$.

Now, I claim:

$\displaystyle\prod_{i \in U_n} f_i(k) \in \mathbb{Z}$

The proof of this is simple by the Fundamental Theorem of Galois Theory or the Fundamental Theorem of Symmetric Polynomials and is thus left to the reader. As a sketch of the elementary second proof, simply let:

$\displaystyle k = \sum_{i=0}^{\varphi(n) - 1} a_i \omega_n^i$

And then note that the product is a symmetric function in $\omega_n^i$ for each $i \in U_n$, and is thus an integer polynomial in terms of the symmetric sums of those numbers. But the symmetric sums are the coefficients of $\Phi_n(x)$, so the result follows.

Now we prove the product is nonzero. Suppose $f_i(k) = 0$ for some $i$. Take the inverse in the automorphism group of $f_i$, call it $f_i^{-1}$. But then $f_i^{-1}(0) = k$, absurd since an automorphism fixes the rationals. Thus the product has modulus $\ge 1$ as it is a nonzero integer.

We now show that each $i$, $|f_i(k)| \le m$. The proof of this is simple; simply remark that:

$\displaystyle|f_i(\zeta_1 + \zeta_2 + ... + \zeta_m)| = \left | \sum_{j=1}^m \zeta_j^i \right |$

$\Longrightarrow |f_i(\zeta_1 + \zeta_2 + ... + \zeta_m)|\le m$

Therefore it follows:

$\displaystyle \left |k \cdot \prod_{\underset{i \neq 1}{i \in U_n}} f_i(k) \right | \ge 1$

$\Longrightarrow \displaystyle|k| \ge \frac{1}{m^{\varphi(n)-1}}$

$\Longrightarrow \displaystyle|k| \ge m^{-\varphi(n)}$

as desired. $\square$

Extensions:
You can find much more sophisticated methods to find much stronger bounds on this mathoverflow thread : http://mathoverflow.net/questions/46068/how-small-can-a-sum-of-a-few-roots-of-unity-be