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}

which is stated in this article : https://www.dropbox.com/s/nlfn8ldodz4uvgb/Cyclotomic%20Polynomials.pdf. Now, let

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

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One Response to Number Theory #1 Sum of Roots of Unity

  1. hi says:

    Excellent. @lawrence, nice article!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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