Let $n$ be a prime number. Consider $\mathbb U_n$ which is the set of roots of unity.

Question

Let $n$ be a prime number. Consider $\mathbb U_n$ which is the set of roots of unity, i.e. the solutions of the equation $z^n=1$, more precisely $\mathbb U_n = ${$1,\epsilon , \epsilon^2 , ... , \epsilon^{n-1}$} , where $\epsilon = \cos \frac{2k \pi}{n}+i\sin{2k \pi}{n} ,k={0,1,2,...,n-1}$. My question is if $n$ is prime , then how many subsets of $\mathbb U_n$ have the property that the sum of all the elements from the subset is 0 ? I think the answer would be 1(the the whole set ), but I'm not sure and there might be more.I need someone's confirmation/contradiction. Thank you all !

Answer 1

If $n$ is prime, then each of $\epsilon^k$, $1\leq k\leq n-1$ is a root of

$$p(z)=z^{n-1}+z^{n-2}+\dots+z+1.$$ This polynomial is irreducible over $\mathbb{Q}$ and hence it is the minimal polynomial of $\epsilon$ over $\mathbb{Q}$.

Suppose for some elements of the set $A=\{\epsilon, \epsilon^2, \dots \epsilon^{n-1}, 1\}$, we get sum equal to zero, say $$\epsilon^{k_1}+\epsilon^{k_2}+\dots+\epsilon^{k_r}=0,$$ where all $k_i$ are distinct and $0\leq k_i\leq n-1$. Assuming without loss of generality that $k_1$ is the highest value, then this implies $\epsilon$ would satisfy another polynomial of degree $k_1 \leq n-1$. Hence, this new polynomial must be $p(z)$ itself as this new polynimial is monic and the minimal polynomial is always unique and hence the above sum must involve all the elements of the set $A$.