Sum of a proper subset of the $p^\text{th}$ roots of unity I know, because I have read it, that if $p$ is prime, no sum of a proper subset of the $p^{\text{th}}$ roots of unity (in $\mathbb{C}$) is zero. I thought I knew how to prove this, but found to my dismay when I tried that I do not. Can someone give me an elementary proof? By "elementary" I mean to say that I'm hoping for something that doesn't go too far off into the nether realms of algebra, since I'm only beginning to learn algebra now. If you leap right into Galois Theory or cyclotomic polynomials I'm lost. :-/ A geometric proof would be awesome...
Muchas gracias!
 A: First recall for a nontrivial $p$-th root of unity $\zeta$, that is $\zeta \neq 1$, you have that $\zeta^j$ with $j=0, \dots, p-1$ is the set of all $p$-th roots of unity. So if there are roots of unity summing to $0$, this means  $\sum_{j \in J}  \zeta^j = 0$ for some  subset $J$ of $\{0, \dots, p-1\}$. 
This means $\zeta$ is a root of $\sum_{j \in J}  X^j$. Yet the minimal polynomial of $\zeta$ is  $\sum_{j = 0}^{p-1}  X^j$. The minimal polynomial of an algebraic number (over the rationals) is the monic polynomial (with rational coefficients) of lowest degree that has this number as a root; equivalently, it is the irreducible monic polynomial having this number as a root. Each polynomial (with rational coefficients) having that algebraic number as a root must be  a multiple of the minimal polynomial. 
Since $\sum_{j \in J}  X^j$ cannot be a multiple of $\sum_{j = 0}^{p-1}  X^j$, except if they are equal, the claim follows.
What remains to check is that  $\sum_{j = 0}^{p-1}  X^j$ is the minimal polynomial. 
Since this is equal to $(X^p-1)/(X-1)$ it is clear that $\zeta$ is a root. It remains to show that it is irreducible. This can be done by applying Eisenstein's criterion to this polynomial shifted by $1$; see page 2 of Section 16 of some Algebra notes by Paul Garrett 
