Algebraic dependence of some squares Let $a_1,...,a_n \in \mathbb{C}$ such that $a_1 + ... + a_n = 0$. Intuitively $a_1^2,...,a_n^2$ should be algebraically dependent over $\mathbb{Q}$ as well. As I understand it, this follows from the fact that $\mathbb{Q}(a_1,...,a_n)$ has transcendence degree at most $n-1$.
However, I would like to find a concrete algebraic relation between such elements, i.e. a polynomial $f$ with rational coefficients such that $f(a_1^2,...,a_{n-1}^2,(a_1 + ... + a_{n-1})^2) = 0$ for any choice of $a_1,...,a_{n-1}$.
For $n=3$, $f(x_1,x_2,x_3)=\sum{x_i^2} - 2 \sum_{i<j}x_ix_j$ works, but the obvious generalization fails for larger $n$. I'm not sure how to proceed (besides brute computation). Is it true that we can choose such a polynomial to be symmetric? If so, can this be used to find a solution?
 A: Let me reformulate your question slightly in a way that turns out to make it much easier: what you are looking for is a polynomial $g(a_1, a_2, \dots a_n)$ which is both


*

*divisible by $a_1 + a_2 + \dots + a_n$, and

*a polynomial in $a_1^2, a_2^2, \dots a_n^2$.



Exercise: $g$ has the second property if and only if it is invariant under changing the sign of each of its inputs, e.g. if and only if $g(a_1, a_2, \dots a_n) = g(\pm a_1, \pm a_2, \dots \pm a_n)$.

This suggests an obvious candidate for $g$ which works and which is in fact minimal, in that every candidate must be divisible by it, namely
$$g = \prod (a_1 \pm a_2 \dots \pm a_n).$$
(Invariance under changing the sign of $a_1$ is guaranteed by the fact that this product has $2^{n-1}$ factors in it, which when $n \ge 2$ is even. The case $n = 1$ is degenerate.)
When $n = 3$ this gives
$$\begin{eqnarray} g &=& (a_1 + a_2 + a_3)(a_1 + a_2 - a_3)(a_1 - a_2 + a_3)(a_1 - a_2 - a_3) \\
 &=& ((a_1 + a_2)^2 - a_3^2)((a_1 - a_2)^2 - a_3)^2) \\
 &=& (a_1^2 + 2 a_1 a_2 + a_2^2 - a_3^2)(a_1^2 - 2a_1 a_2 + a_2^2 - a_3^2) \\
 &=& (a_1^2 + a_2^2 - a_3^2)^2 - 4 a_1^2 a_2^2 \\
 &=& a_1^4 + a_2^4 + a_3^4 - 2 a_1^2 a_2^2 - 2 a_2^2 a_3^2 - 2 a_3^2 a_1^2 \end{eqnarray} $$
which agrees with your answer. A similar trick works for higher powers instead of squares, where instead of $\pm$ signs we have suitable roots of unity. 
