# 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 $$n=3$$, $$f(x_1,x_2,x_3)=\sum{x_i^2} - 2 \sum_{i works, but the obvious generalization fails for larger $$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?

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

1. divisible by $$a_1 + a_2 + \dots + a_n$$, and
2. 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.

• This is extremely neat, thank you! – Dániel G. Nov 6 '18 at 14:17