# Polynomial of squares vs polynomial of square root

Let $$f\in \mathbb{R}[X_1,\ldots,X_n]$$ be a polynomial in $$n$$ variables such that for all $$(a_1,\ldots,a_n),(b_1,\ldots,b_n) \in \mathbb{R}_{>0}^n$$ with $$f(a_1,\ldots,a_n)=0=f(b_1,\ldots,b_n)$$ the following holds $$f(a_1 b_1,\ldots,a_n b_n)=0.$$ Let $$(a_1,\ldots,a_n) \in \mathbb{R}_{>0}^n$$ with $$f(a_1,\ldots,a_n)=0$$. We can follow that $$f(a_1^2,\ldots,a_n^2)=0$$ holds (for $$b=a$$). Is it also true that $$f\left(\sqrt{a_1},\ldots,\sqrt{a_n}\right) = 0\text{ ?}$$

The motivation comes from real algebraic Lie groups, where the entries in a diagonal matrix have to satisfy certain polynomials. An example for such a polynomial comes from the determinant $$=1$$ -condition and is given by $$f(X_1,\ldots,X_n)=\prod_{i=1}^nX_i -1 .$$ I have only managed to prove it for $$n=1$$, but I think it should always be true.

• The case for $n=1$ is so vastly different/simpler than I can't really see it being indicative of much for higher values of $n$. – Fimpellizieri Jan 16 at 14:19
• You are correct. I think that it should be true because algebraic Lie groups should have roots. – Strichcoder Jan 16 at 14:21

For the case $$N=2$$. Write $$f(x,y)=\sum_{i,j} c_{i,j}x^iy^j$$. If $$f(a,b)=0$$, then by assumption $$f(a^n,b^n)=0$$ for all $$n\in \mathbb{N}$$. That is $$\sum_{i,j}c_{i,j} (a^{i}b^{j})^n=0$$ for all $$n\in \mathbb{N}$$. We can rearrange the indices to write the last equation in the simplified form $$\sum_i c_i \alpha_i^n=0.$$ If the $$\alpha_i$$'s are distinct, then the corresponding Vandermonde matrix is invertible, so $$c_i=0$$. Hence $$c_{i,j}=0$$, and consequently $$f=0$$. If the $$\alpha_i$$'s are not distinct, then $$a^ib^j=a^kb^l$$ for some $$i,j,k,l$$. That is $$(a,b)=(a,a^t)$$ for some $$t=p/q\in \mathbb{Q}$$, $$q>0$$. Consider (eventually a Laurent polynomial) $$g(x)=\sum_{i,j} c_{i,j} x^{qi+pj}.$$ We have $$g(a^n)=0$$ for all $$n\in \mathbb{N}$$. If $$a\neq 1$$, then $$g$$ has infinitely many roots, so $$g=0$$. In particular $$f(\sqrt{a},\sqrt{b})=f(\sqrt{a},a^{t/2})=g(a^{1/2q})=0.$$
Thus, in general, if $$f(a^n,b^n)=0$$ for all $$n\in\mathbb{N}$$ then $$f(\sqrt{a},\sqrt{b})=0$$.
I don't see how to generalize this to $$N>2$$. As for the examples of polynomials satisfying the above property, there is the family $$f_{k,l}(x,y)=x^k-y^l$$. Is there any other that is nontrivial?