I noticed this while solving another problem on this site.

Let $P(x)$ be a polynomial in $x$ with integer coefficients, and let the roots of $P(x)=0$ be $r_1, r_2 \ldots ,r_n$, where multiple $r_i$ might be equal if there are roots with multiplicity higher than one. Let Q(x) be some other polynomial in $x$, also with integer coefficients.

Prove that $$ \prod_i Q(r_i) \in \Bbb{Z} $$

For example, if $P(x) = x^5+2x^2+1$ and $Q(x) = x^2-2$ then $\prod Q(r_i) = -7$.

I am pretty sure it is true, because you can express each term in the product of those polynomials in a form like
$$ \sum_{i<j<\ldots <n} r_i^{p_1} r_j^{p_2} \ldots $$ and laboriously express those sums as sums of products of combinations of the roots that match expressions determined by the (integer) coefficients of $P(x)$. But making that constructive proof anything more than hand-waving seems difficult.

I wonder if any ideals in the theory of rings, for instance, can make this proposition easier to prove.

NOTE Afterward

A counterexample would also nicely resolve the question, showing that the conjecture is false.


I agree with Bill that symmetric polynomials should somehow be the standard solution, but I wanted to point out that Galois theory makes this straightforward.

It is easy to show that $s = \prod_i Q(r_i)$ is an algebraic integer, so it is enough to show that $s\in\mathbb{Q}$. By the Galois correspondence, this is the same as checking that $\sigma(s)=s$ for every automorphism $\sigma$ of a splitting field $\mathbb{Q}\subset K$ of $P(X)$.

But any such automorphism just permutes the roots of $P$, and therefore permutes the terms of the product $\prod_i Q(r_i)$.


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