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Let $x_1$, $\ldots$, $x_{\ell} $ complex numbers such that $$h_n \colon =\sum_{k=1}^{\ell} x_k^n \in \mathbb{Z}$$ for all $n \in \mathbb{N}$. To show that $$e_k \colon = \sum x_{i_1} \cdots x_{i_k} $$ are integers for all $1 \le k \le \ell$

Note: The converse is clear from the formulas for the Newton's sums.

An attempt: One sees that $ k ! e_{k}$ are integers for all $1\le k \le \ell$.

Now, since $$e_{\ell} ( x_1^d, \ldots, x_{\ell}^d) = x_1^d \cdots x_{\ell}^d = e_{\ell} ( x_1, \ldots, x_{\ell})^d$$

and

$$h_{n} (x_1^d, \ldots, x_{\ell}^d) = h_{ n d} (x_1, \ldots, x_{\ell})$$

we conclude that

$$\ell ! (x_1 \cdots x_{\ell})^d \in \mathbb{Z}$$ for all $d \ge 1$. This implies $e_{\ell} =x_1\cdots x_{\ell} \in \mathbb{Z}$.

Now, if $\ell = 3$, we can also show that $e_2 \in \mathbb{Z}$. Indeed, $2 e_2 \in \mathbb{Z}$, and then we have

$$h_4 = e_1^4 - 4 e_1^2 e_2 + 4 e_1 e_3 - 2 e_2^2$$

and we conclude $2 e_2^2 \in \mathbb{Z}$, and this implies $e_2 \in \mathbb{Z}$.

$\bf{Added:}$ For $\ell = 4$, one proceeds by showing that $e_4$, then $e_2$, then $e_3$ are integers ( note $e_1 = h_1$, so we don't worry about it). Perhaps this works for small $\ell$'s. Note that if the statement is true for $\ell+1$ it is also true for $\ell$ ( take the last number to be $0$).

Another possible approach: show that all of the $x_k$ are algebraic integers.( Clearly they are algebraic numbers, since we already know $e_k$ rational). So one might ask a related question:

Assume that all of the sums $h_n =\sum x_k^n $ are algebraic integers. Show that the $x_k$'s are algebraic integers.

$\bf{Added:}$ Another possible approach, and a generalization:

Let $x_1$, $\ldots$, $x_{\ell}$ distinct complex numbers, $\alpha_1$, $\ldots$, $\alpha_{\ell}$ non-zero numbers, such that

$$\sum_{k=1}^n \alpha_k x_k^n$$ is an algebraic integer for all $n$. Then $x_k$'s are algebraic integers ( and $\alpha_k$ are algebraic numbers).

Maybe looking at it in this way could provide a solution.

$\bf{Added:}$ I am happy that the answer is positive, as @Aphelli: has shown so neatly.

$\bf{Added:}$ Another similar approach:

Check that $N x_k$ are (algebraic) integers for some fixed $N$ ( in this case it's because the $e_k$'s are rational). But also $N x_k^d$ are integral for all $d$ and this implies $x_k$ integral. This procedure can work in other cases too, not only in the symmetric case. More precisely, suppose we have a polynomial $P$ in $x_1$, $\ldots$, $x_{\ell}$ ( integral coefficients) and $$P(x_1^n, \ldots, x_{\ell}^n)$$ is an integer for all $n$. Can we conclude that $x_k$'s are integers?

Assume that the extension $$\mathbb{Q}[ P[x_1, \ldots, x_{\ell}], P[x^2_1, \ldots, x^2_{\ell}], \ldots] \subset \mathbb{Q}[x_1, \ldots, x_{\ell}]$$

is integral. Then the same method applies.

Note that this will not work for the polynomial $P[x_1, x_2] = x_1 - x_2$. The statement clearly does not hold, since we could have $x_1= x_2$, otherwise arbitrary.

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1 Answer 1

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Actually, your argument is almost enough to conclude. There is a number field $K$ containing all the $x_i$. Consider any prime ideal $\mathfrak{p}$ of $K$ and let $A$ be the valuation ring at $\mathfrak{p}$.

I claim the following: let $z_1,\ldots,z_{\ell} \in K$ be such that for all $n \geq 1$, $h_n(z_1,\ldots,z_{\ell}) \in A$. Then all the $z_i$ are in $A$.

Proof: let $S$ be the set of indices $i$ such that $z_i \notin A$. Then for all $n,d \geq 1$, $\sum_{i \in S}{(z_i^d)^n} \in A$. Thus by your argument (that is, Newton sums), $|S|!\prod_{i \in S}{(z_i)^d} \in A$. But as $d$ goes to infinity, the LHS gets a negative valuation, unless $S$ is empty.

In particular, all the $x_i$ are in $A$. Thus, the $x_i$ have non-negative valuation at any prime of $K$, so they are algebraic integers (and thus their symmetric polynomials are algebraic integers as well).

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  • $\begingroup$ Thank you! What a beautiful solution! $\endgroup$
    – orangeskid
    Sep 27, 2022 at 8:49

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