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Suppose I have a polynomial function $f\in \mathbb{Z}[x_1, \dotsc, x_k],$ such that whenever $r_1, \dotsc, r_k$ are roots of a monic polynomial of degree $k$ with integer coefficients, we have $f(r_1, \dotsc, r_k) \in \mathbb{Z}.$ Is it true that $f$ is a symmetric function of its arguments?

Note: this was asked on MSE, but only resulted in an inconclusive discussion with Qiaochu...

EDIT Fedor Petrov's argument certainly answers the question quite elegantly, but I was thinking of the case where $r_1\dotsc, r_k$ are roots of an irreducible polynomial, so the reduction to $k=2$ is not quite so obvious. If there is an argument in that setting as well, that would be very much of interest.

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3 Answers 3

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Consider the minimal polynomial $P$ of $f$ over the fixed field $\mathbb Q(\sigma_1, \ldots, \sigma_k) = \mathbb Q(x_1, \ldots, x_k)^{S_k}$ (where the $\sigma_j$ are the elementary symmetric polynomials), so $P$ is irreducible and $P(f) = 0$. Since $f$ is a polynomial, the coefficients of $P$ will actually be polynomials in the $\sigma_j$.

By the Hilbert Irreducibility Theorem, $P$ will remain irreducible when we specialize $\sigma_1, \ldots, \sigma_k$ to rational numbers, outside a thin set of such $k$-tuples. But the set of $k$-tuples corresponding to $(x_1, \ldots, x_k)$ forming a Galois orbit is not thin (its complement is thin: it consists of the points $(a_1, \ldots, a_k)$ such that $h(T) = T^k - a_1 T^{k-1} + \ldots \pm a_k$ is reducible; this set can be written as a finite union of images of $\mathbb Q$-rational points under dominant morphisms that correspond to the various possibilities of factoring $h$).

So there are (plenty of) integer (see this Wikipedia entry) specializations $\boldsymbol{a}$ with irreducible $h$ such that the specialized $P_{\boldsymbol{a}}$ is irreducible. But by assumption, $P_{\boldsymbol{a}}$ has a rational root (since $f$ evaluated at the roots $\boldsymbol{r}$ of $h$ is an integer): $P_{\boldsymbol{a}}(f(\boldsymbol{r})) = 0$. So $P$ must have degree 1, and $f$ is in $\mathbb Q[\sigma_1, \ldots, \sigma_k]$.

In more detail: Write $$P(X) = X^n + p_{n-1}(\sigma_1, \ldots, \sigma_k) X^{n-1} + \ldots + p_0(\sigma_1, \ldots, \sigma_k) .$$ Then there are integers $\boldsymbol{a} = (a_1, \ldots, a_k)$ such that $h_{\boldsymbol{a}} = T^n - a_1 T^{n-1} + \ldots \pm a_k$ is irreducible and $$P_{\boldsymbol{a}}(X) = X^n + p_{n-1}(a_1, \ldots, a_k) X^{n-1} + \ldots + p_0(a_1, \ldots, a_k) \in {\mathbb Q}[X]$$ is also irreducible. Let $\boldsymbol{r} = (r_1, \ldots, r_k)$ be the roots of $h_{\boldsymbol{a}}$. Then $f(\boldsymbol{r}) = m \in \mathbb Z$. On the other hand, $\sigma_j(\boldsymbol{r}) = a_j$, so $$0 = P_{\boldsymbol{a}}(f(\boldsymbol{r})) = m^n + p_{n-1}(\boldsymbol{a}) m^{n-1} + \ldots + p_0(\boldsymbol{a}),$$ i.e., $m$ is a root of the irreducible polynomial $P_{\boldsymbol{a}}$.

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  • $\begingroup$ Sorry, I am being dense: how does the "rational root" implication work? There are many domains here, so it is a little confusing.... $\endgroup$
    – Igor Rivin
    Commented Nov 29, 2015 at 11:43
  • $\begingroup$ @IgorRivin : See edit. $\endgroup$ Commented Nov 29, 2015 at 12:04
  • $\begingroup$ Ah, OK, it's all clear now! $\endgroup$
    – Igor Rivin
    Commented Nov 29, 2015 at 12:14
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I think, yes by some boring reasons. At first, we may suppose that $k=2$. Indeed, this partial case implies (if we take all but two $r_i$'s integer) that polynomial is symmetric in any two variables. For $k=2$ we may subtract from $f$ some symmetric polynomial and get a polynomial with monomials $x_1^{a}x_2^b$ only for $b>a$. Now fix integer value of $x_1x_2$ and choose $x_2$ very close to 0.

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    $\begingroup$ As a variant: In the $k = 2$ case, $2f = f_+ + f_-$ where $f_+$ is symmetric and $f_-$ is antisymmetric. Then $f_-$ is divisible by $x_1 - x_2$, and the quotient is symmetric, so $f_-(a+\sqrt{d},a-\sqrt{d}) = 2 \sqrt{d} n$ with an integer $n$, for all integers $a$ and positive integers $d$. Since $\sqrt{d}$ is irrational unless $d$ is a square, the symmetric factor of $f_-$ must vanish at sufficiently many points to force it to be zero. $\endgroup$ Commented Nov 28, 2015 at 21:46
  • $\begingroup$ Let me get this straight: you are saying: let $y = x_1 x_2,$ so your polynomial will be a polynomial in $y$ and $x_2,$ with no constant term. Since $y$ is fixed, this will go to zero as $x_2$ goes to $0,$ which means (since it is always integer) that it is constantly zero, which means the original function was symmetric. Nice argument! But see edit - your argument probably adapts to this case. $\endgroup$
    – Igor Rivin
    Commented Nov 28, 2015 at 21:46
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    $\begingroup$ @MichaelStoll See the edit. Can you adapt your argument to the slightly stronger hypothesis? $\endgroup$
    – Igor Rivin
    Commented Nov 28, 2015 at 22:13
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I would argue this way. For $k=2$ it's true. For general $k$ and $f$, and for any choice of $k-2$ integers $m_1,\dots,m_{k-2}$, the two variables polynomial $f(x_1 ,x_2, m_1,m_2, \dots,m_{k-2})$ is symmetric in $(x_1,x_2)$ because of the case $k=2$. But since this is true for any $m_1,\dots,m_{k-2}$, this implies $f(x_1,x_2,x_3\dots, x_k)=f(x_2,x_1,x_3\dots, x_k)$. The same for any other transposition of a pair of variables $x_i,x_j$.

edit. Details on the last implication. Writing $f(x_1,x_2,\dots, x_k)=\sum_{ij}f_{ij}(x_3,\dots,x_k)x_1^ix_2^j$, the two variables case implies that for any $(i,j)$ the $(k-2)$-variables polynomial $f_{ij}-f_{ji}$ vanishes on $\mathbb{Z}^{k-2}$, so it is the zero polynomial.

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