A condition such that $p\mid\sum_{f(\theta)=0}\theta^n$ for all $n$?

Question

If $f$ is any monic polynomial/$\mathbb{Z}$ with non-zero constant coefficient. I wish to study the quantities

$$t_n=\sum_{i}\theta_i^n\in\mathbb{Z}$$

where $(\theta_i)_{i=1}^{d}$ are the roots of $f$ counted with multiplicity.

The main question I am interested is finding all the primes $p$ such that $p\mid t_n$ for all large enough $n$. I have proved that this can only happen in the case that either $p\mid c_n$ $\forall$ $n$ where $c_n$ are the non-leading coefficients of $f$, or $p\mid t_n$ $\forall$ $n\geq0$.

The condition that $p\mid c_n$ is easy to check, so my interest has turned to some algebraic interpretation of the condition that $p\mid t_n$ for all $n$. My current idea is to let $K=\mathbb{Q}(\theta_i)_{i=1}^{d}$ be the field attached to $f$, and analyse the ideal $I$ of $\mathcal{O}_K$ generated by $(\theta_i)_{i=1}^{d}$. My idea is that it seems as if $p\mid t_n$ $\forall$ $n$ implies that any element of $I \cap \mathbb{Z}$ will be a multiple of $p$. The converse is obvious.

Is the above observation correct? If so, is there some known property of such primes (i.e. $I\cap \mathbb{Z} \subset p\mathbb{Z}$ iff $p$ ramifies in $\mathcal{O}_K$)?

Answer 1

You have $t_k=0$ for all $k$ if and only if $f(x) \bmod p$ is a $p$-th power.

Let $g(x)$ be the image of $f(x)$ in $\mathbb{F}_p[x]$; let $\alpha_1$, $\alpha_2$, ..., $\alpha_n$ be the roots of $g$ (with multiplicity) in $\overline{\mathbb{F}_p}$, let $e_k$ be the $k$-th elementary symmetric function in the roots, and let $p_k = \sum \alpha_i^k$. You want a criterion for when all the $p_k$ are $0$.

If each $\alpha_i$ occurs with multiplicity divisible by $p$ then, clearly, $p_k=0$.

Conversely, suppose all the $p_k$ are $0$. Then, by Newton's identities, $k e_k = 0$ for all $k$. So, whenever $p$ does not divide $k$, we have $e_k=0$. But this means that the coefficient of $x^{(\deg g(x))-k}$ in $g(x)$ vanishes whenever $p$ does not divide $x$, so $g(x)$ is of the form $h(x^p) x^m$, where $m$ is equivalent mod $p$ to the degree of $g$. Since you imposed that $p_0=0$ as well, the degree of $g(x)$ is $0 \bmod p$.