Characterization of the denominator of a polynomial $P \in \mathbb Q[X]$ by looking at roots?

Question

Let $P(X) \in \mathbb Q[X]$ be a monic polynomial of degree $n$ and $a \in \mathbb Z$. We have the implications $$a P(X) \in \mathbb Z[X] \implies a^n P(X/a) \in \mathbb Z[X] \iff \text{the roots of $P$ lie in } \overline{\mathbb Z}/a \,.$$

I would like to add extra conditions to the statements on the right to make this an equivalence.

This question is a bit vague, naturally, because I don't know what conditions to add. Of course, I don't want anything trivial that amounts to saying that $aP \in \mathbb Z[X]$. My motivation below can help to get a feeling for what type of condition I'd like to see. Something like "$aP$ sends integers to integers" would be acceptable.

Motivation. I started thinking about this when answering (or trying to) the following questions:

If $f(x)\in\mathbb Q[x]$ and $f(f(x)),f(f(f(x)))\in\mathbb Z[x]$ prove that $f(x)\in\mathbb Z[x]$

Can composition of integer polynomial and rational polynomial with a non-integer coefficient result in integer polynomial?

Ideas. Requiring that $aP(\mathbb Z) \subset \mathbb Z$ is not sufficient: take $a = 2$, $P = X^3 + \frac{X^2+X}4 = \frac{(2X)^3 + (2X)^2 + 2 (2X)}{8}$, which has roots in $\overline{\mathbb Z}/2$.

Even imposing the same on all derivatives of $aP$ and imposing that $c-d \mid aP(c) - aP(d)$ is not enough: $a = 2$, $P = X^5 + \frac{X^4+X^2}4$.

Answer 1

Maybe you're looking for something like this:

Let $P$ be a primitive polynomial in $\mathbb{Z}[X]$. Then $P$ is monic if and only if all roots of $P$ are in $\overline{\mathbb{Z}}$.

Proof: For $\Rightarrow$, this is true by definition of $\overline{\mathbb{Z}}$. For $\Leftarrow$, let $a \in \mathbb{Z}$ be the leading coefficient of $P$; the coefficients of $\frac{1}{a}P$ are sums of products of elements of $\overline{\mathbb{Z}}$, hence are themselves contained in $\overline{\mathbb{Z}}$; but $\overline{\mathbb{Z}} \cap \mathbb{Q} = \mathbb{Z}$ (i.e. $\mathbb{Z}$ is integrally closed).

Answer 2

Edit: the OP clarified after that the goal was to get $aP\in \Bbb{Z}[x]\iff ...$ not "the roots are in $a^{-1}\Bbb{Z}\iff...$"

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For $P=\sum_{n=0}^d c_n X^n =\prod_j (X-\beta_j)\in \Bbb{Q}[X]$ monic then all the $a\beta_j\in \overline{\Bbb{Z}}$ iff $a^d P(X/a)\in \Bbb{Z}[X]$ iff the denominator of $c_n$ divide $a^{d-n}$.

I'd say there is no simpler criterion, the usual statement

$a \beta\in \overline{\Bbb{Z}}$ if $P(\beta)=0$, $P\in \Bbb{Z}[X]$ with leading coefficient $a$

is weaker and cannot be adapted to a converse.

This leads to the $p$-adic Newton polygon: the minimum power of $p$ such that all the $p^k\beta_j$ are in $\overline{\Bbb{Z}_p}$ is the least $k$ such that $v_p(c_n p^{k(d-n)})\ge 0$, allowing $k$ rational we find $\inf v_p(\beta_j)$.