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

Question

Can we find two polynomials $p(x)$ and $q(x)$, where $p(x)$ is a non-constant monic polynomial over integers and $q(x)$ is a monic polynomial over rationals with at least one non-integer coefficient, such that their composition $p(q(x))$ is a polynomial over integers? If not, how to prove it?

For example let $q(x)=x^2+\frac{1}{2}x+1$ and $p(x)=x^3+a_2x^2+a_1x+a_0$, then $p(q(x))=x^6+\frac{3}{2}x^5+\dots$, so no matter what integers $a_i$ we choose, the resulting polynomial will have a non-integer coefficient. The monic condition is important, since otherwise we could multiply $p(x)$ with such integer that would guarantee all coefficients to be integers. I've tried to look at the coefficient in composition for general polynomials, which I believe should follow this formula:

\begin{align} [x^r]p(q(x))=\sum_{k_1+2k_2+\dots+mk_m=r}\sum_{k_0=0}^{n-(k_1+\dots+k_m)}\binom{k_0+k_1+\dots+k_m}{k_0,k_1,\dots,k_m}a_{k_0+k_1+\dots+k_m}\left(\prod_{j=0}^{m}b_j^{k_j}\right) \end{align} (here $a_i$ and $b_i$ are the coefficients of $p(x)$ and $q(x)$ with degrees $n$ and $m$, respectively). However it is not at all clear on which coefficient to focus to prove it will give the non-integer number.

This arose when trying to solve the Infinitely many solutions leads to existence of a polynomial, but it seems interesting enough by itself.

Answer 1

In fact, we may ignore the assumption that $q$ is monic. The composition $p \circ q$ cannot have all integer coefficients.

For let $r$ be a prime factor of some fully simplified denominator of a coefficient of $q$. Consider the largest $k$ s.t. $r^k$ is a factor of some denominator of a $q$ coefficient. Then write the polynomial $q$ as $x^j w(x) / r^k + s(x)$, where every fully simplified numerator of $w(x)$ is not divisible by $r$ and no fully simplified denominator of $s(x)$ is divisible by $r^k$, and where $w$ has a non-zero constant term. Do this by grouping all terms with denominators divisible by $r^k$, obtaining $x^j w(x) / r^k$, and all terms with denominators not divisible by $r^k$, obtaining $s(x)$.

Let $n$ be the degree of $p$, and consider the coefficient of $x^{jn}$ in $p \circ q$. One of the contributing summands will be $w(0)^n / r^{kn}$, which is fully simplified. And none of the other summands can have a denominator divisible by $r^{kn}$. So this coefficient is not an integer.

Answer 2

One can show this with some elementary properties of algebraic integers:

Fact. If $p \in \mathbb C[x]$ and $q \in \mathbb Q[x]$ are monic, $p(0) \in \mathbb Z$, $\deg p, \deg q > 0$ and $p \circ q \in \mathbb Z[x]$, then $q \in \mathbb Z[x]$.

Proof: We must show that all roots of $q$ are algebraic integers. Let $\alpha$ be a root of $q$. Then $p(q(\alpha)) - p(0) = 0$. The assumptions imply that $r(x) = p(q(x)) - p(0)$ is a monic integer polynomial. Because $\alpha$ is a root of $r$, it is an algebraic integer.

Answer 3

For the non-monic case:

Consider the monic polynomial $p(y) - p(q(x)) \in \mathbb Z[x][y]$. It has a root $q(x) \in \operatorname{Frac}(\mathbb Z[x])$. By the rational root theorem applied to the UFD $\mathbb Z[x]$, it follows that $q(x) \in \mathbb Z[x]$.