$P,R \neq 0$ are polynomials with rational coefficients. Show that there exists a polynomial $Q$ such that $P(X) | Q(R(X))$

Question

Given any non-zero polynomials $P,R$ with rational coefficients, show that there exists a polynomial $Q \neq 0$ with rational coefficients such that $P(X)|Q(R(X))$. I would like to know if my solution is correct.
My solution: We will prove the result by induction on $deg(P)$. For $deg(P) = 0$, we can simply take $Q=P$.
Now, assume that we have proven the result for $deg(P) < n$, and now we will prove it for $deg(P) = n$.
First, if $P$ is reducible (over $\mathbb{Q}$), write $P=H_1H_2$, and by the induction hypothesis we have $Q_1,Q_2$ such that $H_1(X) | Q_1(R(X)), H_2(X)|Q_2(R(X))$ and therefore $P=H_1H_2 | Q_1(R)Q_2(R)$ so taking $Q=Q_1Q_2$ gives the desired result.
Now, assume that $P$ is irreducible. WLOG $P$ is monic (otherwise multiply by a rational constant, and of course it does not change the divisibility). Write, over the complex numbers, $P= (X-\alpha_1)...(X-\alpha_n)$. It is well known that since $P$ is irreducible, then each $\alpha_i$ appears once and only once in the factorization of $P$ over the complex numbers. Now, since $\alpha_1,...,\alpha_n$ are algebraic over $\mathbb{Q}$, then so are $R(\alpha_1),...,R(\alpha_n)$. Therefore, there are polynomials $Q_1,...,Q_n$ over the rational numbers such that $Q_1(R(\alpha_1))=...=Q_n(R(\alpha_n))=0$. So if we take $Q=Q_1...Q_n$ we will have $Q(R(\alpha_1))=...=Q(R(\alpha_n))=0$, therefore over $\mathbb{C}$ we have $P=(X-\alpha_1)...(X-\alpha_n)|Q(R(X))$ (again, because $\alpha_1,...,\alpha_n$ are distinct). Therefore the divisibility must also carry to $\mathbb{Q}$ and we are done. Is this proof correct? If so, is there a more elementary approach?

Answer 1

Looks good to me! If you want, you can avoid the induction and irreducibility argument by simply letting the $\alpha$'s be non-distinct. Then each $R(\alpha_i)$ is a root of $Q = Q_1 \cdots Q_n$ with the correct multiplicity (or higher), so $P$ divides $Q \circ R$.

Edit: As a silly side note, this argument even works when $\operatorname{deg}(P) = 0$: in that case, $P$ is a nonzero constant, hence has no roots, whence $Q = 1$ (the empty product). Then $Q \circ R = 1$ and indeed $P$ divides $1$.