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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?

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    $\begingroup$ There might be some benefit to noting that the $R(\alpha_j)$ are all conjugate under the Galois group of $P$, but it seems correct and pretty consise as it is. $\endgroup$ – Greg Martin May 30 '20 at 23:12
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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 $1$.

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  • $\begingroup$ Do you mean to take $Q_i$ such that $R(\alpha_i)$ is a root of $Q_i$, and then raise $Q_i$ to a large power so that it reaches the desired multiplicity? $\endgroup$ – Omer May 30 '20 at 23:17
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    $\begingroup$ Well that's not what I meant but that would certainly work. But as long as each $Q_i$ has $R(\alpha_i)$ as a root at all, this will be fine. You just need to list out the roots $\alpha_1, \dots, \alpha_n$ of $P$ with multiplicity, so $n = \operatorname{deg}(P)$. Now for any root $\alpha$ of $P$, let $m$ be its multiplicity. Since $\alpha$ appears $m$ times in the list of $\alpha_i$'s, there are $m$ different values of $i$ such that $R(\alpha)$ is a root of $Q_i$. Thus, $(X-R(\alpha))^m$ divides $Q$, so $(X-\alpha)^m$ divides $Q \circ R$. $\endgroup$ – diracdeltafunk May 30 '20 at 23:26
  • $\begingroup$ I understand, thank you! $\endgroup$ – Omer May 30 '20 at 23:32

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