Let $g(x),h(x)\in\mathbb{Z}[x]$ and $h(x)$ is monic. If $h(x)$ divides $g(x)$ in $\mathbb{Q}[x]$ then $h(x)$ divides $g(x)$ in $\mathbb{Z}[x]$

Question

Question: Let $g(x),h(x)\in\mathbb{Z}[x]$ and $h(x)$ is monic. If $h(x)$ divides $g(x)$ in $\mathbb{Q}[x]$ then $h(x)$ divides $g(x)$ in $\mathbb{Z}[x]$

I know that, as $h(x)$ divides $g(x)$ in $\mathbb{Q}[x]$ hence $$g(x)=h(x)q(x)\qquad (*)$$ for some $q(x)\in\mathbb{Q}[x]$.

Let $b$ be lcm of denominators of coefficients of $q(x)$ then from above we have, $bh(x)=g(x)(bq(x))=g(x)q_1(x)$ for some $q_1(x)\in\mathbb{Z}[x]$. As given that $h(x)\in\mathbb{Z}[x]$ and it is monic, hence its content is $1$ i.e. $c(h(x))=1$. So that, $c(bh(x))=b×1=b$. How to proceed further...? Or is there is alternative easier proof... please help

Answer 1

Just apply

Lemma If $f(x),g(x)\in\mathbb{Z}[x]$ with $f$ monic, then there exists (unique) $q(x),r(x)\in\mathbb{Z}[x]$ with $\deg r<\deg f$ such that $g(x)=f(x)q(x)+r(x)$.

Proof of Lemma: Induct on $\deg g$ -- if $\deg g<\deg f$ there is nothing to prove, otherwise $g(x)-f(x)\cdot LC(g)x^{\deg g-\deg f}$ has smaller degree, where $LC(g)$ is the leading coefficient of $g$ (i.e. coefficient of $x^{\deg g}$). QED.