Divisibility in $\mathbb{Q}[x]$ Let $F(x),G(x) \in \mathbb{Z}[x]$ with $G(x)$ being irreducible in $\mathbb{Z}[x]$. (By irreducible, we mean that if $G(x)=A(x)B(x)$ for $A(x),B(x) \in \mathbb{Z}[x]$, then either $A(x)$ or $B(x)$ is a unit. In other words if we have such a factorization, then $A(x)$ or $B(x)$ is necessarily $\pm1$). If $G(x) \big| F(x)$ in $\mathbb{Q}[x]$, then can we conclude that $G(x) \big| F(x)$ in $\mathbb{Z}[x]$?
Here is my naïve attempt. We can write $F(x) = G(x)H(x)$ for $H(x) \in \mathbb{Q}[x]$. We can further write $H(x) = ch(x)$ with $c \in \mathbb{Q}$ and $h(x) \in \mathbb{Z}[x]$. Putting it together we get that $$F(x) = cG(x)h(x).$$ If I can argue that $c \in \mathbb{Z}$ then we are done. I believe this is were the irreducibility of $G(x)$ comes into play. All we know is that since $G(x)$ is irreducible in $\mathbb{Z}[x]$, then it is also irreducible in $\mathbb{Q}[x]$. But where do we go from here?
 A: The desired conclusion for arbitrary $G(x)$ holds if and only if $G(x)$ is primitive; that is, if the content of $G$ is $1$.
Recall that for a polynomial $p(x)\in\mathbb{Z}[x]$, the content of $p(x) = a_nx^n+\cdots + a_1x + a_0$ is defined to be
$$\mathrm{con}(g) = \gcd(a_0,a_1,\ldots,a_n).$$
A polynomial is primitive if its content is $1$.
Gauss's Lemma states that the product of primitive polynomials is primitive. Since every polynomial $q(x)\in \mathbb{Z}[x]$ can be written uniquely (up to sign) as $q(x) = cp(x)$ where $c\in\mathbb{Z}$ and $p(x)$ is primitive, it follows that for any $r(x),s(x)\in\mathbb{Z}[x]$, we have $\mathrm{con}(rs) = \mathrm{con}(r)\mathrm{con}(s)$.
Likewise, every polynomial $f(x)\in\mathbb{Q}[x]$ can be written uniquely (up to sign) in the form $f(x) = rg(x)$, where $g(x)\in\mathbb{Z}[x]$ is primitive and $r\in\mathbb{Q}$.
Assume first that $G(x)$ is primitive. Writing $H(x) = ch(x)$ with $h(x)$ primitive, we have that $F(x) = c(G(x)h(x))$ with $G(x)h(x)$ primitive, hence $\mathrm{con}(F)=c$, so $c$ is an integer; thus, $H(x)\in\mathbb{Z}[x]$, so $G(x)$ divides $F(x)$ in $\mathbb{Z}[x]$.
Conversely, assume that $G(x)$ is not primitive. Write $G(x) = cg(x)$ with $g(x)\in\mathbb{Z}[x]$ primitive. Then $G(x)$ divides $g(x)$ in $\mathbb{Q}[x]$, but does not divide it in $\mathbb{Z}[x]$, since $c$ is not a unit.
Note that we do not require $G(x)$ to be irreducible.
Now, if we add the assumption that $G(x)$ is irreducible, then we see that the only irreducible polynomials that are not primitive are the primes. Thus, for an irreducible polynomial $G(x)$, the result  holds if and only if $G(x)$ is not constant.
You can also prove this using the fact that $\mathbb{Z}[x]$ is a UFD whose irreducibles are the integer primes and the primitive polynomials that are irreducible in $\mathbb{Q}[x]$, but this type of result is usually a stepping stone along the way of proving that, so presumably you cannot invoke that fact.
