Proving that the products GCDs of the coefficients of two polynomials is equal to the GCD of their product's coefficients? Assume that $p(x)=a_nx^n+\dots+a_0$ and $q(x)=b_nx^n+\dots+b_0$ where the coefficients are integers. Let $y$ be the gcd of $a_n,\dots,a_0$ and let $z$ be the gcd of $b_n,\dots,b_0$. How does one prove that the GCD of the coefficients of $p(x)q(x)$ is $yz$?

Write $p(x)=y P(x)$ and $q(x)=zQ(x)$, where $P(x)$ and $Q(x)$ are primitive polynomials (i.e. the g.c.d.  of their coefficients is $1$). Thus 
$$p(x)q(x)=yz P(x)Q(x)$$
and it is enough to show $P(x)Q(x)$  is primitive. . It not, there exists a prime number $a$ that divides all its coefficients. Reduce the coefficients modulo $a$; in the ring $\mathbf Z/a\mathbf Z$, we have $\overline{\!PQ}=\overline{\mkern-1muP\vphantom Q}\,\overline{\mkern-1muQ}$. As  $\mathbf Z/a\mathbf Z$ is an integral domain, this implies $\overline{\mkern-1muP}=0 $ or $\overline{\mkern-1 muQ}=0$, which means $a$ divides all coefficients of $P$ or all coefficients of $Q$, and is impossible since $P$ and $Q$ are primitive.
Note. This proof is valid for polynomials over any UFD, since irreducible elements in such rings are prime.
