Divisibility in a certain ring and divisibility in integers. Divisibility in the ring $\mathbb{Z}[x,y]$ implies divisibility in $\mathbb{Z}$ ?
Let $P(x,y)=Q(x,y)\cdot R(x,y)$ with $P,Q,R$ polynomials with integer coefficients, evaluating in $(x,y)=(a,b)$ with $a,b\in \mathbb{Z}$ clearly gives a divisibility relation in $\mathbb{Z}$. More precisely, my question is:
May I show that a polynomial $P(x,y)$ is not divisible by $Q(x,y)$ assuming that it's, and later to evaluate in a particular pair $(a,b)$ getting not divisibility in integers, and hence a contradiction with assumption?  
 A: Yes, that works fine, e.g. if $\ x\mid y\ $ in $\,\Bbb Z[x,y]\,$ then $\ 0\mid 1\ $ in $\,\Bbb Z\,$ by evaluating at $\,x,y = 0,1.$
Remark $\ $ Often one can deduce much information about divisibility of polynomials by examining the divisibility of their evaluations. For example, the possible factorizations of a polynomial $\in\Bbb Z[x]$ are constrained by the factorizations of the integer values that the polynomial takes. For a simple example, if some integer value has few factorizations (e.g. it is a unit $\,\pm1 $ or prime $p$) then the polynomial must also have few factors, asssuming that that the factors are distinct at the evaluation point. More precisely
If $\, f(x) = f_1(x)\cdots f_k(x)\,$ and $\,f_i\in\Bbb Z[x]\,$ satisfy $\color{#0a0}{f_i(n) \ne f_j(n)}\,$ for $\,i\ne j,$ all $\,n\in \Bbb Z$
$\quad \color{#c00}{f(n) =\pm1}\,\Rightarrow\, k\le 2\ $ else $1$ would have $\rm\,3\,\ \color{#0a0}{distinct}$ factors $\,f_1(n),f_2(n),f_3(n)$  
$\quad f(n) = \pm p\,\Rightarrow\, k\le \color{#c0f}4\ $ since a prime $p$ has at most $\,\color{#c0f}4\,$ distinct factors $\,\pm1,\pm p$
One can push the key idea to the hilt to obtain a simple algorithm  for polynomial factorization using factorization of its integer values and Lagrange interpolation. The ideas behind this algorithm are due in part to Bernoulli, Schubert, Kronecker. See this answer for references.
