Coefficients of polynomials and their summands When multiplying two polynomials over an integral domain, a frequently arising problem is that coefficients of the product can vanish and information on the coefficients of the factors is lost. I feel that in many cases this is the reason why it can be very hard to show that properties of the base ring go over to the polynomial ring. This motivates the following question:
Let $D$ be a domain with quotient field $K$. Let $f = a_0 + a_1x + a_2y \in K[x,y]$ and $g = b_0 + b_1x + b_2y \in K[x,y]$ be two linear polynomials such that $fg \in D[x,y]$. It follows immediately that $a_0b_0$, $a_1b_1$, $a_2b_2$, $a_0b_2 + a_2b_0$, $a_1b_2 + a_2b_1$ and $a_1b_0 + a_0b_1$ are elements of $D$. Can we conclude $a_ib_j \in D$ for all $i,j \in \{0,1,2\}$?
More generally, if $f = \sum a_{i,j}x^iy^j$ and $g = \sum b_{k,l}x^ky^l $ are in $K[x,y]$ such that $fg \in D[x,y]$, does it follow that $a_{i,j}b_{k,l} \in D$ for all $i,j,k,l$?
If there is no positive answer to this question, is there any mathematical work with this kind of flavor done at the moment?
Thank you for your help!
 A: In general, no.  Let $D = \mathbb Z[\sqrt 5]$, $\alpha = \frac{1-\sqrt 5}{2}$, and $\beta = \frac{1+\sqrt 5}{2}$.  Then we have $\alpha \beta = -1 \in D$, but the elements $\alpha^2 = \frac{3-\sqrt 5}{2}$ and $\beta^2 = \frac{3+\sqrt 5}{2}$ are not in $D$.  Now setting $f = \alpha + \beta x$ and $g = \beta + \alpha x$, we have $fg = -1 + 3x - x^2 \in D[x]$, even though not all of the individual products of coefficients are in $D$.
However, if $D$ is a unique factorization domain (or even a GCD domain), then a more general result follows from a version of Gauss's lemma (stated here).  Suppose we have $f, g \in K[x_1, \dots, x_n]$ and $fg \in D[x_1, \dots, x_n]$.  We claim that the product of any coefficient of $f$ and any coefficient of $g$ lies in $D$.
Proof:  Assume $f$ and $g$ are nonzero.  By clearing denominators and factoring out a gcd, we can find $c, d \in K^{\times}$ such that $cf$ and $dg$ lie in $D[x_1, \dots, x_n]$ but have no nontrivial constant factors.  Equivalently, using the notation from the link, we have $\mathrm{gcd}(\mathrm{cont}(cf)) = \mathrm{gcd}(\mathrm{cont}(dg)) = 1$.  It follows by Gauss's lemma that $\mathrm{gcd}(\mathrm{cont}(cdfg)) = 1$; that is, the coefficients of $cdfg$ have no common factor.  Since $\frac{cdfg}{cd} = fg \in D[x_1, \dots, x_n]$, it follows that $1/cd \in D$.
To see why this implies the result, let $a$ be any coefficient of $f$ and $b$ any coefficient of $g$.  Since $cf, dg \in D[x_1, \dots, x_n]$, we have $ac, bd \in D$.  But then $ab = (ac)(bd)(1/cd) \in D$.
