Multivariate polynomial divisibility and Gauss's lemma

Multivariate polynomial divisibility and Gauss's lemma

Let $\mathbb{F}$ be a field and $A(x,y)$ and $B(x,y)$ be polynomials in
$\mathbb{F}[x,y]$. We would like to prove that $A(x,y)$ divides $B(x,y)$.
Will the following approach work? We can interpret $A(x,y)$ and $B(x,y)$
as univariate polynomials in $y$ over the field of rational functions in
$x$ over $\mathbb{F}$, that is $\mathbb{F}(x)$. We then show that $A(x,y)$
divides $B(x,y)$ as univariate polynomials in $y$ over $\mathbb{F}(x)$
(that is, as polynomials in $\mathbb{F}(x)[y]$), and then use Gauss's
lemma to imply that therefore $A(x,y)$ divides $B(x,y)$ as polynomials in
$\mathbb{F}[x][y]$.
I noticed this in a paper, and am unable to see how Gauss's lemma
trivially establishes this implication (Assume the other step can be done
and is irrelevant, namely the univariate divisibility proof. I am only
interested in the implication part where it goes from univariate over the
rational function field to univariate over $\mathbb{F}[x][y]$ using Gauss
lemma). Any clarification would be appreciated.
For instance, suppose $A(x,y)=yx^2$ and $B(x,y)=y^2x$. Then $A$ does
divide $B$ as polynomials in $y$ over $\mathbb{F}(x)$, but clearly not as
bivariate polynomials in $\mathbb{F}[x,y]$.
Gauss's lemma states that if a univariate polynomial in $\mathbb{D}[x]$
(where $\mathbb{D}$ is a UFD) is reducible as a polynomial in
$\mathbb{K}[x]$ (where $\mathbb{K}$ is the field of fractions of
$\mathbb{D}$), then it is reducible in in $\mathbb{D}[x]$ itself.
The paper can be looked at here. Check out the first couple of paragraphs
in Section 5 "Piecing it together". (The paper as such is an important
work in the field of probabilistically checkable proofs in complexity
theory, and the constructions use algebraic tools).