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).
Sometimes (I learned this the hard way), you just have to keep reading. In the text, it is stated that one first divides by the $\gcd(A,B)$ to make sure that $A$ and $B$ share no factor. We now want to show that $A$ is constant.
Now what you are saying is correct, of course. If $A$ divides $B$ in $\mathbb F(x)[y]$, then all you get is $A\cdot C=B$ with $C\in\mathbb F(x)[y]$. You can multiply by some $D\in \mathbb F[x]$ of minimal degree to have $\tilde C := CD\in \mathbb F[x,y]$, though. This yields $A\cdot \tilde C=B\cdot D$. Now if $A$ and $B$ share no common factor, this means $A\mid D$. Since we chose $D$ to be of minimal degree, it shares no factor with $\tilde C$, so actually $A=D\in \mathbb F[x]$. However, this did not depend on our choice of $x$ or $y$, so we can just as well prove $A\in\mathbb F[y]$. Hence, $A\in \mathbb F[x]\cap \mathbb F[y]=\mathbb F$, so $A$ is a constant.
Regarding your example: It is important to have the statement not just for one of the variables, but for both. Otherwise, it's false as your example shows.