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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 $\mathbb{F}[x,y]$.

  1. 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.
  2. 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).
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up vote 1 down vote accepted

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.

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Thanks! I got caught up at that point since the paper made it sound like it was an immediate consequence of Gauss Lemma (though I do not see the need for the lemma at all). While your entire answer was helpful, I think the key point is the last line, that we need the univariate divisibility for both variables and not just one. Also, "Now if $A$ and $B$ share no common factor, this means $A\mid D$." Doesn't this itself imply that $A \in \mathbb F[x]$?. Do we really need "Since we chose $D$ to be of minimal degree, it shares no factor with $\tilde C$, so actually $A=D\in \mathbb F[x]$."? – BharatRam Aug 26 '13 at 12:18
You're right about the last part. In a first version of the answer, I had prematurely concluded $A=D$ right away and then noticed that one can only get $A\mid D$. I fixed that instead of realizing that $A\mid D$ is actually enough. – Jesko Hüttenhain Aug 26 '13 at 12:24

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