Using resultants to check if multivariate polynomials have a common factor - is my proof correct? Proposition: Let $f, g \in \mathbb R[x,y,z]$. Then the condition that $f, g$ have a common polynomial factor is an algebraic condition on their coefficients.
By algebraic condition, I mean there is a finite set of algebraic (polynomial) expressions in the coefficients, such that all expressions are zero if and only if $f, g$ have a common factor.
My proof: Let $\operatorname{Res}_x(f,g)$ be the resultant of $f, g$ when viewed as univariate polynomials in $(\mathbb R[y,z])[x]$.
Then $\operatorname{Res}_x(f,g) \in \mathbb R[y,z]$ is zero if and only if $f, g$ have a common factor that depends on $x$. As a polynomial in $y,z$, it is zero if and only if all of its coefficients are zero. But the coefficients of the resultant are simply polynomial expressions in the coefficients of $f, g$.
Therefore there is an algebraic condition on the coefficients of $f,g$ that determines when $f,g$ have a common factor that depends on $x$.
Repeating the argument, there are two more algebraic conditions on the coefficients of $f,g$, which determine when they have a common factor that depends on $y$ and when they have one that depends on $z$. The union of all 3 conditions is what we're after.
Questions:


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*Is my proof correct? I'm especially concerned about the claim that the resultant is the zero polynomial if and only if there is a common factor that depends on $x$. (Does the resultant behave ok for polynomials over rings instead of over fields?)

*Do you have comments or suggestions for improvement? Any recommended books on this subject?

 A: Your proof seems to work, but it looks incomplete because you need to use more explicitly the fact that the ring of polynomials is an UFD.
The condition that the resultant $\mathrm{Res}(f,g)=0$ means that there are some polynomials $p,q$ such that $\deg p<\deg f$, $\deg q<\deg g$ and 
$fq=gp$ (so $p/q=f/g$). In a UFD this is equivalent to having a common factor of $f$ and $g$.
To complete your approach you must replace the coefficient ring by its quotient field. You need the unique factorization at the end when you simplify the coefficients to be ring elements again.
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For zero divisor free rings, without unique factorization, it will not work. For example, consider the polynomials $x^2-4y^2$ and $x^2+4xy+4y^2$. In some rings, like $(\mathbb{R}[y])[x]$, they have the common factor $x+2y$. In other rings, for example in $(\mathbb{Z}[4y,4y^2])[x]$, they don't.
However, the resultants will be the same zero polynomial in all rings.
So this method does not work if the coefficients are from the ring
$\mathbb{Z}[4y,4y^2]$.
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Another possible approach is building a single (but huge) resultant. If $f,g\in\mathbb{R}[x_1,\dots,x_n]$ are nonzero polynomials then take all polynomials of the form $x_1^{a_1}\cdots x_n^{a_n}f$ with $a_1+\cdots+a_n<\deg g$ and $x_1^{b_1}\cdots x_n^{b_n}g$  with $b_1+\cdots+b_n<\deg f$. These polynomials are linearly dependent if and only if there are some polynomials $p,q$ with $\deg p<\deg f$, $\deg q<\deg g$ and $p/q=f/g$.
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Sorry, I cannot recommend you books other than I can find by internet search...
:-(
