Factor theorem for multiple variables I know that similar questions have been asked often. I have had a look at many of them, and I have found the following claim over polynomials in two variables in an answer here:

$$x-(by+c) \ | \ f(x,y) \Longleftrightarrow f(by+c,y)=0 \in K[y]$$

where $K$ is a field. However, I was unable to find a proof, and it seems strange to me that this fact should be true, because one of the proofs I know for the standard factor theorem (in one variable) uses that $K[x]$ is a Euclidean domain, but $K[x,y]$ is not. So here are my questions:


*

*Does anybody know a proof / counterexample to the above statement?

*If it is true, are there further generalizations? For instance with more variables or with factors with degree higher than linear or with $K$ not a field.


Edit: The "only if" implication is clear even if instead of $by+c$ we have any polynomial in any number of variables (different from $x$), and also for any ring $K$. The difficult implication is the "if" part, because we need some form of Euclidean division, and it is not clear to me when we have it.
 A: The standard proof of the factor theorem using long division still works over $K[y]$ or any other ring, as discussed elsewhere on MSE, because long division can be performed the usual way in any ring as long as the leading coefficient is invertible (a “unit”): walk through the argument yourself or look it up on MSE.  In the factor theorem, you’re dividing by $x - a$, $a\in R$, which as an element of $R[x]$ has (invertible) leading coefficient $1$; thus the factor theorem works over any $R$ whatsoever, including $R = K[y]$.
Why can’t you split arbitrary polynomials in two variables into linear factors, then?  If you think of polynomials as functions, to apply the factor theorem, you need to find $g\in K[y]$ such that $f(g(y), y) = 0$ for all values of $y$ simultaneously.  But this is not possible, for example, for $xy - 1$ over $\mathbf C$.
P. S. There is an extension of polynomial long division for non-invertible leading coefficients as well, but it’s necessarily weaker than the standard result.
