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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:

  1. Does anybody know a proof / counterexample to the above statement?
  2. 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.

57Jimmy
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  • Math Gems gives a fairly easy proof there, right? Just rewrite it as $x-a\mid f(x)$ iff $f(a)=0$ in $K[y]$. – Dietrich Burde Feb 26 '18 at 16:17
  • @DietrichBurde It is not a proof, it is just writing the statement differently. But why does it hold? The one-variable case holds because $K$ is a field, hence $K[x]$ is a Euclidean domain. But in the two variable case $K[y]$ is not a field, and $K[x,y]$ is not a Euclidean domain. Is there anything I missed? – 57Jimmy Feb 26 '18 at 16:22
  • @DietrichBurde Moreover, the same reasoning would also work for higher degree polynomials in $y$. But is the claim true also for them? – 57Jimmy Feb 26 '18 at 16:26

1 Answers1

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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.

Alex Shpilkin
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