9

Given a polynomial $P(x,y)$ I would like to know what the criteria are for factoring out linear factors.

For instance, in one variable, if $Q(a) = 0$, then one may say $Q(x) = (x-a)R(x)$. In two variables this is not true, as shown by $P(x,y) = x^2+y^2$ one has $P(0,0)=0$ but one cannot factor out anything.

When can one factor out a linear factor from a polynomial in two variables?

nullUser
  • 27,877

2 Answers2

6

I have a reference which cites this theorem for real polynomials, but if I go through the proof perhaps this also works over other fields/euclidean domains possibly. Here it is :

Theorem. Let $f(x,y)$ be a polynomial with real coefficients and degree $d$. Let $a,b,c$ be real numbers, and let $L$ denote the set of points $(x,y)$ for which $ax + by + c = 0$. If the curve $f(x,y) = 0$ and $L$ have strictly more than $d$ distinct points in common, then there exists $k(x,y)$ with real coefficients such that $$ f(x,y) = (ax+by+c)k(x,y). $$ This is in Niven & Zuckerman's Introduction to the theory of numbers, so it's not focused on algebra. I'll edit this answer later if I see a reason why this would hold over some other fields, but the proof uses mainly the Taylor expansion of the polynomial (which can be done formally, without using differentiation in the "limit" sense).

Hope that helps,

Patrick Da Silva
  • 41,413
  • 1
    Formal Taylor expansion (if you mean what I think you mean by this) can only be done in general in characteristic zero. – Qiaochu Yuan Feb 16 '12 at 19:30
  • 2
    Dear Patrick, yes the result holds for every field. However the proof you alllude to won't work in characteristic $p$, because of the denominators in Taylor's expansion. An amusing caveat is that the theorem is true but empty in many cases: yes, if a polynomial of degree 17 has at least 18 zeros on a line over a field with 2 elements, the equation of the line divides the polynomial.However no line has more than 2 points (nor less) over such a field... – Georges Elencwajg Feb 16 '12 at 19:39
3

The Factor Theorem remains applicable, for example

$\rm\qquad x-a\ |\ f(x) \iff f(a) = 0\ \ $ for $\rm\:a = by+c \in R[y]\:$ is

$\rm\qquad x-by-c\ |\ f(x,y)\iff f(by+c,\:y) = 0$

e.g. $\rm\ x-y\ |\ f(x) - f(y)\ \:$ since $\rm\ f(y)-f(y) = 0 $

Math Gems
  • 19,574
  • Well this is not the most general linear factor, now is it. =P I think OP expected this to be true. ^_^ – Patrick Da Silva Feb 17 '12 at 22:10
  • @Patrick But it is, since over a field any linear factor is associate to a monic one. Otoh, in the presence of zero-divisors one cannot say so much. In my experience, many students don't immediately notice this and related consequences of the Factor Theorem, esp. when they are obfuscated by less-trivial contexts. – Math Gems Feb 17 '12 at 22:46
  • Sorry, my comment was junk, I didn't even bother reading your entire proof when I saw that $x-a$ was the first thing you wrote ; but you're proof has a wrong step in it ; we can't say that $p(x,y) , | , f(x,y)$ just because $p(x,a) , | , f(x,a)$ for any value of $a \in R$, now can we? – Patrick Da Silva Feb 17 '12 at 22:57
  • 1
    @Patrick How does that pertain to the above? – Math Gems Feb 17 '12 at 23:00