3
$\begingroup$

Let $F$ be an algebraic closed field of char(F)=0 and let $(a,b)\in F^2$ be a non-zero vector. Suppose I have two polynomials $f,g\in F[x]$ with $\deg f\neq \deg g$ and $1\leq \deg f, \deg g$. I want to show that \begin{equation} \gcd(f(x)-f(y)-a, g(x)-g(y)-b)=1. \end{equation} For instance, by Eisenstein criterion, it is easy to verify this for an arbitrary non-zero vector $(a,b)$ and $f(x)=x^n$ and $g(x)=x^m$ when $n\neq m$. I have calculated several other examples which again verify the claim but still I am not sure whether it is true in general.

I would be thankful if you share with me your ideas.

Improve this question
$\endgroup$
10
  • $\begingroup$ Are you hoping this claim is true for any polynomials of different degree? If so, you're out of luck for the silly example: $f(x) = const$. Maybe if you require that neither function is constant? $\endgroup$
    – Pat Devlin
    Commented Dec 30, 2016 at 17:44
  • 1
    $\begingroup$ You may need $gcd(a,b)=1$. $\endgroup$
    – T. Amdeberhan
    Commented Dec 30, 2016 at 17:45
  • $\begingroup$ @T.Amdeberhan $a,b$ are constants so automatically have $gcd(a,b) = 1$. $\endgroup$
    – Felipe Voloch
    Commented Dec 30, 2016 at 17:48
  • 1
    $\begingroup$ @FelipeVoloch On American playgrounds, if two children say the same thing at the same time, one child gets to say a special phrase, and then the other child has to remain silent until somebody says the release phrase :) $\endgroup$
    – Jason Starr
    Commented Dec 30, 2016 at 17:54
  • 1
    $\begingroup$ Great minds think alike and fools seldom differ. $\endgroup$
    – Felipe Voloch
    Commented Dec 30, 2016 at 17:56

1 Answer 1

Reset to default
3
$\begingroup$

Not a complete argument but here are some ideas. Suppose $f$ has degree $d$ and let $c_1,\ldots,c_{d-1}$ be the zeros of $f'(x)$. Consider the curve $X: f(x)-f(y)-a$ in the projective plane. It can only be singular at points of the form $(c_i,c_j)$ with $a=f(c_i)-f(c_j)$. So if $a$ is not one of those $(d-1)(d-2)$ values then $X$ is smooth, hence irreducible and the gcd is $1$ for every $g, \deg g < \deg f$. Even if $a$ is one of the "bad" values, as long as the bad value does not repeat for many pairs $(c_i,c_j)$ we can still conclude that $X$ is irreducible. Namely, if $X$ is the union of two curves of degrees $m,d-m$ it has at least $m(d-m)>d-1$ singularities where the two curves meet. (One needs to be careful with intersections of high multiplicity here, I am ignoring these subtleties). I think it should be easy to rule out $m=1$ so you might even improve the number of coincidental $(c_i,c_j)$ to $2(d-2)$. Is it even possible to construct an $f$ such that $f(c_i)-f(c_j)$ takes the same value for at least $2(d-2)$ pairs $(c_i,c_j)$ of roots of $f'$?

Improve this answer
$\endgroup$
1
  • $\begingroup$ @Voloch: Many thanks for your answer. $\endgroup$
    – MO B
    Commented Dec 31, 2016 at 20:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .