-1

I come across the following terminology: Chebyshev polynomials over finite fields, they are used in graph theory. They have the same recurrence equation and other properties. I am confused about the variable $x$ in their definition. Is it the ordinary real variable in $[-1,1]$ or the $x$ is an element of the finite field.

Safwane
  • 3,840

1 Answers1

1

If $\mathbb F$ is a field, and $p(x)\in \mathbb F[x]$ is any polynomial with coefficients in $\mathbb F$ then you can certainly evaluate $p(x)$ at any element in that field or in any extension of that field.

Thus, if $p(x)=x^2+1$ is defined over $\mathbb Q$, it would make sense to evaluate it on any complex number as $\mathbb Q$ is a subfield of $\mathbb C$. Similarly, if $p(x)$ is defined over a finite field it would make sense to evaluate it on an element of that field or any extension of that field.

lulu
  • 70,402
  • So, $x$ in those polynomials of Chebyshev is in the finite field. – Safwane Jan 04 '20 at 15:56
  • Well, if $a$ is an element of that field (or any extension of it) then $p(a)$ is well defined. $x$ of course is a variable, it is not itself an element of that field. But don't ignore the part about extension fields. It is quite important that we can evaluate polynomials on extensions of the field of definition. – lulu Jan 04 '20 at 15:57
  • Sorry, I don't want to read that paper. But I'm sure they just mean the standard thing. If you have a polynomial $p(x)\in \mathbb F[x]$ then it makes sense to consider $p(a)$ for $a\in \mathbb F$ $\textit {or}$ for $a$ in any extension field of $\mathbb F$. Generally of course one is interested in the algebraic properties of the polynomial (or collection of polynomials) and not in particular values of it (resp. them). – lulu Jan 04 '20 at 16:02