[tildedave]

Even if the number field is finite, there are an infinite number of polynomials for it. For example the finite field F2 has two elements {0, 1} but an infinite number of polynomials 0, 1, x, x + 1, x^2, x^2 + x + 1, x^2 + 1, «x^2 + x, and on and on. Some of these have nontrivial factors (x^2 + x = x(x+1)) and some don't (x^2 + x + 1 has no factors beyond itself and 1).

(The other answer to your question is more complete but also a bit more advanced, figured this was worth surfacing.)

[klodolph]

This is not correct, there are a finite number of polynomials for any finite field. This is using the usual definition for polynomial equivalence, where two polynomials are equal if they take the same values over all points in the domain.

You are talking about the polynomial ring, which is not a field.

The finite field F2 has four polynomials: 0, 1, x, and x+1. Other polynomials do not exist, because x^2=x.

[yarg]

It might help if you explain what is meant by F2.

As far as I can tell you're referring to the integer finite field containing the values {0, 1}.

  2 % 2 = 0
  3 % 2 = 1
  x^n % 2 = x
  2x % 2 = 0
  (2n + 1)x % 2 = x

So there are no (distinct) constants other than 0 or 1 and no multiples or exponents of x other than 1.

These facts might be obvious to someone who understands the jargon and theory of the mathematics in question, but probably need a bit more clarity when the target is the general populace.

[tildedave]

Appreciate what you're saying, but the linked paper is talking about results in F_p[T], the polynomial ring.

[edflsafoiewq]

Generally given a polynomial in R[x], you want to be able to evaluate it not just when x is in R, but when it's in any R-algebra. So we wouldn't automatically identify x^2 and x.