4 ms·
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 pol
by klodolph 7y ago
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.
- deleted 7y ago[deleted]
- yarg 7y agoIt 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 7y agoAppreciate what you're saying, but the linked paper is talking about results in F_p[T], the polynomial ring.
- deleted 7y ago[deleted]
- edflsafoiewq 7y agoGenerally 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.