2017-04-19, 20:33 | #1 |
"Sam"
Nov 2016
2·163 Posts |
Polynomials defining the same field as cyclotomic polynomial order 5
I haven't been on here a while, I was studying further information and knowledge on monic polynomials P(x) = x^4 + a*x^3 + b*x^2 + c*x + d defining the same field as cyclotomic polynomial 5 C_5(x) = x^4 + x^3 + x^2 + x + 1. A conjecture I made a while back reverting from this states that there are infinitely many polynomials P(x) defining the same field as C_n(x) for fixing the largest (n+1)/2 coefficients (a_k) (for prime n) as long as they satisfy conditions
a (mod n) = t^(n-2), a_2 (mod n) = t^(n-3), a_3 (mod n) = t^(n-4),...... a_((n-1)/2) = t^((n+1)/2)). This seems to be proven for C_5(x) and the proof for C_7(x) might require more knowledge about cubic polynomials and fields. Proof: Choose any two integers; A and B such that A^2 (mod 5) = B (mod 5). and solve -C^2 + D^2 + T^2 = (8*B-3*A^2)/5 [1] If D + A + C + T = 2 (mod 4), (if not then change the sign of T so it is). then solve [D = a - b - c + d, A = a + b + c + d, C = - a - b + c + d, T = a - b + c - d] [2] with the solution set (a, b, c, d) write; a*x + b*x^2 + c*x^3 + d*x^4 [3] then take the minimum polynomial of [a*x + b*x^2 + c*x^3 + d*x^4, x^4 + x^3 + x^2 + x + 1] Since we can choose infinitely many values of T, we will get infinitely many polynomials defining the same field as C_5(x). Example; take A = 9, B = 21, 9^2 (mod 5) = 21 (mod 5), -C^2 + D^2 + T^2 = (8*21-3*9^2)/5 = -15 choosing T = 1, D^2 - C^2 = 16 D = 5 C = 3 then we check 5 + 9 + 3 + 1 = 18 = 2 (mod 4). then solve [5 = a - b - c + d, 9 = a + b + c + d, 3 = - a - b + c + d, 1 = a - b + c - d] a = 3, b = 0, c = 2, d = 4 Replacing the coefficients into [3]: 3*x + 0*x^2 + 2*x^3 + 4*x^4 = 3*x + 2*x^3 + 4*x^4 take the minimum polynomial of [3*x + 2*x^3 + 4*x^4, x^4 + x^3 + x^2 + x + 1] = x^4 + 9*x^3 + 21*x^2 + 19*x + 131 I am making efforts in researching the field of C_7(x) and finding a similar equation and system of solutions for fixing the first four coefficients (1, a, b, c) of x^6 + a*x^5 + b*x^4 + c*x^3 + d*x^2 + f*x + g. Fixing the coefficient of x^3 requires more knowledge about cubic forms, and I don't have the knowledge or theories for this. Anyone else care to investigate this problem? I spent a lot of time and effort trying to find and come up with these assertions. Thanks for help, comments, suggestions. |
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