20170223, 06:18  #1 
"Sam"
Nov 2016
2×163 Posts 
Polynomials with constant terms replaced defining the same field
I don't know weather this is of interest or not, for the nth cyclotomic polynomials for prime n, namely 3, 5, 7 we have the cyclotomic polynomials:
x^2+x+1 = S(3) x^4+x^3+x^2+x+1 = S(5) x^6+x^5+x^4+x^3+x^2+x+1 = S(7) If the constant w is added to each of these, we can ask what w makes S(n)+w define the same field as S(n). For S(3), w = 6, 60, 546, 4920 The polynomials defining the same field as x^2+x+1 are: x^2+x+7 x^2+x+61 x^2+x+547 x^2+x+4921 In fact the w has a closed form: w = ((3^n+1)/4)1 However for S(5) and S(7) there are no such w > 0 known and believed that no w exists. Is someone able to go on about and find the truth of this conjecture. Hopefully you understand the point here. Last fiddled with by carpetpool on 20170223 at 06:41 
20170223, 18:36  #2  
Feb 2017
Nowhere
2·3^{3}·7·11 Posts 
Quote:
x^2 + x + (3*k^2 + 3*k + 1), with discriminant D = 3*(2*k + 1)^2 determines the same extension of Q as does x^2 + x + 1, for any rational value of k except 1/2; in particular, nonnegative integer values of k give every possible integer constant term that fulfills your condition, without repetition. Applying the same standard methods leads to rapidlyworsening algebraic messes for the terms past the nexthighest degree term of polynomials of degree greater than 2. For polcyclo(5) it wasn't hard to get polynomials determining the same extension of Q in which the x^4, x^3 and x^2 terms all match up, e.g. x^4 + x^3 + x^2 + 61541*x + 1579481 or x^4 + x^3 + x^2 + 3219741*x + 318038281 but, without any reason of mathematical interest to consider the problem further, I'm dropping it. 

20170225, 00:50  #3  
"Sam"
Nov 2016
2×163 Posts 
Quote:
Prove that for any polynomial P(x), of degree n, there exists a polynomial Q(x) defining the same field as P(x), with the same terms for degrees n to n/2 (or degrees n to (n+1)/2 if the highest degree n of P(x) is odd) meaning without replacing the coefficients for the terms of degrees from n to n/2 or (n+1)/2 depending on if n is odd or even, and there are infinitely many sets of new coefficients for degrees (n/2)1 to 0, or ((n+1)/2)1 to 0, depending on weather the highest degree n of P(x) is even or odd, for Q(x) to define same field as P(x), Q(x) not necessarily having the same discriminant as P(x). For example: P(x) = 5x^4+12x^36x^2+49x+3 There exists a polynomial Q(x) = 5x^4+12x^36x^2+ax+b (since the highest degree of P(x) is 4 in this case, and we do not need to vary the terms of degrees from 4 to 4/2 for P(x), and all other degrees from (4/2)1 to 0, we must vary, and call the solution pair to all varied coefficients) defining the same field as P(x), same or not the same discriminant as P(x), and there are also infinitely many (a, b) such that this is true. A more easier example: P(x) = 5x^216x+7, There are infinitely many solutions a for Q(x) = 5x^216x+a such that Q(x) defines the same field as P(x). Again, this is a (reasonable) conjecture which depends on a proof or counterexample, not necessarily a statement yet, though I seem to be treating it like one. Last fiddled with by carpetpool on 20170225 at 00:51 

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