20171226, 18:35  #1 
"Sam"
Nov 2016
2^{2}·83 Posts 
Construction of polynomials with same discriminants
Hi all,
For the monic polynomial P(x) = x^n+a_1*x^(n1)+a_2*x^(n2)+......+a_(n2)*x^2+a_(n1)*x+r where r = p*q, there exists a polynomial Q(x) = p*x^n+a_1*x^(n1)+a_2*x^(n2)+......+a_(n1)*x^2+a_n*x+q with the same discriminant as P(x) and defines the same field as P(x). In fact, there should be a simple method to perform the construction of Q(x) from P(x) for any degree n. Does anyone know an easy method for constructing Q(x)? For quadratics n = 2, it is easy to construct. The quadratic polynomial f = a*x^2 + b*x + c has discriminant d = b^24*a*c. Hence the polynomials c*x^2 + b*x + c and x^2 + b*x + a*c have the same discriminant as f. For instance, let P(x) = x^2 + 9*x + 21. P(x) has discriminant 9^24*1*21 = 3 and 21 = 3*7. The construction of Q(x) would use 3 and 7 as the leading coefficients and or ending coefficients. Q(x) = 3*x^2 + 9*x + 7 has the same discriminant as P(x): 9^24*3*7 = 3. What about cubic polynomials a*x^3 + b*x^2 + c*x + d? When we are given P(x) = x^3 + a*x^2 + b*x + r where r = p*q, how can one construct Q(x) = p*x^3 + a_2*x^2 + b_2*x + q with the same discriminant as P(x) and defines the same field as P(x)? For instance, take the cubic polynomial P(x) = x^3 + 4*x^2  x + 15 which has discriminant d = 10975. What is the polynomial Q(x) = 3*x^3 + a*x^2 + b*x + 5 (or reversed possibly) with discriminant d = 10975 and defining the same field as P(x)? More specifically, is one able to show the work for the construction of Q(x)? Is this construction easy to do for say, 200degree polynomials or higher? (I do believe it is possible, I don't know the complexity of it however.) Thanks for help, comments, and suggestions. Last fiddled with by carpetpool on 20171226 at 18:36 
20171226, 18:49  #2 
Mar 2016
3·5^{3} Posts 
For prime generators with use of quadratic polynomials i recomand my page http://www.devalco.de/#106
It is useful to classify the quadratic polynomials concerning the determinant b²4ac because polynomials with the same determinant "describes" the same primes Greetings from the primes Bernhard 
20171226, 21:30  #3 
"Sam"
Nov 2016
2^{2}×83 Posts 
Thanks for that page, Bernhard. I find it useful for quadratic fields with class number h > 1. The point of starting this thread, was to investigate the determinants, leading coefficients and constants for higher degree polynomials, such as 200degree or higher.
The discriminant of the cubic polynomial a*x^3 + b*x^2 + c*x + d is b^2*c^2  4*a*c^3  4*b^3*d  27*a^2*d^2 + 18*a*b*c*d. I tried one example with P(x) = X^3 + X + 9 with discriminant d = 0^2*1^2  4*1*1^3  4*9^3*0  27*1^2*9^2 + 18*1*0*1*9 = 2191 = 7*313 Using the p*q constant product I was talking about earlier, I solved for Q(x) discriminant = 2191. Q(x) = 3*X^3 + b*X^2 + c*X + 3 2191 = b^2*c^2  4*3*c^3  4*b^3*3  27*3^2*3^2 + 18*3*b*c*3 2191 = b^2*c^2  12*c^3  12*b^3  2187 + 162*b*c I wasn't able to find any straightup solutions unfortunately  any help? 
20171226, 22:14  #4 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 

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