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Old 2017-12-26, 18:35   #1
carpetpool
 
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"Sam"
Nov 2016

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Post Construction of polynomials with same discriminants

Hi all,

For the monic polynomial P(x) = x^n+a_1*x^(n-1)+a_2*x^(n-2)+......+a_(n-2)*x^2+a_(n-1)*x+r where r = p*q, there exists a polynomial Q(x) = p*x^n+a_1*x^(n-1)+a_2*x^(n-2)+......+a_(n-1)*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^2-4*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 9^2-4*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): 9^2-4*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, 200-degree 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 2017-12-26 at 18:36
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Old 2017-12-26, 18:49   #2
bhelmes
 
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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 b²-4ac because polynomials with the same determinant
"describes" the same primes

Greetings from the primes
Bernhard
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Old 2017-12-26, 21:30   #3
carpetpool
 
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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 200-degree 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 straight-up solutions unfortunately --- any help?
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Old 2017-12-26, 22:14   #4
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http://mathworld.wolfram.com/Polynom...criminant.html
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