 mersenneforum.org Trying to solve discriminant of a cubic polynomial by product method
 Register FAQ Search Today's Posts Mark Forums Read 2018-11-24, 03:47 #1 carpetpool   "Sam" Nov 2016 52·13 Posts Trying to solve discriminant of a cubic polynomial by product method If P(x) = x^2 + a*x + b is a monic quadratic polynomial with b = c*d and the discriminant D = a^2-4*b, then Q(x) = c*x^2 + a*x + d is the corresponding polynomial to P(x) with the same discriminant as P(x) and defining the same number field as P(x). What is the case with cubic polynomials? Suppose that P(x) = x^3 + a*x^2 + b*x + c and c = f*g. D = -4*a^3*c + a^2*b^2 + 18*a*b*c - 4*b^3 - 27*c^2 is the discriminant of P(x). The problem is to find the corresponding polynomial Q(x) = f*x^3 + d*x^2 + e*x + g such that D2 = D (where D2 is the discriminant of Q(x)) and the fields defined by P(x) and Q(x) are isomorphic. Thus, this leads to finding a solution [d, e] for the equations: (D2=-4*d^3*g + e^2*d^2 + 18*e*d*f*g - 27*f^2*g^2 - 4*e^3*f, is also the discriminant of any general cubic polynomial) -4*d^3*g + e^2*d^2 + 18*e*d*f*g - 27*f^2*g^2 - 4*e^3*f = -4*a^3*c + a^2*b^2 + 18*a*b*c - 4*b^3 - 27*c^2 (implies that D2=D) c = f*g Using some basic algebra (substitution) we find that: -4*d^3*g + e^2*d^2 + 18*e*d*f*g - 27*f^2*g^2 - 4*e^3*f = -4*a^3*c + a^2*b^2 + 18*a*b*c - 4*b^3 - 27*c^2 -4*d^3*g + e^2*d^2 + 18*e*d*f*g - 27*f^2*g^2 - 4*e^3*f = -4*a^3*f*g + a^2*b^2 + 18*a*b*f*g - 4*b^3 - 27*f^2*g^2 -4*d^3*g + e^2*d^2 + 18*e*d*f*g - 4*e^3*f = -4*a^3*f*g + a^2*b^2 + 18*a*b*f*g - 4*b^3 -4*d^3*g + e^2*d^2 + 18*e*d*f*g - 4*e^3*f = -4*a^3*f*g + a^2*b^2 + 18*a*b*f*g - 4*b^3 Since we are only solving for d and e, the right hand side of this equation is a constant, while the left had side remains a two variable cubic expression, thus getting d and e in terms of a, b, f, and g seems extremely difficult. Any help or ideas? Thanks   2018-11-24, 09:51   #2
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts Quote:
 Originally Posted by carpetpool If P(x) = x^2 + a*x + b is a monic quadratic polynomial with b = c*d and the discriminant D = a^2-4*b, then Q(x) = c*x^2 + a*x + d is the corresponding polynomial to P(x) with the same discriminant as P(x) and defining the same number field as P(x). What is the case with cubic polynomials? Suppose that P(x) = x^3 + a*x^2 + b*x + c and c = f*g. D = -4*a^3*c + a^2*b^2 + 18*a*b*c - 4*b^3 - 27*c^2 is the discriminant of P(x). The problem is to find the corresponding polynomial Q(x) = f*x^3 + d*x^2 + e*x + g such that D2 = D (where D2 is the discriminant of Q(x)) and the fields defined by P(x) and Q(x) are isomorphic. Thus, this leads to finding a solution [d, e] for the equations: (D2=-4*d^3*g + e^2*d^2 + 18*e*d*f*g - 27*f^2*g^2 - 4*e^3*f, is also the discriminant of any general cubic polynomial) -4*d^3*g + e^2*d^2 + 18*e*d*f*g - 27*f^2*g^2 - 4*e^3*f = -4*a^3*c + a^2*b^2 + 18*a*b*c - 4*b^3 - 27*c^2 (implies that D2=D) c = f*g Using some basic algebra (substitution) we find that: -4*d^3*g + e^2*d^2 + 18*e*d*f*g - 27*f^2*g^2 - 4*e^3*f = -4*a^3*c + a^2*b^2 + 18*a*b*c - 4*b^3 - 27*c^2 -4*d^3*g + e^2*d^2 + 18*e*d*f*g - 27*f^2*g^2 - 4*e^3*f = -4*a^3*f*g + a^2*b^2 + 18*a*b*f*g - 4*b^3 - 27*f^2*g^2 -4*d^3*g + e^2*d^2 + 18*e*d*f*g - 4*e^3*f = -4*a^3*f*g + a^2*b^2 + 18*a*b*f*g - 4*b^3 -4*d^3*g + e^2*d^2 + 18*e*d*f*g - 4*e^3*f = -4*a^3*f*g + a^2*b^2 + 18*a*b*f*g - 4*b^3 Since we are only solving for d and e, the right hand side of this equation is a constant, while the left had side remains a two variable cubic expression, thus getting d and e in terms of a, b, f, and g seems extremely difficult. Any help or ideas? Thanks
if f is 1, a=d;b=e works without solving for g

Last fiddled with by science_man_88 on 2018-11-24 at 09:52   2018-11-24, 16:35 #3 Batalov   "Serge" Mar 2008 Phi(4,2^7658614+1)/2 17·563 Posts Rewriting one cubic equation into another is not a big help to find a solution. If you cannot solve x^3 - 3*x -1 = 0 properly, then you cannot solve it with your coefficient juggling.   2018-11-24, 17:39   #4
carpetpool

"Sam"
Nov 2016

32510 Posts Quote:
 Originally Posted by Batalov Rewriting one cubic equation into another is not a big help to find a solution. If you cannot solve x^3 - 3*x -1 = 0 properly, then you cannot solve it with your coefficient juggling.
I recall there being a cubic formula (its like the quadratic formula for solving quadratic equations except for cubic equations). One of the solutions is given by

x = l/l2

where

l = -(2 - 2*sqrt(-3) + 2^(1/3) (1 + sqrt(-3))^(5/3))
l2 = (2*2^(2/3)*(1 + sqrt(-3))^(1/3))

^^^
I had to look that one up using Wolfram Alpha   2018-11-24, 17:47 #5 Batalov   "Serge" Mar 2008 Phi(4,2^7658614+1)/2 17·563 Posts Forget Wolfram Alpha. Take your own post #1 and demonstrate that it does any good for this easy equation (a=1, b=0, ...etc).   2018-11-27, 16:37 #6 Dr Sardonicus   Feb 2017 Nowhere 33×5×37 Posts There's another issue. Although all cubic fields are "isomorphic" as far as field operations are concerned, there can be distinct cubic fields having defining polynomials in Z[x] with the same field discriminant D and same Galois group. In some cases, it is possible to rewrite polynomials to get defining polynomials with the same polynomial discriminant d, sometimes not. Luckily, with cubic fields, the issue of "signature" (number of real roots, and of pairs of complex-conjugate roots) does not further complicate matters. Evenly many pairs of complex-conjugate roots means a positive discriminant, oddly many pairs a negative discriminant. With a cubic in Z[x], the only possibilities are 3 real roots and no (evenly many) pairs of complex-conjugate roots (positive discriminant), or 1 real root and one pair (oddly many) of complex-conjugate roots (negative discriminant). With degree 4, either 4 real roots and no complex-conjugate pairs, or no real roots and two pairs of complex-conjugate roots mean a positive discriminant, while two real roots and one pair of complex-conjugate roots means a negative discriminant. Here are examples (culled from tables) of cubic discriminants of "multiplicity" greater than 1, in each possible case. G = A3, D = 3969 (different d's, though) x^3 - 21*x - 28 x^3 - 21*x - 35 (In this case, the field discriminants are the same, but good luck rewriting defining polynomials to get equal polynomial discriminants. Although there are oodles of examples of multiple cubic fields with the same field discriminant and G = A3, I was too lazy to devote serious effort to finding examples where I could match up polynomial discriminants in this case.) G=S3, D = -972, d = -3888 x^3 + 12*x^2 + 12*x + 4 x^3 - 12 G = S3, D = d = -1228 x^3 - x^2 + x - 7 x^3 - x^2 + 7*x - 1 x^3 + 4*x - 6 G = S3, D = d = 22356 x^3 + 3*x^2 - 15*x - 23 x^3 + 3*x^2 - 51*x - 203 x^3 + 3*x^2 - 33*x - 113   2018-11-28, 13:16   #7
Dr Sardonicus

Feb 2017
Nowhere

10011100000112 Posts Quote:
 Originally Posted by Dr Sardonicus There's another issue. Although all cubic fields are "isomorphic" as far as field operations are concerned, there can be distinct cubic fields having defining polynomials in Z[x] with the same field discriminant D and same Galois group.
Aw, geez! That's gotta be one of the dumbest things I've posted! Any two cubic field extensions are isomorphic as Q-vector spaces -- not as fields! If you define the extensions via polynomial modulo arithmetic, and define a correspondence by equating the quadratic remainders, addition works just fine, and multiplication by rational numbers works just fine. But when you multiply by field elements that aren't in Q, the party's over...

The part about different cubic fields having the same (field) discriminant is of course quite correct, as shown by example.

You can get field isomorphism via algebraic conjugacy. Sometimes, with a defining polynomial f, algebraic conjugates of Mod(x, f) can be expressed as Mod(r(x), f) where r(x) is a polynomial (other than the identity polynomial x) of degree less than that of f. Sometimes not.

You can also "rewrite" a defining polynomial using a "Tschirnhausen transformation." Pari's function poltschirnhaus() applies a "random" Tschirnhausen transformation. If you pick one yourself -- and for a cubic polynomial f, this would look like t = Mod(a + b*x + c*x^2, f) with a, b, c rational, you can get the new polynomial by applying charpoly(t). Of course if b and c are 0, this will not be irreducible
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