[math-fun] 12th degree radicals?
In http://mathworld.wolfram.com/SquareDissection.html, x₂ satisfies 6 - 4 √2 + (-24 + 16 √2) x + (28 - 20 √2) x^2 + (-8 + 8 √2) x^3 - 2 x^4 + x^6, whereat my sextic solver chokes on the √2s. Does anybody know if x comes out in radicals? --rwg
Bill Gosper <billgosper@gmail.com> wrote:
In http://mathworld.wolfram.com/SquareDissection.html, x_2 satisfies [...]
These fancy \sqrt signs just make it hard to plug into a software package that might do this. Do you mean the polynomial with coefficients 1, 0, -2, -8+8*w, 28-20*w, -24+16*w, 6-4*w where w^2 = 2? That seems to have Galois group S_6 over Q(sqrt(2)). It satisfies an irreducible polynomial of degree 12 over Q, too large for gp's "polgalois"; but reducing modulo some primes finds factors of degrees 5,1 (for a prime above 23), degree 6 (both primes above 47) and degrees 4,2 (a prime above 71), which I think is enough to exclude all proper subgroups of S_6, let alone solvability by radicals. Also the discriminant has a huge prime factor of norm 218206319 with multiplicity 1. NDE
On Thu, Feb 27, 2014 at 11:03 AM, Noam Elkies <elkies@math.harvard.edu>wrote:
Bill Gosper <billgosper@gmail.com> wrote:
In http://mathworld.wolfram.com/SquareDissection.html, x_2 satisfies [...]
These fancy \sqrt signs just make it hard to plug into a software package that might do this. Do you mean the polynomial with coefficients
1, 0, -2, -8+8*w, 28-20*w, -24+16*w, 6-4*w
where w^2 = 2?
Yes, sorry.
That seems to have Galois group S_6 over Q(sqrt(2)).
<Gloom>
It satisfies an irreducible polynomial of degree 12 over Q, too large for gp's "polgalois"; but reducing modulo some primes finds factors of degrees 5,1 (for a prime above 23), degree 6 (both primes above 47) and degrees 4,2 (a prime above 71), which I think is enough to exclude all proper subgroups of S_6, let alone solvability by radicals. Also the discriminant has a huge prime factor of norm 218206319 with multiplicity 1.
NDE
Thanks!
Dear Bill, I used the online Magma calculator at http://magma.maths.usyd.edu.au/calc/, and tested your 12-deg (6-deg over sqrt ext). Z := Integers(); P < x > := PolynomialRing(Z); f := 4 - 32*x + 80 *x^2 - 32 *x^3 - 168 *x^4 + 288 *x^5 - 164 *x^6 - 16 *x^7 + 60 *x^8 - 16 *x^9 - 4 *x^10 + x^12; G, R := GaloisGroup(f); G; It gave the order as 1036800. This is 12T299, and unfortunately is a non-solvable group. Sincerely, Tito On Thu, Feb 27, 2014 at 2:13 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Thu, Feb 27, 2014 at 11:03 AM, Noam Elkies <elkies@math.harvard.edu>wrote:
Bill Gosper <billgosper@gmail.com> wrote:
In http://mathworld.wolfram.com/SquareDissection.html, x_2 satisfies [...]
These fancy \sqrt signs just make it hard to plug into a software package that might do this. Do you mean the polynomial with coefficients
1, 0, -2, -8+8*w, 28-20*w, -24+16*w, 6-4*w
where w^2 = 2?
Yes, sorry.
That seems to have Galois group S_6 over Q(sqrt(2)).
<Gloom>
It satisfies an irreducible polynomial of degree 12 over Q, too large for gp's "polgalois"; but reducing modulo some primes finds factors of degrees 5,1 (for a prime above 23), degree 6 (both primes above 47) and degrees 4,2 (a prime above 71), which I think is enough to exclude all proper subgroups of S_6, let alone solvability by radicals. Also the discriminant has a huge prime factor of norm 218206319 with multiplicity 1.
NDE
Thanks!
Tito Piezas <tpiezas@gmail.com> writes:
Dear Bill,
I used the online Magma calculator at http://magma.maths.usyd.edu.au/calc/, and tested your 12-deg (6-deg over sqrt ext). [...]
It gave the order as 1036800. This is 12T299, and unfortunately is a non-solvable group.
a.k.a. the subgroup of S_{12} that stabilizes a partition of the 12 roots into two sets of 6, which is the largest it could be given that it comes from a 6th-degree polynomial over a quadratic number field. NDE
participants (3)
-
Bill Gosper -
elkies＠math.harvard.edu -
Tito Piezas