I wrote:
There's an extensive theory of "complex multiplication" that shows (among many other things) that if a,b are rational numbers with b>0 then DedekindEta(a+b*I) / DedekindEta(I) is algebraic, and that some power of it (at most the 24th, but maybe one never has to go that far) generates an abelian extension of the Gaussian numbers.
Bill Gosper <billgosper@gmail.com> replies:
So THAT's what Hilbert was raving about!
Yes, in part. (Kronecker's Jugendtraum ~ Hilbert's 12th problem.)
That means that there's always an expression "in radicals", but you probably don't want to write it that way past the first few cases. You don't even want to see a solution in radicals of x^5 - 2*x^4 + 2*x^3 - x^2 + 1 . . .
polySolve[x^5 - 2*x^4 + 2*x^3 - x^2 + 1]
{{x -> 2/5 + (-597 - 215 Sqrt[5] + 5 Sqrt[94 (265 + 101 Sqrt[5])])^(1/5) / [ . . . . . ]
Can't say I didn't warn you!
(somehow turning 1s and 2s into 265 and 597.-)
That much is not a surprise. The solution of x^3 = x + 1 (and thus the closed form for one version of "Tribonacci numbers") already involves the square root of 69 (no joke), and x^4 = x + 1 is even worse. The quintic polynomial I gave generates a field with dihedral Galois group and discriminant 47^2, and figures in the "complex multiplication" picture, connected with the values of elliptic modular functions at suitable points whose imaginary part is a multiple of sqrt(47).
Actually, for 1 + 110287494 x + 70410440943 x^2 - 83149258860 x^3 + 380834340687 x^4 - 313810596090 x^5 + 282429536481 x^6, the sextic behind eta(i + 1/3), polysolve returned an expression so large as to defeat even one step of simplification.
Using gp's "polredabs" I find that this sextic field (again with dihedral Galois group over the rationals) contains a root of r^6 - 4*r^3 + 1. You can solve for this r, and then recover x = (36*r^5 - 8*r^4 - 8*r^3 - 144*r^2 + 44*r + 31) / 81.
Many thanks for all of the above. —rwg
You're welcome, --NDE