How to solve a cyclic quintic in radicals?
Galois theory tells us that
can be solved in radicals because its group is solvable. Actually performing the calculation is beyond me, though - here what I have got so far:
Let the roots be , following Gauss we can split the problem into solving quintics and quadratics by looking at subgroups of the roots. Since 2 is a generator of the group [2,4,8,5,10,9,7,3,6,1] we can partition into the five subgroups of conjugate pairs [2,9],[4,7],[8,3],[5,6],[10,1].
Once one has one easily gets . It's easy to find . The point is that takes to and so takes to . Thus can be written down in terms of rationals (if that's your starting field) and powers of . Alas, here is where the algebra becomes difficult. The coefficients of powers of in are complicated. They can be expressed in terms of a root of a "resolvent polynomial" which will have a rational root as the equation is cyclic. Once one has done this, you have as a fifth root of a certain explicit complex number. Then one can express the other in terms of . The details are not very pleasant, but Dummit skilfully navigates through the complexities, and produces formulas which are not as complicated as they might be. Alas, I don't have the time nor the energy to provide more details.
If U is a set, let