View Single Post
 2020-04-12, 01:04 #3 Dr Sardonicus     Feb 2017 Nowhere 2·32·199 Posts I believe your cubic t should be x^3 + y*x^2 + (y-3)*x - 1 (not +1). Then, replacing y with -y gives the usual formulation for the simplest cubic fields. The discriminant of your cubic T is 4*(y^2 + 27)^2. If you replace y with -y in your quartic T, it will coincide with the usual form for the "simplest quartic fields." You might want to look up Washington's cyclic quartic fields. I haven't tried comparing with yours, but there is a family of cyclic degree-5 polynomials known as Emma Lehmer's quintics. There is a family of "simplest" degree-6 number fields, defined by a one-parameter family of polynomials of degree 6 which have cyclic Galois groups.