I believe your cubic t should be x^3 + y*x^2 + (y3)*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 degree5 polynomials known as Emma Lehmer's quintics.
There is a family of "simplest" degree6 number fields, defined by a oneparameter family of polynomials of degree 6 which have cyclic Galois groups.
