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Old 2020-04-12, 01:04   #3
Dr Sardonicus
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Feb 2017

2·3·11·59 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.
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