2018-07-25, 05:12 | #1 |
"Sam"
Nov 2016
2^{2}·79 Posts |
Families of cyclic cubic fields... and more
It is well known that if K is the splitting field of the polynomial t = x^3 + y*x^2 + (y-3)*x - 1, then K must be a cyclic cubic field, as explained here. This result is known as the "simplest cubic fields". However, I was interested in finding more cyclic cubic families, and I did happen to spot another occurence:
If K is the splitting field of the polynomial T = x^3 - y*x^2 - 9*x + y (assume t is irreducible), then K must be a cyclic cubic field. The proof for this is similar to the one illustrated in the pdf link. The discriminant of t is d = (y^2 - 3*y + 9)^2 while the discriminant of T is D = (y^2 + 27)^2. Both splitting fields of y^2 - 3*y + 9 and y^2 + 27 are isomorphic to the field of Einstein Integers --- a field obtained by adjoining a root of y^2 + y + 1 to the field of real numbers, known as the third cyclotomic field. Besides these two sole examples, are there any more unique cyclic cubic field "families" or is there perhaps a finite number of instances, and it might be the case that these are the only two instances? For quartic cyclic field families, by trial and error, I found that when K is the splitting field of the polynomial T = x^4 + y*x^3 - 6*x^2 - y*x + 1, then K must be a cyclic quartic field. t has the discriminant D = (y^2 + 16)^3. The same question related to cubic fields is addressed to quartic fields. Last fiddled with by carpetpool on 2020-04-14 at 21:56 |
2020-04-11, 02:19 | #2 |
"Sam"
Nov 2016
2^{2}·79 Posts |
I've been back on researching the task again. I thought I might share some new results, if they are not known already:
x^5 + (n^2 + n + 4)*x^4 + (-2*n^2 - 2*n + 2)*x^3 + (-n^4 - 2*n^3 - 10*n^2 - 9*n - 5)*x^2 + (-3*n^2 - 3*n - 2)*x + n^2 + n + 1 x^7 + (n^3 - 3*n^2 - 4*n - 1)*x^6 + (-3*n^5 + 9*n^3 + 3*n^2 + 3*n - 12)*x^5 + (3*n^7 + 6*n^6 - 4*n^5 + n^4 - 7*n^3 + 28*n^2 + 15*n + 7)*x^4 + (-n^9 - 5*n^8 - 2*n^7 - 5*n^6 + 6*n^5 - 11*n^4 - 21*n^3 + 12*n^2 - n + 28)*x^3 + (n^10 + n^9 + 2*n^8 - n^7 - 9*n^6 + 4*n^5 - 32*n^4 - 6*n^3 - 49*n^2 + 5*n - 14)*x^2 + (-n^9 + 3*n^8 - n^7 + 11*n^6 + 4*n^5 + 4*n^4 - 10*n^3 - 30*n^2 - 20*n - 9)*x - 3*n^7 + 3*n^6 - 7*n^5 + 14*n^4 - 5*n^3 + 12*n^2 - 6*n - 1 As far as I know, there is only one other known cyclic septic besides the one I have listed here. |
2020-04-12, 01:04 | #3 |
Feb 2017
Nowhere
2·5^{2}·71 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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