Question

Given a monic cubic function `x^3+bx^2+cx+d` with zeros `alpha, beta, gamma` how can you construct a set of `6xx6` rational matrices that form a field isomorphic to `QQ[alpha, beta, gamma]` ?

Assume that `QQ[alpha] != QQ[alpha, beta, gamma]`, `QQ[beta] != QQ[alpha, beta, gamma]` and `QQ[gamma] != QQ[alpha, beta, gamma]`, i.e. the field generated by all three zeros is larger than the fields generated by any one of them individually.

Answer 1

See explanation...

Explanation:

Given the conditions of the question,

`x^3+bx^2+cx+d`

is the minimum polynomial of `alpha` over `QQ`.

Also, note that the minimum polynomial of `beta` or `gamma` over `QQ[alpha]` is:

`(x^3+bx^2+cx+d)/(x-alpha) = x^2+(b+alpha)x+(c+alphab+alpha^2)`

The companion matrix of `alpha` over `QQ` is:

`M = ((0, 0, -d), (1, 0, -c), (0, 1, -b))`

Then:

`M^2 = ((0, -d, b d), (0, -c, b c - d), (1, -b, b^2 - c))`

While the companion matrix of `beta` over `QQ[alpha]` is:

`((0, -c-alphab-alpha^2), (1, -b-alpha))`

We find:

`-cI-bM-M^2`

`= ((-c, 0, 0),(0, -c, 0),(0, 0, -c))+((0, 0, bd), (-b, 0, bc), (0, -b, b^2))- ((0, -d, b d), (0, -c, b c - d), (1, -b, b^2 - c))`

`= ((-c, d, 0), (-b, 0, d), (-1, 0, 0))`

`-bI-M`

`=((-b,0,0),(0,-b,0),(0,0,-b))-((0, 0, -d), (1, 0, -c), (0, 1, -b))`

`=((-b, 0, d), (-1, -b, c), (0, -1, 0))`

Hence we can represent `alpha` with the `6xx6` matrix:

`M_alpha = ((alpha, 0), (0, alpha)) = ((0, 0, -d, 0, 0, 0), (1, 0, -c, 0, 0, 0), (0, 1, -b, 0, 0, 0), (0, 0, 0, 0, 0, -d), (0, 0, 0, 1, 0, -c), (0, 0, 0, 0, 1, -b))`

and `beta` with the `6xx6` matrix:

`M_beta = ((0, -c-balpha-alpha^2), (1, -b-alpha)) = ((0, 0, 0, -c, -d, 0), (0, 0, 0, -b, 0, -d), (0, 0, 0, -1, 0, 0), (1, 0, 0, -b, 0, d), (0, 1, 0, -1, -b, c), (0, 0, 1, 0, -1, 0))`

The field generated by these two matrices is isomorphic to `QQ[alpha, beta, gamma]`