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.

1 Answer
George C.
Mar 10, 2018

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]

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