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]# ?

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]#