quartic polynomial with Galois group
which factors over into a linear and an irreducible quadratic. Additionally, remains irreducible over , since none of the roots of lie in this field and the discriminant of , regarded as a quadratic in , does not lie in this field either, so cannot factor as a product of two quadratics. So according to the article on the Galois group of a quartic polynomial, should indeed have Galois group isomorphic to . We show that this is the case by explicitly examining the structure of its splitting field.
Let be the splitting field of over , and let .
Let and . Clearly contains both and and thus contains . But obviously splits in , so that . We next determine the degree of over .
Note that since is a real field while is not. Thus . Clearly , so . But
so . Hence ; call this field .
Since , we also have and ; thus is a quadratic extension of each and .
Putting these results together, we see that
so that has order .
Now, neither nor is Galois over (since the Galois closure of either one is ), so that the subgroup of fixing (say) is a nonnormal subgroup of . Thus must be nonabelian, so must be isomorphic to either or (the quaternions). But the subgroups of corresponding to and are distinct subgroups of order in , and has only one subgroup of order . Thus . (Alternatively, note that all subgroups of are normal, so since it has a nonnormal subgroup).
|Title||quartic polynomial with Galois group|
|Date of creation||2013-03-22 17:44:09|
|Last modified on||2013-03-22 17:44:09|
|Last modified by||rm50 (10146)|