1. The polynomail is irreducible in . Find the number if real roots of
2. Let M be the splitting field for over Show that is a Galois extension
and determine its Galois group Is solvable by root?
In the following it can be used without a proof that the Discrimentant of
is
3. Show that is contained in and determine that
is irreducible in
4. Determine the Galois group
Using analytic methods (or just graphing it) we see that has three real roots, the other two roots are complex conjugates.
Obviously, is a Galois extension since it is a splitting field over a perfect field. We want to compute . First of all we can identify the Galois group as a subgroup of because each automorphism of is also a permutation of the roots of . To show that all we need to do is show that contains a permutation of order 5 and a transpotion.* Let be a root of in then . Notice that , therefore divides by the fundamental theorem of Galois theory. Thus, by Cauchy's theorem there is an element in of order i.e. a -cycle if viewed as an element in . And surly there is a transposition in , just take defined by i.e. complex conjugation. Then is a transposition if viewed as an element of . Therefore, . This symmetric group is not solvable therefore the polynomial is not solvable by radicals.2. Let M be the splitting field for over Show that is a Galois extension
and determine its Galois group Is solvable by root?
*)A subgroup of generated by a 5-cycle and a 2-cycle must be itself.
Let then corresponds to subgroup . Notice that , therefore under this Galois correspondence.3. Show that is contained in and determine that
is irreducible in
4. Determine the Galois group