## Math 503: Abstract Algebra

Homework 10 <me@tylerlogic.com>
April 18, 2014
http://coursework.tylerlogic.com/courses/upenn/math503/homework10

### 1 Determine the Galois group of the splitting field of T4- 3

Since the polynomial T4 - 3 can be broken down via the difference of squares as

then the roots are ±,±i. So the splitting field of T4 - 3 over is (,i). This yields

 (1.1)

as the possible elements of the Galois group of K, leaving us with a potential Galois group of size 8.

Now since T4 - 3 is irreducible over (by Eisenstein’s criteria), then () is an extension of degree four. In (), T4 - 3 factors as T4 - 3 = (T -)(T + )(T2 + ) where (T2 + ) is irreducible with roots ±i. Then K = (,i) is an extension of degree two over (). Therefore K has degree 8 over , and this implies that the 8 automorphisms above are all elements of the Galois group of K, and furthermore are the only elements.

Note that every automorphism σ of the Galois group of K will take to an element in K of the form (-1)min and therefore take i to an element of the form (-1)rin+1 for m,n,r ∈{0,1}. This then defines the mapping of i as

Conversly if we were to define the the mapping of only and i, mapping them to (-1)min and (-1)ri, respectively, we can recover where i is mapped to as (-1)ri(-1)min = (-1)r+min+1, resulting in one of the automorphisms in equation 1.1. Hence, we can simply define an automorphism in Gal(K∕ ) by its mapping of to any other root of T4 - 3 and its mapping of i to ±i.

So let’s define

form which we see |σ| = 4 and |τ| = 2. Furthermore we see that σ and τ generate Gal(K∕ ) since any automorphism

in the Galois group, where m ∈{0,1,2,3} and n ∈{0,1} is equal to σmτn. Finally we have

so that στ = τσ3. Since |σ| = 4 then this can be restated as στ = τσ. Hence Gal(K∕ ) = σ,τ|σ4 = τ2 = 1,στ = τσ-1, or in other words, Gal(K∕ ) is the dihedral group, D8, of size eight.

### 3

#### (a)

Using the fact that ζ2 + ζ + 1 = 0, α + β + γ = 0, R(ζ;α,β,γ) = α + ζ β + ζ2 γ, and R(ζ2;α,β,γ) = α + ζ2 β + ζ γ, we have
Furthermore, since ζ3 = 1 then
and finally