## Math 503: Abstract Algebra

Homework 10
Lawrence Tyler Rush

<me@tylerlogic.com>

April 18, 2014

http://coursework.tylerlogic.com/courses/upenn/math503/homework10
### 1 Determine the Galois group of the splitting field of T^{4} - 3

Since the polynomial T^{4} - 3 can be broken down via the difference of squares
as
then the roots are ±,±i. So the splitting field of T^{4} - 3 over ℚ is ℚ(,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 T^{4} - 3 is irreducible over ℚ (by Eisenstein’s criteria), then ℚ() is an extension of degree four. In ℚ(),
T^{4} - 3 factors as T^{4} - 3 = (T -)(T + )(T^{2} + ) where (T^{2} + ) is irreducible with roots ±i. Then
K = ℚ(,i) is an extension of degree two over ℚ(). Therefore K has degree 8 over ℚ, 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)^{m}i^{n} and
therefore take i to an element of the form (-1)^{r}i^{n+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)^{m}i^{n} and (-1)^{r}i, respectively,
we can recover where i is mapped to as (-1)^{r}i(-1)^{m}i^{n} = (-1)^{r+m}i^{n+1}, resulting in one of the automorphisms in
equation 1.1. Hence, we can simply define an automorphism in Gal(K∕ ℚ) by its mapping of to any other root of T^{4} - 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∕ ℚ) since any
automorphism

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

so that στ = τσ^{3}. Since |σ| = 4 then this can be restated as στ = τσ. Hence Gal(K∕ ℚ) = ⟨σ,τ|σ^{4} = τ^{2} = 1,στ = τσ^{-1}⟩, or
in other words, Gal(K∕ ℚ) is the dihedral group, D_{8}, of size eight.

### 2

#### (a)

#### (b)

#### (c) Extra Credit

### 3

#### (a)

Using the fact that ζ^{2} + ζ + 1 = 0, α + β + γ = 0, R(ζ;α,β,γ) = α + ζ ⋅β + ζ^{2} ⋅γ,
and R(ζ^{2};α,β,γ) = α + ζ^{2} ⋅ β + ζ ⋅ γ, we have Furthermore, since ζ^{3} = 1 then and finally

#### (b)

#### (c)

#### (d)

#### (e)

### 4 Extra Credit