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 T⁴ - 3

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

√ - √- √ - √- √- √- √ - √- T4 - 3 = (T2 - 3)(T 2 + 3) = (T 2 - 3)(T2 - (- 3)) = (T - 43)(T + 43)(T - i 43)(T + i 43)

then the roots are ± √43 ,±i 4√3 . So the splitting field of T⁴ - 3 over ℚ is ℚ( 4√3 ,i 4√3 ). This yields

{ √ - √ - { √- √- 4√3 ↦→ ± 4√3- 4√3- ↦→ ±i√43 i 43 ↦→ ±i 43 i 43 ↦→ ± 4 3

(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 ℚ (by Eisenstein’s criteria), then ℚ( 4√3 ) is an extension of degree four. In ℚ( 4√3 ), T⁴ - 3 factors as T⁴ - 3 = (T - 4√3 )(T + 4√3 )(T² + ) where (T² + ) is irreducible with roots ±i 4√3 . Then K = ℚ( 4√3- ,i √43- ) is an extension of degree two over ℚ( √43- ). 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 4√3- to an element in K of the form (-1)^miⁿ 4√3 and therefore take i √43 to an element of the form (-1)^riⁿ⁺¹ 4√3 for m,n,r ∈{0,1}. This then defines the mapping of i as

σ(i) = σ-1(√43-)σ (i 4√3-) = ((- 1)min 4√3)-1(- 1)rin+1 4√3 = (- 1)r- mi

Conversly if we were to define the the mapping of only √ -
4 3 and i, mapping them to (-1)^miⁿ √-
43 and (-1)^ri, respectively, we can recover where i √ -
43 is mapped to as (-1)^ri(-1)^miⁿ √-
43 = (-1)^r+miⁿ⁺¹ √-
43 , 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 √-
43 to any other root of T⁴ - 3 and its mapping of i to ±i.

So let’s define

{ √ - √- { √ - √- σ = 43 ↦→ i 43 τ = 43 ↦→ 43 i ↦→ i i ↦→ - i

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

{ √43- ↦→ im 4√3 i ↦→ (- 1)ni

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

√ - √- √ - √ - √ - τσ3( 43) = τ(- i 43) = i 43 = σ ( 43) = στ ( 43) τσ3(i) = τ(i) = - i = σ(- i) = σ τ(i)

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

2

(a)

(b)

(c) Extra Credit

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

3α = 2α- β - γ = 2α+ (ζ2 + ζ)β + (ζ2 + ζ)γ = α+ ζ2β +ζβ + α + ζ2γ + ζγ 2 2 = (α + ζβ + ζ γ)+ (α + ζβ + ζγ) = R(ζ;α,β,γ)+ R(ζ2;α,β,γ)

Furthermore, since ζ³ = 1 then

3β = 2β - α- γ = 2β + (ζ2 + ζ)α + (ζ2 + ζ)γ = β + ζ2α+ ζα + β + ζ2γ +ζγ 2 2 = (ζ α +β + ζγ)+ (ζα+ β + ζ γ) = (ζ2α +ζ3β + ζ4γ)+ (ζα + ζ3β + ζ2γ) = ζ2(α +ζ1β + ζ2γ)+ ζ(α + ζ2β + ζγ) = ζ2R (ζ;α,β,γ) +ζR (ζ2;α,β,γ)

and finally

3γ = 2γ - α- β 2 2 = 2γ + (ζ2 + ζ)α + (ζ +2 ζ)β = γ + ζ α+ ζα + γ + ζ β + ζβ = (ζα + ζ2β + γ)+ (ζ2α+ ζβ + γ) = (ζα + ζ2β + ζ3γ)+ (ζ2α + ζ4β + ζ3γ) 2 2 2 = ζ(α + ζβ + ζ γ)+2 ζ (α2+ ζ β +ζγ ) = ζR (ζ;α,β,γ) + ζR (ζ;α,β,γ)