sites.wcsu.edu€¦ · Fundamental Theorem of Algebra (Examples) Consider: x4 9 = (x2 3)(x2 +3) = (x p 3)(x + p 3)(x2 +3) = (x p 3)(x + p 3)(x i p 3)(x +i p 3) This could be partially

Great Big Galois Example

Dr. Chuck [email protected]

http://sites.wcsu.edu/roccacC. F. Rocca Jr. (WCSU) Great Big Galois Example 1 /26

[email protected]

http://sites.wcsu.edu/roccac

IntroductionPolynomials and Fundamental Theorem of AlgebraVisualizing Complex NumbersPermuting RootsFixed Fields and Groups

C. F. Rocca Jr. (WCSU) Great Big Galois Example 2 /26

Introduction

Quadratic Equations (Area) ∼ 2000 BCE

Cubic and Quartic Equations ∼ 1500 CEQuintic Equations and Above ∼ 1800 CEPermutations of Roots and the Birth of Group Theory


Introduction

Quadratic Equations (Area) ∼ 2000 BCECubic and Quartic Equations ∼ 1500 CE

Quintic Equations and Above ∼ 1800 CEPermutations of Roots and the Birth of Group Theory


Introduction

Quadratic Equations (Area) ∼ 2000 BCECubic and Quartic Equations ∼ 1500 CEQuintic Equations and Above ∼ 1800 CE

Permutations of Roots and the Birth of Group Theory


Introduction

Quadratic Equations (Area) ∼ 2000 BCECubic and Quartic Equations ∼ 1500 CEQuintic Equations and Above ∼ 1800 CEPermutations of Roots and the Birth of Group Theory




Fundamental Theorem of Algebra

Fundamental Theorem of Algebra: If f (x) is a polynomial ofdegree n with complex coe�cients, then over the complexnumbers f (x) factors into a product of n factors, not necessarilyall distinct.


Fundamental Theorem of Algebra (Examples)

Consider:

3x4 + 5x3 − 5x2 − 5x + 2 =

(x − 1)(3x3 + 8x2 + 3x − 2)= (x − 1)(x + 1)(3x2 + 5x − 2)= (x − 1)(x + 1)(x + 2)(3x − 1)= 3(x − 1)(x + 1)(x + 2)(x − 1/3)

So, this could be factored using just integers/rationals



Consider:

3x4 + 5x3 − 5x2 − 5x + 2 = (x − 1)(3x3 + 8x2 + 3x − 2)

= (x − 1)(x + 1)(3x2 + 5x − 2)= (x − 1)(x + 1)(x + 2)(3x − 1)= 3(x − 1)(x + 1)(x + 2)(x − 1/3)




Consider:

3x4 + 5x3 − 5x2 − 5x + 2 = (x − 1)(3x3 + 8x2 + 3x − 2)= (x − 1)(x + 1)(3x2 + 5x − 2)

= (x − 1)(x + 1)(x + 2)(3x − 1)= 3(x − 1)(x + 1)(x + 2)(x − 1/3)




Consider:

3x4 + 5x3 − 5x2 − 5x + 2 = (x − 1)(3x3 + 8x2 + 3x − 2)= (x − 1)(x + 1)(3x2 + 5x − 2)= (x − 1)(x + 1)(x + 2)(3x − 1)

= 3(x − 1)(x + 1)(x + 2)(x − 1/3)




Consider:

3x4 + 5x3 − 5x2 − 5x + 2 = (x − 1)(3x3 + 8x2 + 3x − 2)= (x − 1)(x + 1)(3x2 + 5x − 2)= (x − 1)(x + 1)(x + 2)(3x − 1)= 3(x − 1)(x + 1)(x + 2)(x − 1/3)




Consider:

3x4 + 5x3 − 5x2 − 5x + 2 = (x − 1)(3x3 + 8x2 + 3x − 2)= (x − 1)(x + 1)(3x2 + 5x − 2)= (x − 1)(x + 1)(x + 2)(3x − 1)= 3(x − 1)(x + 1)(x + 2)(x − 1/3)




Consider:

x4 − 12x2 + 27 =

(x2 − 9)(x2 − 3)= (x − 3)(x + 3)(x2 − 3)= (x − 3)(x + 3)(x −

√3)(x +

√3)

This could be partially factored using integers, but required realnumbers to finish the job



Consider:

x4 − 12x2 + 27 = (x2 − 9)(x2 − 3)

= (x − 3)(x + 3)(x2 − 3)= (x − 3)(x + 3)(x −

√3)(x +

√3)




Consider:

x4 − 12x2 + 27 = (x2 − 9)(x2 − 3)= (x − 3)(x + 3)(x2 − 3)

= (x − 3)(x + 3)(x −√3)(x +

√3)




Consider:

x4 − 12x2 + 27 = (x2 − 9)(x2 − 3)= (x − 3)(x + 3)(x2 − 3)= (x − 3)(x + 3)(x −

√3)(x +

√3)




Consider:

x4 − 12x2 + 27 = (x2 − 9)(x2 − 3)= (x − 3)(x + 3)(x2 − 3)= (x − 3)(x + 3)(x −

√3)(x +

√3)




Consider:

x4 − 9 =

(x2 − 3)(x2 + 3)= (x −

√3)(x +

√3)(x2 + 3)

= (x −√3)(x +

√3)(x − i

√3)(x + i

√3)

This could be partially factored using integers and reals, butrequired imaginary numbers to factor it completely. Note thatwhen the coe�cients of the polynomial are real, any complexroots come in conjugate pairs.



Consider:

x4 − 9 = (x2 − 3)(x2 + 3)

= (x −√3)(x +

√3)(x2 + 3)

= (x −√3)(x +

√3)(x − i

√3)(x + i

√3)




Consider:

x4 − 9 = (x2 − 3)(x2 + 3)= (x −

√3)(x +

√3)(x2 + 3)

= (x −√3)(x +

√3)(x − i

√3)(x + i

√3)




Consider:

x4 − 9 = (x2 − 3)(x2 + 3)= (x −

√3)(x +

√3)(x2 + 3)

= (x −√3)(x +

√3)(x − i

√3)(x + i

√3)




Consider:

x4 − 9 = (x2 − 3)(x2 + 3)= (x −

√3)(x +

√3)(x2 + 3)

= (x −√3)(x +

√3)(x − i

√3)(x + i

√3)




Consider:

x3 − 2 =

(x − 3√2)(x2 + x 3√2+

3√22)= (x − 3√2)

(x −

(1+ i

√3

2

)3√2)(

x −(1− i

√3

2

)3√2)

Well, that’s ugly. Let’s see if we can do better.



Consider:

x3 − 2 = (x − 3√2)(x2 + x 3√2+

3√22)

= (x − 3√2)(x −

(1+ i

√3

2

)3√2)(

x −(1− i

√3

2

)3√2)




Consider:

x3 − 2 = (x − 3√2)(x2 + x 3√2+

3√22)= (x − 3√2)

(x −

(1+ i

√3

2

)3√2)(

x −(1− i

√3

2

)3√2)




Consider:

x3 − 2 = (x − 3√2)(x2 + x 3√2+

3√22)= (x − 3√2)

(x −

(1+ i

√3

2

)3√2)(

x −(1− i

√3

2

)3√2)





Complex Numbers

C = {z = a+ bi : a,b ∈ R}

(Called an Extension)

z0 = a+ bi

b

a

z1 = c+ di

z0 + z1 = a+ c+ (b+ d)i

z0 · z1 = ac− bd+ (ad+ bc)i


Complex Numbers

R[i] = {z = a+ bi : a,b ∈ R}

(Called an Extension)

z0 = a+ bi

b

a

z1 = c+ di

z0 + z1 = a+ c+ (b+ d)i



Complex Numbers

R[i] = {z = a+ bi : a,b ∈ R} (Called an Extension)

z0 = a+ bi

b

a

z1 = c+ di

z0 + z1 = a+ c+ (b+ d)i



Complex Numbers


z0 = a+ bi

b

a

z1 = c+ di

z0 + z1 = a+ c+ (b+ d)i



Complex Numbers


z0 = a+ bi

b

a

z1 = c+ di

z0 + z1 = a+ c+ (b+ d)i



Complex Numbers


z0 = a+ bi

b

a

z1 = c+ di

z0 + z1 = a+ c+ (b+ d)i



Complex Numbers


z0 = a+ bi

b

a

z1 = c+ di

z0 + z1 = a+ c+ (b+ d)i



Complex Numbers


z0 = a+ bi

b

a

z1 = c+ di

z0 + z1 = a+ c+ (b+ d)i



Complex Numbers


z0 = a+ bi

b

a

z1 = c+ di

z0 + z1 = a+ c+ (b+ d)i



Complex Numbers


z0 = a+ bi

b

a

z1 = c+ di

z0 + z1 = a+ c+ (b+ d)i



Complex Numbers


z0 = a+ bi

b

a

z1 = c+ di

z0 + z1 = a+ c+ (b+ d)i



Roots of Unity:

xn − 1 = (x − 1)(xn−1 + xn−2 + xn−3 + · · ·+ 1) = 0

De Moivre’s Theorem: (cos(θ) + i sin(θ))n = cos(n θ) + i sin(n θ)ω = cos (2π/n) + i sin (2π/n)ωn = (cos (2π/n) + i sin (2π/n))n = cos (2π) + i sin (2π) = 1


Roots of Unity:

xn − 1 = (x − 1)(xn−1 + xn−2 + xn−3 + · · ·+ 1) = 0De Moivre’s Theorem: (cos(θ) + i sin(θ))n = cos(n θ) + i sin(n θ)

ω = cos (2π/n) + i sin (2π/n)ωn = (cos (2π/n) + i sin (2π/n))n = cos (2π) + i sin (2π) = 1


Roots of Unity:

xn − 1 = (x − 1)(xn−1 + xn−2 + xn−3 + · · ·+ 1) = 0De Moivre’s Theorem: (cos(θ) + i sin(θ))n = cos(n θ) + i sin(n θ)ω = cos (2π/n) + i sin (2π/n)

ωn = (cos (2π/n) + i sin (2π/n))n = cos (2π) + i sin (2π) = 1


Roots of Unity:

xn − 1 = (x − 1)(xn−1 + xn−2 + xn−3 + · · ·+ 1) = 0De Moivre’s Theorem: (cos(θ) + i sin(θ))n = cos(n θ) + i sin(n θ)ω = cos (2π/n) + i sin (2π/n)ωn = (cos (2π/n) + i sin (2π/n))n = cos (2π) + i sin (2π) = 1


Roots of Unity:

All the solutions to x1 − 1 = 0

ω1=1

ω = cos (2π/1) + i sin (2π/1)


ω1ω2=1

ω = cos (2π/2) + i sin (2π/2)


ω1

ω2

ω3=1

ω = cos (2π/3) + i sin (2π/3)


ω1

ω2

ω3

ω4=1

ω = cos (2π/4) + i sin (2π/4)


ω1

ω2

ω3

ω4

ω5=1

ω = cos (2π/5) + i sin (2π/5)


ω1ω2

ω3

ω4 ω5

ω6=1

ω = cos (2π/6) + i sin (2π/6)


Roots of Unity:


ω1=1

ω = cos (2π/1) + i sin (2π/1)


ω1ω2=1

ω = cos (2π/2) + i sin (2π/2)


ω1

ω2

ω3=1

ω = cos (2π/3) + i sin (2π/3)


ω1

ω2

ω3

ω4=1

ω = cos (2π/4) + i sin (2π/4)


ω1

ω2

ω3

ω4

ω5=1

ω = cos (2π/5) + i sin (2π/5)


ω1ω2

ω3

ω4 ω5

ω6=1

ω = cos (2π/6) + i sin (2π/6)


Roots of Unity:


ω1=1

ω = cos (2π/1) + i sin (2π/1)


ω1ω2=1

ω = cos (2π/2) + i sin (2π/2)


ω1

ω2

ω3=1

ω = cos (2π/3) + i sin (2π/3)


ω1

ω2

ω3

ω4=1

ω = cos (2π/4) + i sin (2π/4)


ω1

ω2

ω3

ω4

ω5=1

ω = cos (2π/5) + i sin (2π/5)


ω1ω2

ω3

ω4 ω5

ω6=1

ω = cos (2π/6) + i sin (2π/6)


Roots of Unity:


ω1=1

ω = cos (2π/1) + i sin (2π/1)


ω1ω2=1

ω = cos (2π/2) + i sin (2π/2)


ω1

ω2

ω3=1

ω = cos (2π/3) + i sin (2π/3)


ω1

ω2

ω3

ω4=1

ω = cos (2π/4) + i sin (2π/4)


ω1

ω2

ω3

ω4

ω5=1

ω = cos (2π/5) + i sin (2π/5)


ω1ω2

ω3

ω4 ω5

ω6=1

ω = cos (2π/6) + i sin (2π/6)


Roots of Unity:


ω1=1

ω = cos (2π/1) + i sin (2π/1)


ω1ω2=1

ω = cos (2π/2) + i sin (2π/2)


ω1

ω2

ω3=1

ω = cos (2π/3) + i sin (2π/3)


ω1

ω2

ω3

ω4=1

ω = cos (2π/4) + i sin (2π/4)


ω1

ω2

ω3

ω4

ω5=1

ω = cos (2π/5) + i sin (2π/5)


ω1ω2

ω3

ω4 ω5

ω6=1

ω = cos (2π/6) + i sin (2π/6)


Roots of Unity:


ω1=1

ω = cos (2π/1) + i sin (2π/1)


ω1ω2=1

ω = cos (2π/2) + i sin (2π/2)


ω1

ω2

ω3=1

ω = cos (2π/3) + i sin (2π/3)


ω1

ω2

ω3

ω4=1

ω = cos (2π/4) + i sin (2π/4)


ω1

ω2

ω3

ω4

ω5=1

ω = cos (2π/5) + i sin (2π/5)


ω1ω2

ω3

ω4 ω5

ω6=1

ω = cos (2π/6) + i sin (2π/6)


Other Roots

xn − k = 0

z = ωin√k, with ω = cos (2π/n) + i sin (2π/n)

zn = (ωi)n (n√k)n = (ωn)i k = k

ω1n√k

ω2n√k

ω3n√k

ω4n√k

ωnn√k =

n√k


Other Roots

xn − k = 0z = ωi

n√k, with ω = cos (2π/n) + i sin (2π/n)

zn = (ωi)n (n√k)n = (ωn)i k = k

ω1n√k

ω2n√k

ω3n√k

ω4n√k

ωnn√k =

n√k


Other Roots

xn − k = 0z = ωi

n√k, with ω = cos (2π/n) + i sin (2π/n)zn = (ωi)n (

n√k)n = (ωn)i k = k

ω1n√k

ω2n√k

ω3n√k

ω4n√k

ωnn√k =

n√k


Other Roots

xn − k = 0z = ωi

n√k, with ω = cos (2π/n) + i sin (2π/n)zn = (ωi)n (

n√k)n = (ωn)i k = k

ω1n√k

ω2n√k

ω3n√k

ω4n√k

ωnn√k =

n√k




Permuting Roots Visually

3√2

ω3√2

ω2 3√2

α

β

Roots of x3 − 2

Cycling the RootsSwapping Two RootsSwapping theOthers?Swap and Cycle vs.Cycle and SwapWhat does thisremind us of?Only move the roots!



3√2

ω3√2

ω2 3√2

α

β

Roots of x3 − 2Cycling the Roots

Swapping Two RootsSwapping theOthers?Swap and Cycle vs.Cycle and SwapWhat does thisremind us of?Only move the roots!



3√2

ω3√2

ω2 3√2

α

β

Roots of x3 − 2Cycling the RootsSwapping Two Roots

Swapping theOthers?Swap and Cycle vs.Cycle and SwapWhat does thisremind us of?Only move the roots!



3√2

ω3√2

ω2 3√2

α

β

Roots of x3 − 2Cycling the RootsSwapping Two RootsSwapping theOthers?

Swap and Cycle vs.Cycle and SwapWhat does thisremind us of?Only move the roots!



3√2

ω3√2

ω2 3√2

α

β

Roots of x3 − 2Cycling the RootsSwapping Two RootsSwapping theOthers?Swap and Cycle vs.Cycle and Swap

What does thisremind us of?Only move the roots!



3√2

ω3√2

ω2 3√2

α

β

Roots of x3 − 2Cycling the RootsSwapping Two RootsSwapping theOthers?Swap and Cycle vs.Cycle and SwapWhat does thisremind us of?

Only move the roots!



3√2

ω3√2

ω2 3√2

α

β

Roots of x3 − 2Cycling the RootsSwapping Two RootsSwapping theOthers?Swap and Cycle vs.Cycle and SwapWhat does thisremind us of?Only move the roots!


Permuting Roots with Automorphisms

1st - Find an automorphism of Q[ω,

3√2]which “fixes” Q [ω]

α(

3√2)3

=

α

((3√2)3)

= α (2)= 2,

α(

3√2)= 3√2, ω 3√2, or ω2 3√2

α(

3√2)= ω

3√2

Note α3 = e





α(

3√2)3

=

α

((3√2)3)

= α (2)= 2,

α(

3√2)= 3√2, ω 3√2, or ω2 3√2

α(

3√2)= ω

3√2

Note α3 = e


Permuting Roots with Automorphisms1st - Find an automorphism of Q

[ω,


ζ = a+ b 3√2+ c 3√22+ dω + eω 3√2+ fω 3√22

+ gω2 + hω2 3√2+ kω2 3√22

α(

3√2)3

=

α

((3√2)3)

= α (2)= 2,

α(

3√2)= 3√2, ω 3√2, or ω2 3√2

α(

3√2)= ω

3√2

Note α3 = e





α(

3√2)3

=

α

((3√2)3)

= α (2)= 2,

α(

3√2)= 3√2, ω 3√2, or ω2 3√2

α(

3√2)= ω

3√2

Note α3 = e





α(

3√2)3

= α

((3√2)3)

= α (2)= 2,

α(

3√2)= 3√2, ω 3√2, or ω2 3√2

α(

3√2)= ω

3√2

Note α3 = e





α(

3√2)3

= α

((3√2)3)

= α (2)

= 2,

α(

3√2)= 3√2, ω 3√2, or ω2 3√2

α(

3√2)= ω

3√2

Note α3 = e





α(

3√2)3

= α

((3√2)3)

= α (2)= 2,

α(

3√2)= 3√2, ω 3√2, or ω2 3√2

α(

3√2)= ω

3√2

Note α3 = e





α(

3√2)3

= α

((3√2)3)

= α (2)= 2,

α(

3√2)= 3√2, ω 3√2, or ω2 3√2

α(

3√2)= ω

3√2

Note α3 = e





α(

3√2)3

= α

((3√2)3)

= α (2)= 2,

α(

3√2)= 3√2, ω 3√2, or ω2 3√2

3√2 7→ 3√2 =⇒ α = e

α(

3√2)= ω

3√2

Note α3 = e





α(

3√2)3

= α

((3√2)3)

= α (2)= 2,

α(

3√2)= 3√2, ω 3√2, or ω2 3√2

3√2 7→ ω3√2 is the inverse of 3√2 7→ ω2

3√2

α(

3√2)= ω

3√2

Note α3 = e





α(

3√2)3

= α

((3√2)3)

= α (2)= 2,

α(

3√2)= 3√2, ω 3√2, or ω2 3√2

α(

3√2)= ω

3√2

Note α3 = e





α(

3√2)3

= α

((3√2)3)

= α (2)= 2,

α(

3√2)= 3√2, ω 3√2, or ω2 3√2

α(

3√2)= ω

3√2

Note α3 = e



2nd - Find an automorphism of Q[ω,

3√2]which “fixes” Q

[3√2]

β (ω)3 =

β(ω3)

= β (1)= 1,

β (ω) = 1, ω, or ω2β (ω) = ω2

Note β2 = e





[3√2]

β (ω)3 =

β(ω3)

= β (1)= 1,

β (ω) = 1, ω, or ω2β (ω) = ω2

Note β2 = e





[3√2]

β (ω)3 = β(ω3)

= β (1)= 1,

β (ω) = 1, ω, or ω2β (ω) = ω2

Note β2 = e





[3√2]

β (ω)3 = β(ω3)

= β (1)

= 1,

β (ω) = 1, ω, or ω2β (ω) = ω2

Note β2 = e





[3√2]

β (ω)3 = β(ω3)

= β (1)= 1,

β (ω) = 1, ω, or ω2β (ω) = ω2

Note β2 = e





[3√2]

β (ω)3 = β(ω3)

= β (1)= 1,

β (ω) = 1, ω, or ω2

β (ω) = ω2

Note β2 = e





[3√2]

β (ω)3 = β(ω3)

= β (1)= 1,

β (ω) = 1, ω, or ω2

ω 7→ 1 =⇒ β is not 1-1 and not an automorphism

β (ω) = ω2

Note β2 = e





[3√2]

β (ω)3 = β(ω3)

= β (1)= 1,

β (ω) = 1, ω, or ω2ω 7→ ω =⇒ β = e

β (ω) = ω2

Note β2 = e





[3√2]

β (ω)3 = β(ω3)

= β (1)= 1,

β (ω) = 1, ω, or ω2β (ω) = ω2

Note β2 = e





[3√2]

β (ω)3 = β(ω3)

= β (1)= 1,

β (ω) = 1, ω, or ω2β (ω) = ω2

Note β2 = e



3rd - How do α and β e�ect elements of Q[ω,

3√2]?

α fixes Q [ω]

β fixes Q[

3√2]

What about α ◦ β?

α(β(ω)) = ω2

α(β(3√2)) = ω

3√2

What about β ◦ α2?What about α2 ◦ β?How many di�erent possibilities are there?




3√2]?

α fixes Q [ω]

β fixes Q[

3√2]


α(β(ω)) = ω2

α(β(3√2)) = ω

3√2





3√2]?

α fixes Q [ω]

β fixes Q[

3√2]


α(β(ω)) = ω2

α(β(3√2)) = ω

3√2





3√2]?

α fixes Q [ω]

β fixes Q[

3√2]


α(β(ω)) = ω2

α(β(3√2)) = ω

3√2





3√2]?

α fixes Q [ω]

β fixes Q[

3√2]

What about α ◦ β?α(β(ω)) = ω2

α(β(3√2)) = ω

3√2





3√2]?

α fixes Q [ω]

β fixes Q[

3√2]


α(β(3√2)) = ω

3√2




But, where does a general element, ζ , go?

α ◦ β(ζ) =

α(β(a+ b 3√2+ c 3√22

+ dω + eω 3√2

+fω 3√22+ gω2 + hω2 3√2+ kω2 3√22))

= a+ bω 3√2+ cω2 3√22+ dω2 + e 3√2

+ fω 3√22+ gω + hω2 3√2+ k 3√22

Supposing ζ is fixed ...

a = a b = e c = kd = g f = f h = h




α ◦ β(ζ) =

α(β(a+ b 3√2+ c 3√22

+ dω + eω 3√2

+fω 3√22+ gω2 + hω2 3√2+ kω2 3√22))

= a+ bω 3√2+ cω2 3√22+ dω2 + e 3√2

+ fω 3√22+ gω + hω2 3√2+ k 3√22


a = a b = e c = kd = g f = f h = h




α ◦ β(ζ) = α(β(a+ b 3√2+ c 3√22

+ dω + eω 3√2

+fω 3√22+ gω2 + hω2 3√2+ kω2 3√22))

= a+ bω 3√2+ cω2 3√22+ dω2 + e 3√2

+ fω 3√22+ gω + hω2 3√2+ k 3√22


a = a b = e c = kd = g f = f h = h




α ◦ β(ζ) = α(β(a+ b 3√2+ c 3√22

+ dω + eω 3√2

+fω 3√22+ gω2 + hω2 3√2+ kω2 3√22))

= a+ bω 3√2+ cω2 3√22+ dω2 + e 3√2

+ fω 3√22+ gω + hω2 3√2+ k 3√22


a = a b = e c = kd = g f = f h = h




α ◦ β(ζ) = α(β(a+ b 3√2+ c 3√22

+ dω + eω 3√2

+fω 3√22+ gω2 + hω2 3√2+ kω2 3√22))

= a+ bω 3√2+ cω2 3√22+ dω2 + e 3√2

+ fω 3√22+ gω + hω2 3√2+ k 3√22


a = a b = e c = kd = g f = f h = h




α ◦ β(ζ) = α(β(a+ b 3√2+ c 3√22

+ dω + eω 3√2

+fω 3√22+ gω2 + hω2 3√2+ kω2 3√22))

= a+ bω 3√2+ cω2 3√22+ dω2 + e 3√2

+ fω 3√22+ gω + hω2 3√2+ k 3√22


a = a b = e c = kd = g f = f h = h




a = a b = e c = kd = g f = f h = h

So ...

ζ = a+ b 3√2+ c 3√22+ dω + eω 3√2

+ fω 3√22+ gω2 + hω2 3√2+ kω2 3√22

And thus ζ ∈ Q[ω2 3√2

]




a = a b = e c = kd = g f = f h = h

So ...

ζ = a+ b 3√2+ c 3√22+ dω + bω 3√2

+ fω 3√22+ dω2 + hω2 3√2+ cω2 3√22


]



So ...

ζ = a+ b(

3√2+ ω3√2)+ c

(3√22

+ ω23√22)

+ d(ω + ω2

)+ fω 3√22

+ hω2 3√2


]



So ...

ζ = (a− d) + b 3√2 (1+ ω) + c 3√22 (1+ ω2)+ fω 3√22

+ hω2 3√2

= (a− d)− bω2 3√2− cω 3√22+ fω 3√22

+ hω2 3√2

= (a− d) + (h− b)ω2 3√2+ (f − c)ω 3√22

= (a− d) + (h− b)ω2 3√2+ (f − c)(ω2

3√2)2


]



So ...

ζ = (a− d) + b 3√2 (1+ ω) + c 3√22 (1+ ω2)+ fω 3√22

+ hω2 3√2

= (a− d)− bω2 3√2− cω 3√22+ fω 3√22

+ hω2 3√2

= (a− d) + (h− b)ω2 3√2+ (f − c)ω 3√22

= (a− d) + (h− b)ω2 3√2+ (f − c)(ω2

3√2)2


]



So ...

ζ = (a− d) + b 3√2 (1+ ω) + c 3√22 (1+ ω2)+ fω 3√22

+ hω2 3√2

= (a− d)− bω2 3√2− cω 3√22+ fω 3√22

+ hω2 3√2Why does 1+ ω = −ω2 and 1+ ω2 = −ω and ω + ω2 = −1?

= (a− d) + (h− b)ω2 3√2+ (f − c)ω 3√22

= (a− d) + (h− b)ω2 3√2+ (f − c)(ω2

3√2)2


]



So ...

ζ = (a− d) + b 3√2 (1+ ω) + c 3√22 (1+ ω2)+ fω 3√22

+ hω2 3√2

= (a− d)− bω2 3√2− cω 3√22+ fω 3√22

+ hω2 3√2

= (a− d) + (h− b)ω2 3√2+ (f − c)ω 3√22

= (a− d) + (h− b)ω2 3√2+ (f − c)(ω2

3√2)2


]



So ...

ζ = (a− d) + b 3√2 (1+ ω) + c 3√22 (1+ ω2)+ fω 3√22

+ hω2 3√2

= (a− d)− bω2 3√2− cω 3√22+ fω 3√22

+ hω2 3√2

= (a− d) + (h− b)ω2 3√2+ (f − c)ω 3√22

= (a− d) + (h− b)ω2 3√2+ (f − c)(ω2

3√2)2


]



So ...

ζ = (a− d) + b 3√2 (1+ ω) + c 3√22 (1+ ω2)+ fω 3√22

+ hω2 3√2

= (a− d)− bω2 3√2− cω 3√22+ fω 3√22

+ hω2 3√2

= (a− d) + (h− b)ω2 3√2+ (f − c)ω 3√22

= (a− d) + (h− b)ω2 3√2+ (f − c)(ω2

3√2)2


]



3√2

ω3√2

ω2 3√2

α

β

Roots of x3 − 2Cycling the RootsSwapping Two RootsSwapping theOthers?Swap and Cycle vs.Cycle and SwapWhat does thisremind us of?Only move the roots!




3√2]?

α fixes Q [ω]

β fixes Q[

3√2]


α(β(3√2)) = ω

3√2

What about β ◦ α2?

What about α2 ◦ β?How many di�erent possibilities are there?




3√2]?

α fixes Q [ω]

β fixes Q[

3√2]


α(β(3√2)) = ω

3√2

What about β ◦ α2?What about α2 ◦ β?

How many di�erent possibilities are there?




3√2]?

α fixes Q [ω]

β fixes Q[

3√2]


α(β(3√2)) = ω

3√2





Fixed Fields and Groups

D3 = S3 = 〈α, β〉

〈α〉〈β〉〈αβ〉⟨α2β

⟩

e

Q

Q [ω] Q[

3√2]

Q[ω2 3√2

]Q[ω

3√2]

Q[ω,

3√2]



D3 = S3 = 〈α, β〉

〈α〉

〈β〉〈αβ〉⟨α2β

⟩

e

Q

Q [ω]

Q[

3√2]

Q[ω2 3√2

]Q[ω

3√2]

Q[ω,

3√2]



D3 = S3 = 〈α, β〉

〈α〉〈β〉

〈αβ〉⟨α2β

⟩

e

Q

Q [ω] Q[

3√2]

Q[ω2 3√2

]Q[ω

3√2]

Q[ω,

3√2]



D3 = S3 = 〈α, β〉

〈α〉〈β〉〈αβ〉

⟨α2β

⟩

e

Q

Q [ω] Q[

3√2]

Q[ω2 3√2

]

Q[ω

3√2]

Q[ω,

3√2]



D3 = S3 = 〈α, β〉


⟩

e

Q

Q [ω] Q[

3√2]

Q[ω2 3√2

]Q[ω

3√2]

Q[ω,

3√2]



D3 = S3 = 〈α, β〉


⟩

e

Q

Q [ω] Q[

3√2]

Q[ω2 3√2

]Q[ω

3√2]

Q[ω,

3√2]



D3 = S3 = 〈α, β〉


⟩

e

Q

Q [ω] Q[

3√2]

Q[ω2 3√2

]Q[ω

3√2]

Q[ω,

3√2]

Notes:

Groups get smaller the corresponding fields get larger.Subgroup indices match the degree of the “minimal polynomial” for theadjoined root.The minimal polynomial only factors completely if the subgroup is normal.



D3 = S3 = 〈α, β〉


⟩

e

Q

Q [ω] Q[

3√2]

Q[ω2 3√2

]Q[ω

3√2]

Q[ω,

3√2]

Notes:Groups get smaller the corresponding fields get larger.

Subgroup indices match the degree of the “minimal polynomial” for theadjoined root.The minimal polynomial only factors completely if the subgroup is normal.



D3 = S3 = 〈α, β〉


⟩

e

Q

Q [ω] Q[

3√2]

Q[ω2 3√2

]Q[ω

3√2]

Q[ω,

3√2]

Notes:Groups get smaller the corresponding fields get larger.Subgroup indices match the degree of the “minimal polynomial” for theadjoined root.

The minimal polynomial only factors completely if the subgroup is normal.



D3 = S3 = 〈α, β〉


⟩

e

Q

Q [ω] Q[

3√2]

Q[ω2 3√2

]Q[ω

3√2]

Q[ω,

3√2]

Notes:Groups get smaller the corresponding fields get larger.Subgroup indices match the degree of the “minimal polynomial” for theadjoined root.The minimal polynomial only factors completely if the subgroup is normal.


Great Big Galois Example

Dr. Chuck [email protected]

http://sites.wcsu.edu/roccacC. F. Rocca Jr. (WCSU) Great Big Galois Example 26 /26

[email protected]

http://sites.wcsu.edu/roccac

Documents

sites.wcsu.edu€¦ · Fundamental Theorem of Algebra (Examples) Consider: x4 9 = (x2 3)(x2 +3) = (x p 3)(x + p 3)(x2 +3) = (x p 3)(x + p 3)(x i p 3)(x +i p 3) This could be partially