sites.wcsu.edu€¦ · fundamental theorem of algebra (examples) consider: x4 9 = (x2 3)(x2 +3) =...
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Great Big Galois Example
Dr. Chuck [email protected]
http://sites.wcsu.edu/roccacC. F. Rocca Jr. (WCSU) Great Big Galois Example 1 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 3 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 3 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 3 /26
Introduction
Quadratic Equations (Area) ∼ 2000 BCECubic and Quartic Equations ∼ 1500 CEQuintic Equations and Above ∼ 1800 CEPermutations of Roots and the Birth of Group Theory
C. F. Rocca Jr. (WCSU) Great Big Galois Example 3 /26
IntroductionPolynomials and Fundamental Theorem of AlgebraVisualizing Complex NumbersPermuting RootsFixed Fields and Groups
C. F. Rocca Jr. (WCSU) Great Big Galois Example 4 /26
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.
C. F. Rocca Jr. (WCSU) Great Big Galois Example 5 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 6 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 6 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 6 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 6 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 6 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 6 /26
Fundamental Theorem of Algebra (Examples)
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 7 /26
Fundamental Theorem of Algebra (Examples)
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 7 /26
Fundamental Theorem of Algebra (Examples)
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 7 /26
Fundamental Theorem of Algebra (Examples)
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 7 /26
Fundamental Theorem of Algebra (Examples)
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 7 /26
Fundamental Theorem of Algebra (Examples)
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.
C. F. Rocca Jr. (WCSU) Great Big Galois Example 8 /26
Fundamental Theorem of Algebra (Examples)
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.
C. F. Rocca Jr. (WCSU) Great Big Galois Example 8 /26
Fundamental Theorem of Algebra (Examples)
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.
C. F. Rocca Jr. (WCSU) Great Big Galois Example 8 /26
Fundamental Theorem of Algebra (Examples)
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.
C. F. Rocca Jr. (WCSU) Great Big Galois Example 8 /26
Fundamental Theorem of Algebra (Examples)
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.
C. F. Rocca Jr. (WCSU) Great Big Galois Example 8 /26
Fundamental Theorem of Algebra (Examples)
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.
C. F. Rocca Jr. (WCSU) Great Big Galois Example 9 /26
Fundamental Theorem of Algebra (Examples)
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.
C. F. Rocca Jr. (WCSU) Great Big Galois Example 9 /26
Fundamental Theorem of Algebra (Examples)
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.
C. F. Rocca Jr. (WCSU) Great Big Galois Example 9 /26
Fundamental Theorem of Algebra (Examples)
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.
C. F. Rocca Jr. (WCSU) Great Big Galois Example 9 /26
IntroductionPolynomials and Fundamental Theorem of AlgebraVisualizing Complex NumbersPermuting RootsFixed Fields and Groups
C. F. Rocca Jr. (WCSU) Great Big Galois Example 10 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 11 /26
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
z0 · z1 = ac− bd+ (ad+ bc)i
C. F. Rocca Jr. (WCSU) Great Big Galois Example 11 /26
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
z0 · z1 = ac− bd+ (ad+ bc)i
C. F. Rocca Jr. (WCSU) Great Big Galois Example 11 /26
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
z0 · z1 = ac− bd+ (ad+ bc)i
C. F. Rocca Jr. (WCSU) Great Big Galois Example 11 /26
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
z0 · z1 = ac− bd+ (ad+ bc)i
C. F. Rocca Jr. (WCSU) Great Big Galois Example 11 /26
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
z0 · z1 = ac− bd+ (ad+ bc)i
C. F. Rocca Jr. (WCSU) Great Big Galois Example 11 /26
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
z0 · z1 = ac− bd+ (ad+ bc)i
C. F. Rocca Jr. (WCSU) Great Big Galois Example 11 /26
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
z0 · z1 = ac− bd+ (ad+ bc)i
C. F. Rocca Jr. (WCSU) Great Big Galois Example 11 /26
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
z0 · z1 = ac− bd+ (ad+ bc)i
C. F. Rocca Jr. (WCSU) Great Big Galois Example 11 /26
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
z0 · z1 = ac− bd+ (ad+ bc)i
C. F. Rocca Jr. (WCSU) Great Big Galois Example 11 /26
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
z0 · z1 = ac− bd+ (ad+ bc)i
C. F. Rocca Jr. (WCSU) Great Big Galois Example 11 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 12 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 12 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 12 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 12 /26
Roots of Unity:
All the solutions to x1 − 1 = 0
ω1=1
ω = cos (2π/1) + i sin (2π/1)
All the solutions to x2 − 1 = 0
ω1ω2=1
ω = cos (2π/2) + i sin (2π/2)
All the solutions to x3 − 1 = 0
ω1
ω2
ω3=1
ω = cos (2π/3) + i sin (2π/3)
All the solutions to x4 − 1 = 0
ω1
ω2
ω3
ω4=1
ω = cos (2π/4) + i sin (2π/4)
All the solutions to x5 − 1 = 0
ω1
ω2
ω3
ω4
ω5=1
ω = cos (2π/5) + i sin (2π/5)
All the solutions to x6 − 1 = 0
ω1ω2
ω3
ω4 ω5
ω6=1
ω = cos (2π/6) + i sin (2π/6)
C. F. Rocca Jr. (WCSU) Great Big Galois Example 13 /26
Roots of Unity:
All the solutions to x1 − 1 = 0
ω1=1
ω = cos (2π/1) + i sin (2π/1)
All the solutions to x2 − 1 = 0
ω1ω2=1
ω = cos (2π/2) + i sin (2π/2)
All the solutions to x3 − 1 = 0
ω1
ω2
ω3=1
ω = cos (2π/3) + i sin (2π/3)
All the solutions to x4 − 1 = 0
ω1
ω2
ω3
ω4=1
ω = cos (2π/4) + i sin (2π/4)
All the solutions to x5 − 1 = 0
ω1
ω2
ω3
ω4
ω5=1
ω = cos (2π/5) + i sin (2π/5)
All the solutions to x6 − 1 = 0
ω1ω2
ω3
ω4 ω5
ω6=1
ω = cos (2π/6) + i sin (2π/6)
C. F. Rocca Jr. (WCSU) Great Big Galois Example 13 /26
Roots of Unity:
All the solutions to x1 − 1 = 0
ω1=1
ω = cos (2π/1) + i sin (2π/1)
All the solutions to x2 − 1 = 0
ω1ω2=1
ω = cos (2π/2) + i sin (2π/2)
All the solutions to x3 − 1 = 0
ω1
ω2
ω3=1
ω = cos (2π/3) + i sin (2π/3)
All the solutions to x4 − 1 = 0
ω1
ω2
ω3
ω4=1
ω = cos (2π/4) + i sin (2π/4)
All the solutions to x5 − 1 = 0
ω1
ω2
ω3
ω4
ω5=1
ω = cos (2π/5) + i sin (2π/5)
All the solutions to x6 − 1 = 0
ω1ω2
ω3
ω4 ω5
ω6=1
ω = cos (2π/6) + i sin (2π/6)
C. F. Rocca Jr. (WCSU) Great Big Galois Example 13 /26
Roots of Unity:
All the solutions to x1 − 1 = 0
ω1=1
ω = cos (2π/1) + i sin (2π/1)
All the solutions to x2 − 1 = 0
ω1ω2=1
ω = cos (2π/2) + i sin (2π/2)
All the solutions to x3 − 1 = 0
ω1
ω2
ω3=1
ω = cos (2π/3) + i sin (2π/3)
All the solutions to x4 − 1 = 0
ω1
ω2
ω3
ω4=1
ω = cos (2π/4) + i sin (2π/4)
All the solutions to x5 − 1 = 0
ω1
ω2
ω3
ω4
ω5=1
ω = cos (2π/5) + i sin (2π/5)
All the solutions to x6 − 1 = 0
ω1ω2
ω3
ω4 ω5
ω6=1
ω = cos (2π/6) + i sin (2π/6)
C. F. Rocca Jr. (WCSU) Great Big Galois Example 13 /26
Roots of Unity:
All the solutions to x1 − 1 = 0
ω1=1
ω = cos (2π/1) + i sin (2π/1)
All the solutions to x2 − 1 = 0
ω1ω2=1
ω = cos (2π/2) + i sin (2π/2)
All the solutions to x3 − 1 = 0
ω1
ω2
ω3=1
ω = cos (2π/3) + i sin (2π/3)
All the solutions to x4 − 1 = 0
ω1
ω2
ω3
ω4=1
ω = cos (2π/4) + i sin (2π/4)
All the solutions to x5 − 1 = 0
ω1
ω2
ω3
ω4
ω5=1
ω = cos (2π/5) + i sin (2π/5)
All the solutions to x6 − 1 = 0
ω1ω2
ω3
ω4 ω5
ω6=1
ω = cos (2π/6) + i sin (2π/6)
C. F. Rocca Jr. (WCSU) Great Big Galois Example 13 /26
Roots of Unity:
All the solutions to x1 − 1 = 0
ω1=1
ω = cos (2π/1) + i sin (2π/1)
All the solutions to x2 − 1 = 0
ω1ω2=1
ω = cos (2π/2) + i sin (2π/2)
All the solutions to x3 − 1 = 0
ω1
ω2
ω3=1
ω = cos (2π/3) + i sin (2π/3)
All the solutions to x4 − 1 = 0
ω1
ω2
ω3
ω4=1
ω = cos (2π/4) + i sin (2π/4)
All the solutions to x5 − 1 = 0
ω1
ω2
ω3
ω4
ω5=1
ω = cos (2π/5) + i sin (2π/5)
All the solutions to x6 − 1 = 0
ω1ω2
ω3
ω4 ω5
ω6=1
ω = cos (2π/6) + i sin (2π/6)
C. F. Rocca Jr. (WCSU) Great Big Galois Example 13 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 14 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 14 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 14 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 14 /26
IntroductionPolynomials and Fundamental Theorem of AlgebraVisualizing Complex NumbersPermuting RootsFixed Fields and Groups
C. F. Rocca Jr. (WCSU) Great Big Galois Example 15 /26
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!
C. F. Rocca Jr. (WCSU) Great Big Galois Example 16 /26
Permuting Roots Visually
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!
C. F. Rocca Jr. (WCSU) Great Big Galois Example 16 /26
Permuting Roots Visually
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!
C. F. Rocca Jr. (WCSU) Great Big Galois Example 16 /26
Permuting Roots Visually
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!
C. F. Rocca Jr. (WCSU) Great Big Galois Example 16 /26
Permuting Roots Visually
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!
C. F. Rocca Jr. (WCSU) Great Big Galois Example 16 /26
Permuting Roots Visually
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!
C. F. Rocca Jr. (WCSU) Great Big Galois Example 16 /26
Permuting Roots Visually
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!
C. F. Rocca Jr. (WCSU) Great Big Galois Example 16 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 17 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 17 /26
Permuting Roots with Automorphisms1st - Find an automorphism of Q
[ω,
3√2]which “fixes” 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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 17 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 17 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 17 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 17 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 17 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 17 /26
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
3√2 7→ 3√2 =⇒ α = e
α(
3√2)= ω
3√2
Note α3 = e
C. F. Rocca Jr. (WCSU) Great Big Galois Example 17 /26
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
3√2 7→ ω3√2 is the inverse of 3√2 7→ ω2
3√2
α(
3√2)= ω
3√2
Note α3 = e
C. F. Rocca Jr. (WCSU) Great Big Galois Example 17 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 17 /26
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 17 /26
Permuting Roots with Automorphisms
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 18 /26
Permuting Roots with Automorphisms
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 18 /26
Permuting Roots with Automorphisms
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 18 /26
Permuting Roots with Automorphisms
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 18 /26
Permuting Roots with Automorphisms
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 18 /26
Permuting Roots with Automorphisms
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 ω2
β (ω) = ω2
Note β2 = e
C. F. Rocca Jr. (WCSU) Great Big Galois Example 18 /26
Permuting Roots with Automorphisms
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 ω2
ω 7→ 1 =⇒ β is not 1-1 and not an automorphism
β (ω) = ω2
Note β2 = e
C. F. Rocca Jr. (WCSU) Great Big Galois Example 18 /26
Permuting Roots with Automorphisms
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 ω2ω 7→ ω =⇒ β = e
β (ω) = ω2
Note β2 = e
C. F. Rocca Jr. (WCSU) Great Big Galois Example 18 /26
Permuting Roots with Automorphisms
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 18 /26
Permuting Roots with Automorphisms
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 18 /26
Permuting Roots with Automorphisms
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?
C. F. Rocca Jr. (WCSU) Great Big Galois Example 19 /26
Permuting Roots with Automorphisms
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?
C. F. Rocca Jr. (WCSU) Great Big Galois Example 19 /26
Permuting Roots with Automorphisms
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?
C. F. Rocca Jr. (WCSU) Great Big Galois Example 19 /26
Permuting Roots with Automorphisms
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?
C. F. Rocca Jr. (WCSU) Great Big Galois Example 19 /26
Permuting Roots with Automorphisms
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?
C. F. Rocca Jr. (WCSU) Great Big Galois Example 19 /26
Permuting Roots with Automorphisms
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?
C. F. Rocca Jr. (WCSU) Great Big Galois Example 19 /26
Permuting Roots with Automorphisms
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 20 /26
Permuting Roots with Automorphisms
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 20 /26
Permuting Roots with Automorphisms
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 20 /26
Permuting Roots with Automorphisms
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 20 /26
Permuting Roots with Automorphisms
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 20 /26
Permuting Roots with Automorphisms
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
C. F. Rocca Jr. (WCSU) Great Big Galois Example 20 /26
Permuting Roots with Automorphisms
Supposing ζ is fixed ...
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
]
C. F. Rocca Jr. (WCSU) Great Big Galois Example 21 /26
Permuting Roots with Automorphisms
Supposing ζ is fixed ...
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
And thus ζ ∈ Q[ω2 3√2
]
C. F. Rocca Jr. (WCSU) Great Big Galois Example 21 /26
Permuting Roots with Automorphisms
So ...
ζ = a+ b(
3√2+ ω3√2)+ c
(3√22
+ ω23√22)
+ d(ω + ω2
)+ fω 3√22
+ hω2 3√2
And thus ζ ∈ Q[ω2 3√2
]
C. F. Rocca Jr. (WCSU) Great Big Galois Example 21 /26
Permuting Roots with Automorphisms
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
And thus ζ ∈ Q[ω2 3√2
]
C. F. Rocca Jr. (WCSU) Great Big Galois Example 21 /26
Permuting Roots with Automorphisms
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
And thus ζ ∈ Q[ω2 3√2
]
C. F. Rocca Jr. (WCSU) Great Big Galois Example 21 /26
Permuting Roots with Automorphisms
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
And thus ζ ∈ Q[ω2 3√2
]
C. F. Rocca Jr. (WCSU) Great Big Galois Example 21 /26
Permuting Roots with Automorphisms
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
And thus ζ ∈ Q[ω2 3√2
]
C. F. Rocca Jr. (WCSU) Great Big Galois Example 21 /26
Permuting Roots with Automorphisms
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
And thus ζ ∈ Q[ω2 3√2
]
C. F. Rocca Jr. (WCSU) Great Big Galois Example 21 /26
Permuting Roots with Automorphisms
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
And thus ζ ∈ Q[ω2 3√2
]
C. F. Rocca Jr. (WCSU) Great Big Galois Example 21 /26
Permuting Roots Visually
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!
C. F. Rocca Jr. (WCSU) Great Big Galois Example 22 /26
Permuting Roots with Automorphisms
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?
C. F. Rocca Jr. (WCSU) Great Big Galois Example 23 /26
Permuting Roots with Automorphisms
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?
C. F. Rocca Jr. (WCSU) Great Big Galois Example 23 /26
Permuting Roots with Automorphisms
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?
C. F. Rocca Jr. (WCSU) Great Big Galois Example 23 /26
IntroductionPolynomials and Fundamental Theorem of AlgebraVisualizing Complex NumbersPermuting RootsFixed Fields and Groups
C. F. Rocca Jr. (WCSU) Great Big Galois Example 24 /26
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]
C. F. Rocca Jr. (WCSU) Great Big Galois Example 25 /26
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]
C. F. Rocca Jr. (WCSU) Great Big Galois Example 25 /26
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]
C. F. Rocca Jr. (WCSU) Great Big Galois Example 25 /26
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]
C. F. Rocca Jr. (WCSU) Great Big Galois Example 25 /26
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]
C. F. Rocca Jr. (WCSU) Great Big Galois Example 25 /26
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]
C. F. Rocca Jr. (WCSU) Great Big Galois Example 25 /26
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]
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.
C. F. Rocca Jr. (WCSU) Great Big Galois Example 25 /26
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]
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.
C. F. Rocca Jr. (WCSU) Great Big Galois Example 25 /26
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]
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.
C. F. Rocca Jr. (WCSU) Great Big Galois Example 25 /26
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]
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.
C. F. Rocca Jr. (WCSU) Great Big Galois Example 25 /26
Great Big Galois Example
Dr. Chuck [email protected]
http://sites.wcsu.edu/roccacC. F. Rocca Jr. (WCSU) Great Big Galois Example 26 /26