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Galois Theory and. The Radically insolvable Quintic. Phil Pollard. To Be Spoken:. The purpose of this project is to present a brief introduction to Galois Theory and to illustrate it’s uses in focusing infinite situations into a finite workspace.

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to be spoken
To Be Spoken:

The purpose of this project is to present a brief introduction to Galois Theory

and to illustrate it’s uses in focusing infinite situations into a finite workspace.

These slides will focus mainly on the use of Galois Theory to prove the insolvability

of the quintic polynomial by a general equation using radicals.

Most of the descriptions of theorems and properties are paraphrased interpretations

rather than rigorous proofs.

things you should know by now
Things you should know by now

(And if you don’t, you won’t know what the hell I’m talking about.)

  • Permutation Groups
  • Extension Fields
  • Splitting Fields

(Unless otherwise stated)

  • F := An arbitrary Field (assume char = 0)
  • E := An extension Field over F
  • ℚ := Field of Rational Numbers
  • F[x] := Polynomial Ring over F
  • F(a1,a2,…) := Extension of F by a1, a2,…
  • e := Identity (of current discussed group)
  • p.g. := “Proof Gist” expedited proof.
review of terms
Review of Terms

Degree of Extension

Denoted [E:F] = n

Where n is the dimension of E as a vector space over F.


A ring isomorphism from a Field onto itself

Simple Group

A group with no proper, non-trivial normal subgroup

Field of Rational Functions

f(x1, x2,…,xn)

g(x1, x2,…,xn)

Denoted F(x1, x2,…,xn) = { | f,g ∈ F[x1, x2,…,xn] }

new definitions
New Definitions

Galois Group

Denoted Gal(E/F)[read Galois Group of E fixing F]

The set of all automorphisms of E that “fix” F

(take each element of F to itself)

The Galois Group is a group under function composition.

Fixed Field

Denoted EH for H ⊆ Gal(E/F)

EH = {x ∈ E | Φ(x)=x ∀ Φ∈H}

Note: EGal(E/F) = F

Intermediate Field

F ≤ K ≤ E and K is a field

then K is an intermediate field and

Ģ(K) := Gal(E/K)

Note: EĢ(K) = K

theorem 1
Theorem 1

If f(x) is irreducible in F[x] and E is a splitting field of f(x),

Φ∈Gal(E/F) permutes the roots of f(x)


If ai is a root of f(x), then

f(ai) = Φ (f(ai)) = 0 because 0 is fixed by Φ

= f(Φ(ai)) Operation Preserving

But since ai ∉ F, it is not necessarily fixed

Obvious Corollary 1

If f(x) ∈ F[x] has degree n, then Gal(E/F) is isomorphic to a subgroup of Sn

(E being a splitting field of F)

example 1
Example 1:

Consider E = ℚ(ω,3√2) where ω = -1+i√3


Then the elements of Gal(E/ℚ) are described as follows:

Note: Gal(E/ ℚ ) is not Abelian. αβ ≠ βα


Subgroups of Gal(E/ ℚ)

{e, α, β, β2, αβ, αβ2}

2 3 3 3

{e, β, β2} {e, α} {e, αβ} {e, αβ2}

3 2 2 2


Subfields of E

E = ℚ (ω,3√2)

3 2 2 2

ℚ (ω)

ℚ (ω2 · 3√2)

ℚ (3√2)

ℚ (ω · 3√2)

2 3 3 3

fundamental theorem of galois theory para gallien
Fundamental Theorem of Galois Theory (Para.Gallien)
  • Given E, a splitting field over F, F being finite or with character 0
  • For every subfield K of E containing F
  • There is a subgroup Gal(E/K) of Gal(E/F), and
  • [E:K] = |Gal(E/K)| and [K:F] = |Gal(E/F)| / |Gal(E/K)|
  • If K is a splitting field of some f(x)∈F[x] then Gal(E/K) ◁ Gal(E/F)
    • and |Gal(K/F)| = |Gal(E/F)| / |Gal(E/K)|
  • K = EGal(E/K)
  • For H ≤ Gal(E/F), H = Gal(E/EH)

Proof by Smart Dead Guy. ■

(not para. Gallien)

solvable by radicals
Solvable by Radicals

Pure Extension

An extension F(a)/F where am ∈ F (referred to as “type m”)

Radical Extension

An extension E/F such that

F = K0 ⊆ K1 ⊆ ··· ⊆ Kn-1 ⊆ Kn = E

creates a tower of fields in which each extension Ki+1/Ki is pure

Solvable by Radicals

f(x) ∈ F[x] is Solvable by Radicals if it has a splitting field E

such that E/F forms a radical extension.

solvable group
Solvable Group

A group G that has a series of normal subgroups

{e} = H0◁ H1◁ ··· ◁ Hn-1◁ Hn = G

where for each 0 ≤ i ≤ n, Hi+1/Hi is Abelian

Note 1: The following are easy to show:

· Abelian Groups are solvable

· Non-Abelian Simple Groups are not solvable

· Subgroups of solvable groups are also solvable

(slightly less easy)

Note 2: We will eventually need the following Theorem:

Theorem 2

For E, a splitting field for xn – a over F, Gal(E/F) is solvable.

Proof by Smart Dead Guy ■

suspiciously convenient example
Suspiciously Convenient Example

S5 is not a solvable group


Through an exhaustive orders argument ,

we find that A5 is a simple group.

This means that A5 itself is not a solvable group.

And since subgroups of solvable groups are solvable,

S5 and Sn for n ≥ 5 cannot be solvable. ■

(The viewer is spared the details,

but a full proof of the simplicity of A5

can be found online or in almost any Algebra text.)

solvable by radicals implies solvable group
Solvable by Radicals impliesSolvable group

For E = F(a1,…,an), a Radical Extension for f(x) ∈ F[x], Gal(E/F) is solvable

This can be shown by induction.

For the Base Case,

E = F(a1), E splits some function f(x) = xn – a,

and Gal(E/F) is therefore solvable by Theorem 2.

For the complete induction proof, the viewer is referred to Gallien, page 556.

Proof by Smart Dead Guy ■

Obvious Contraposition 1

If Gal(E/F) is not solvable, then f(x) is not solvable by radicals.

the galois group for certain polynomials of n th power para herstein
The Galois Group for Certain Polynomials of nth power (para. Herstein)

Consider F(x1,…,xn) and

let S be the field of symmetric rational functions over F.

Consider the polynomial:

f(y) = yn – a1yn-1 + a2yn-2 – ··· + (-1)n an in S[y] with a1,…,an ∈ S

The roots of f(y) are not contained in S because

a1,…,an must have independent, permutable variables

But we find that F(x1,…,xn) is a splitting field of f(x) over S

and f(y) = yn – a1 yn-1 + a2 yn-2 – ··· + (-1)n an

= (y – t1)(y – t2)···(y – tn) ∈ F(x1,…,xn) [y] (ti in terms of x’s)

And the roots t1, … ,tn are distinct.

Thus, |Gal(F(x1,…,xn) / S)| = n!

And by Obvious Corollary 1, Gal( F(x1,…,xn) / S) is isomorphic to a subgroup of Sn

Hence, Gal(F(x1,…,xn) / S) is isomorphic to Sn

the insolvability of the quintic by radicals
The Insolvability of the Quintic by Radicals

We just found out that certain n-power Functions produce a Galois Group

isomorphic to Sn

And we know by our Suspiciously Convenient Example

that Sn is not solvable for n ≥ 5

And we know by our Obvious Contraposition that if a Galois Group Gal(E/F)

is not a solvable group, the equation split by E is not solvable by radicals.

Thus, the Quintic Polynomial is not generally solvable by radicals.



Contemporary Abstract Algebra, 7th Edition, Joseph Gallian. 2010, Brooks/Cole

Advanced Modern Algebra, Joseph J. Rotman. 2003, Prentice Hall

Galois Theory, 2nd Edition, Ian Stewart. 1989, Chapman & Hall/CRC

Topics in Algebra, I.N. Herstein. 1965, Blaisdell Publishing

Chapter 6 Notes – Galois Theory, University of Illinois, Urbana-Champaign